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There are several well-known mathematical statements that are 'obvious' but false (such as the negation of the Banach--Tarski theorem). There are plenty more that are 'obvious' and true. One would naturally expect a statement in the latter category to be easy to prove -- and they usually are. I'm interested in examples of theorems that are 'obvious', and known to be true, but that lack (or appear to lack) easy proofs.

Of course, 'obvious' and 'easy' are fuzzy terms, and context-dependent. The Jordan curve theorem illustrates what I mean (and motivates this question). It seems 'obvious', as soon as one understands the definition of continuity, that it should hold; it does in fact hold; but all the known proofs are surprisingly difficult.

Can anyone suggest other such theorems, in any areas of mathematics?

Ricky
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    Perhaps the isoperimetric inequality. – Péter Komjáth Jan 09 '11 at 11:28
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    Perhaps the Kepler Conjecture: http://en.wikipedia.org/wiki/Kepler_conjecture – Aaron Meyerowitz Jan 09 '11 at 11:57
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    The product of nonempty sets is nonempty, and the negation of Vopěnka's principle. ;-) – Jason Jan 09 '11 at 12:14
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    A former colleague of mine used to say (to students), "A theorem is obvious if a proof instantly springs to mind," a maxim I like a lot. I think what you are talking about is theorems where a plausible argument instantly springs to mind but falls short of being a proof. – gowers Jan 09 '11 at 15:12
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    I am tempted to vote to close as subjective and argumentative given the comments on the existing answers. Can we narrow the definition of 'obvious' being used? Something like gowers' definition is good, but depends a lot on one's training. Perhaps something like "if you asked an undergraduate if it were true, they'd bet yes." – Qiaochu Yuan Jan 09 '11 at 16:46
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    I agree with Qiaochu. I think Gowers's definition seems more apt, but still what is instant to him is probably not to others...so obviously, obvious does not have an obvious meaning, and this question as such will remain subjective and argumentative. – Suvrit Jan 09 '11 at 17:02
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    The question does indeed belong in 'community wiki'. To clarify the (still subjective) definition of 'obvious', I'd like to propose a combination of gowers's idea above ('a plausible argument instantly springs to mind'), and Mark Bennett's below ('intuitively obvious at a significantly lower level of mathematical sophistication than is required for the proof'). So, I want theorems for which a plausible argument springs to mind at the level of sophistication required to understand the statement, but for which a proof requires a higher level of sophistication. There are some nice examples here. – Alec Edgington Jan 09 '11 at 19:19
  • (Naturally, the plausibility of the argument that springs to mind will diminish once the actual difficulties become apparent! But it is this tension that interests me.) – Alec Edgington Jan 09 '11 at 19:22
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    Negation of Banach-Tarski theorem is obvious and true, if appropriate axioms choosen. – Anixx Jan 09 '11 at 20:02
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    With reference to Alec's comment on my idea - there is an issue about how we train mathematical intuition which I think I was getting to. It is wrestling with some of these intuitively obvious things which has sharpened definitions, and helped mathematicians to identify fruitful concepts. The actual mathematics helps us distinguish between true intuitions and false ones - and importantly also helps us to understand why those false intuitions are wrong - and perhaps helps to recover constraints which make the intuition true. – Mark Bennet Jan 10 '11 at 11:21
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    I disagree that the Jordan curve theorem is "obvious" but admits a surprisingly difficult proof. The proof for curves with reasonable regularity is not difficult, while the truth of the theorem for wild curves is not so obvious, I think. (At least, I think it is reasonable to argue that most people's sense of this being intuitively clear comes from imagining a rather regular curve in the plane, not a wild one.) – Emerton Jan 10 '11 at 16:46
  • I really think that "obvious" is a poor choice of words for this question. Would "intuitive" work better? Or would it fail to capture the OP's intent? – Thierry Zell Feb 21 '11 at 20:12
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    I agree that 'obvious' isn't quite right, and put it in scare-quotes for that reason. Perhaps 'intuitive' is better, but it feels weaker: there are statements I'd describe as intuitive but not obvious (for example, that the densest 3-dimensional sphere packing is the one that greengrocers use to stack oranges). – Alec Edgington Feb 22 '11 at 06:49
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    The questions seems to have run its course. Thus, I voted to close as no longer relevant. –  Feb 09 '13 at 19:12
  • I protected the question as it got reactivated by a not so great (IMO) answer. Please comment notify me if you disagree with the protection. –  Mar 14 '14 at 20:17
  • To students who don't like mathematics, a proposition is "obvious" if they've always been told by teachers and textbooks that it's true. "Why is $a\times b$ always equal to $b\times a\text{?}$" "Because that's the commutative law, you moron! Didn't you listen when the teacher said that?" $\qquad$ – Michael Hardy Aug 08 '23 at 18:57

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If $I_1,I_2,\dots$ are intervals of real numbers with lengths that sum to less than 1, then their union cannot be all of $[0,1]$. It is quite common for people to think this statement is more obvious than it actually is. (The "proof" is this: just translate the intervals so that the end point of $I_1$ is the beginning point of $I_2$, and so on, and that will clearly maximize the length of interval you can cover. The problem is that this argument works just as well in the rationals, where the conclusion is false.)

gowers
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    Can you expand on this? – Lennart Meier Jan 09 '11 at 20:40
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    @Lennart: enumerate the rationals, and take an interval of length $\epsilon / 2^n$ around the $n$th rational. You get a countable collection of intervals with lengths summing to $\epsilon$ whose union contains all the rationals (never mind just those in $[0,1]$). – Chris Eagle Jan 09 '11 at 21:07
  • Well, you can fix the proof by repeating the argument transfinitely many times. :P – Mark Jan 12 '11 at 07:21
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    I don't see why the "proof" is not actually a proof. You can line them up via axiom of choice. If they don't sum to 1, they can't be the unit interval since the measure of their union is less than or equal to the sum of their measures. – Joe Johnson Jan 12 '11 at 23:23
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    Yeah - it seems that does maximize the measure covered, so if it's less than 1, can't cover. By contrast rationals have measure 0, so the measure covered is already maximized; measure simply isn't the relevant obstacle. Or is the idea to do it without having introduced the machinery of measure? That's the only way I can make sense of this. – Harry Altman Jan 12 '11 at 23:42
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    @Joe and @Harry, it is of course trivial if you know that there is a countably additive measure that extends lengths of intervals. But that result is not trivial, and many people are tempted to think that the simple statement about intervals is just plain obvious. – gowers Jan 13 '11 at 12:16
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    Using Lebesgue measure, you can do it. However, you probably need to know this fact (or something equivalent) in order to construct Lebesgue measure. The proof in my book uses compactness and reduces to the case of finitely many intervals.

    (Measure, Topology, and Fractal Geometry, Lemma 5.1.1)

    – Gerald Edgar Mar 04 '11 at 17:50
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    The proof (any proof) of Heine Borel can be adapted to simultaneously prove the statement here (along with Heine Borel at the same time). For example, consider a covering of $[0,1]$ by open intervals and let $S$ be the set of $x$ so that $[0,x]$ is covered by a finite union of these intervals such that $\sum_{k=1}^n |I_k \cap (- \infty,x]| > x$. $S$ is nonempty since $0 \in S$, and the supremum of $S$ can't be less than $1$. The other proof of Heine Borel (shrinking closed intervals intersect) can also be adapted to give a direct proof of this fact (and countless other basic facts). – Phil Isett Aug 27 '11 at 03:39
  • I love this. We were set an open ended holiday problem last christmas to prove a statement that if a sequence of positive functions, bounded above, tended pointwise to 0, then the integral of the functions tended to zero. I remember a very similar step being quite fiddly in my proof. – riemann_lebesgue Apr 09 '20 at 21:02
  • It's still near-trivial. Cover each interval by a longer open interval that is just slightly longer, then use the finite cover theorem. – Reflecting_Ordinal Jan 18 '24 at 21:48
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$\mathbb R^n$ is not homeomorphic to $\mathbb R^m$ unless $m = n$.

Arend Bayer
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    1+. I think this is the key point of the answer of Georges Elencwajg. It's not so really about one specific topological dimension and its computation, but rather that we can distinguish the affine spaces at all. I wonder how many decades (centuries?) mathematicians were convinced of this fact without having a proof? – Martin Brandenburg Jan 12 '11 at 08:01
  • Until Brauwer provided a tremendous complicated proof in 1921 without using homology, I guess the mathmaticans believe it as sth geometrically trvial. – Henry.L May 04 '13 at 04:07
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    The only proof I know is by puncturing each space and calculating relative homology. Is there a simpler proof? – Uzu Lim Aug 15 '22 at 19:58
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The most deadly example I know is the Hauptvermutung in dimensions 2 and 3 (in dimension $>3$, it is the ultimate "obvious but false" theorem). The Hauptvermutung, or "Main Conjecture" states that any two triangulations of a polyhedron are combinatorially equivalent, i.e. they become isomorphic after subdivision.
The Hauptvermutung is so obvious that it gets taken for granted everywhere, and most of us learn algebraic topology without ever noticing this huge gap in its foundations (of the text-book standard simplicial approach). It is implicit every time one states that a homotopy invariant of a simplicial complex, such as simplicial homology, is in fact a homotopy invariant of a polyhedron, unless one also proves independence relative to triangulation.
The Hauptvermutung for 2-manifolds was proven by Radó, and for 3-manifolds by Moïse in 1953. It is a genuinely deep, difficult theorem.
Edit: This answer is essentially taken from Page 4 of The Hauptvermutung Book.

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    At least since the introduction of singular homology, there is no gap in the foundations of algebraic topology. But I agree that the Hauptvermutung has a feeling like being obvious. – Lennart Meier Jan 10 '11 at 18:46
  • Answer editted to clarify what I meant. – Daniel Moskovich Jan 11 '11 at 01:02
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    By "subdivision," I believe you mean PL-subdivision, making the Hauptvermutung a little less obvious. An essentially equivalent statement to its falsity in dimension 5 that I find very unintuitive is that there exists a 4-dimensional simplicial complex $K$ which is not a triangulation of a manifold, but whose suspension $\Sigma K$ is a triangulation of a 5-sphere. – Richard Stanley Jan 12 '11 at 00:59
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How about the fact that a sphere is the surface of minimal area that bounds a given volume? (BTW, if it isn't geometrically obvious to you, and you understand a little physics and about surface tension, then the roundness of bubbles is a "proof".)

Dick Palais
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    This is my idea of an "obvious" answer that is hard to prove, i.e. one that has a physically persuasive justification. – roy smith Jan 10 '11 at 05:17
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    The fact that bubbles are round only demonstrates a local, not gloabal, minimum, surely? The bubble can't explore arbitrary regions of phase space: it can only go downhill. – Max Jan 10 '11 at 09:10
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    @Max: If it was only a local minimum then some bubbles would be round, and others would form different shapes. – George Lowther Jan 19 '11 at 23:31
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    @George: suppose you have a double-well model with one well lower than the other. Also suppose that the physical states are (for whatever reason) always created in the higher well and that the barrier between the wells is higher than the fluctuations in the system at the given temperature. Surely then the physical system wouldn't see the global minimum. So this argument doesn't work even at the physical level (never mind rigorous treatment)$\ldots$ – Marek Jul 29 '11 at 21:09
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    Ice 9 is a popular account of such a local min not being the global min – Steven Gubkin Mar 14 '14 at 19:19
  • @StevenGubkin - This comment made my week. – benblumsmith Apr 07 '17 at 16:58
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The Jordan curve theorem! Of course in this case the real problem is the meaning of "closed curve".

Ricky
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  • Or other similar considerations that show the plane is not homeomorphic to 3-space. – Gerald Edgar Jan 09 '11 at 14:47
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    Wasn't this already mentioned as the motivating example in the OP? – Todd Trimble Jan 09 '11 at 16:10
  • Isn't there a proof using the Mayer-Vietoris sequence that is simple? – Sean Tilson Jan 09 '11 at 17:18
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    @Sean: I think "easy" here should be taken to mean "without machinery." With enough machinery, everything is simple... – Qiaochu Yuan Jan 09 '11 at 17:30
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    @Qiaochu: I see, but I disagree with your last comment, it is frequently true, but there are some things that are just hard. – Sean Tilson Jan 09 '11 at 17:52
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    @Sean: Maybe they are hard just because the machinery to make it simple hasn't been invented yet :) – José Figueroa-O'Farrill Jan 09 '11 at 18:14
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    @Jose: I still think it plausible that some things are just hard. I think that when machinery makes a previously hard problem easy, for example $K$-theory and vector fields on spheres, we have a good concrete reason for using a tool. However, I do not believe that hard problems necessitate tools that make the problem simple. I was at a talk about more modern proofs of Carleson's Thm (among other things, i can not find the title) by Michael Lacey. Lacey had a conversation with Carleson where he asked Carleson if he felt that there would ever be a "simple" proof of his thm, Carleson replied... – Sean Tilson Jan 09 '11 at 22:31
  • ... "Some things are just hard." I certainly think that there are facts out there that resist absolute simplifications. For example, the odd order theorem of Feit and Thompson. I know it has been shortened, but it is still a hard result. – Sean Tilson Jan 09 '11 at 22:33
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    @Sean: I guess I should have explained the joke. The joke is that anything that is hard is incorporated into the machinery. – Qiaochu Yuan Jan 09 '11 at 23:51
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    @Qiaochu: I see. – Sean Tilson Jan 10 '11 at 02:32
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    "Some things are just hard." This is guaranteed to be true by a silly counting argument, as long as "being readable within a human lifetime" is a requirement for a proof to be "easy." – Daniel Litt Jan 10 '11 at 04:03
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    It is not guaranteed by a silly counting argument that there are short simple statements with irreducibly hard proofs. – gowers Jan 10 '11 at 07:50
  • @gowers: I suppose I'm assuming that there are infinitely many "interesting" statements, but this seems uncontroversial to me. – Daniel Litt Jan 10 '11 at 18:45
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    But there aren't infinitely many that are readable within a human lifetime, the criterion you were using for proofs. – gowers Jan 10 '11 at 20:13
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    Although it doesn't follow from a simple counting argument, it does follow from Godel and Turing's results. Fix a formal system, say PA or ZFC. Suppose that there were a computable function $f$ such that any result of length $n$, which had a valid proof, had a proof of length $\leq f(n)$. Then there would be an algorithm which, given a statement in our system, checked whether it was provable: Just try all proofs of length $\leq f(n)$. By a result of Turing, building on Godel's work, there is no such algorithm. – David E Speyer Jan 20 '11 at 14:42
  • I think that counting / non-halting arguments miss the non-mathematical part of gowers's statement, which is that the statement also be simple. – LSpice Jun 16 '11 at 15:52
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There are a number of facts in multivariable calculus that are obvious but hard to prove. For instance, the change-of-variables formula in a multiple integral is very easy to justify heuristically by talking about little parallelepipeds but troublesome (as I discovered to my cost in a course I once gave) to justify rigorously. And the same goes for the inverse function theorem: although the proof can be made quite transparent and the need for continuous differentiability makes good intuitive sense, there seems to be an irreducible core of actual work needed (in particular, the use of a fixed-point theorem to replace the use of the intermediate-value theorem in the 1D case).

I'd be quite glad to be told that this answer was wrong. If anyone knows of a link to an exposition of these results, particularly the first, that does proper justice to their intuitive obviousness, I'd be very pleased to hear about it.

gowers
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    I'm afraid you're right. The change of variables formula for multiple integrals is a notorious "Is it really this hard?" moment in mathematical exposition. – Pete L. Clark Jan 11 '11 at 18:14
  • Did you mean the inverse function theorem or the implicit function theorem? I thought the former only required the multivariate chain rule. – Qiaochu Yuan Jan 12 '11 at 20:44
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    The inverse and implicit function theorems are actually equivalent. – Paul Siegel Jan 15 '11 at 15:57
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It takes Russell and Whitehead several hundred pages to prove that $1+1=2$ in Principia Mathematica. They then say that "the above proposition is occasionally useful."

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    I've always wondered about the next sentence: "It is used at least three times, in $\ast 113 \cdot 66$ and $\ast 120 \cdot 123 \cdot 472$." [my emphasis]. Does this express some sort of dry humour or is it meant seriously? – Theo Buehler Jan 12 '11 at 03:58
  • The end of the proof is here: $\ast 110 \cdot 643$. http://quod.lib.umich.edu/cgi/t/text/pageviewer-idx?c=umhistmath&cc=umhistmath&idno=aat3201.0002.001&frm=frameset&view=image&seq=126 – Theo Buehler Jan 12 '11 at 04:03
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    Then Poincare made a great comment to this book: To those who doesn't even know 1+1=2. – Henry.L May 04 '13 at 04:09
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    Unlike other answers, that's an artifact of their particular formalization—modern formal proofs of 1+1=2 are immediate, at least in type theory. In fairness, it's cool enough they managed to formalize math at all. – Blaisorblade May 06 '17 at 22:37
  • @Henry.L: Interesting quote. Reference? – C.F.G Oct 09 '20 at 20:22
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That $\mathbb R^n$ has topological dimension $n$. In a similar vein that affine space $\mathbb A^n_k$ over a field $k$ has Zariski dimension $n$.

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    Isn't the problem how to define topological dimension? With the "right" definition, is the proof hard? – Daniel Moskovich Jan 09 '11 at 15:21
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    Dear Daniel, indeed defining topological dimension required amazing ingenuity, from Lebesgue among others. He later explained that his intuition came from contemplating a brick wall and noticing that some points had to be covered by 3(=2+1) bricks! However even granting this, proving that $\mathbb R^n$ has dimension $n$ remains a difficult problem needing techniques of algebraic topology to be solved, even though the definition is on the level of general topology. (To be continued) – Georges Elencwajg Jan 09 '11 at 16:11
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    (Continuation) This is confirmed by Munkres in his well-known text-book Topology. After 10 pages (§50) devoted to dimension theory, he concludes "we do not ask you to prove...that the topological dimension of an $m$-manifold is precisely $m$. And for good reason; the proof requires the tools of algebraic topology." Another indication of hardness is the "invariance of domain" theorem: non-empty open subsets of $\mathbb R^n$ and $\mathbb R^m$ are never homeomorphic unless $n=m$.This does not involve the definition of "dimension" but is quite a difficult theorem (first proved by Brouwer). – Georges Elencwajg Jan 09 '11 at 16:32
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    It's easy to prove that $dim(\mathbb R^n)\le n$. The opposite inequality, $dim(\mathbb R^n)\ge n$, is equivalent and about as difficult to prove as the fixed point property for the n-ball (or n-cube $\mathbb I^n$). – Włodzimierz Holsztyński May 04 '13 at 20:39
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There is a whole class of examples of the following general form: There is an obvious candidate for the solution to an optimization problem, and the obvious candidate is in fact best, but it's very hard to prove that it's best. Two of the examples mentioned in the comments—isoperimetric inequalities and sphere packing—fall into this class. Lower bounds in computational complexity furnish other examples, although our knowledge in this area is so pitiful that the best examples are still conjectural.

I like these examples better than the topological ones like the Jordan curve theorem and the invariance of domain, because there is room to argue that (for example) what makes the Jordan curve theorem hard is that modern mathematics has an exceedingly general definition of a Jordan curve that includes monsters that are non-rectifiable, nowhere differentiable, etc. The "man in the street" doesn't have these monsters in mind when judging that the Jordan curve theorem is obvious. In contrast, if we take something like "the kissing number of the sphere is 12," the man-in-the-street's conception of a counterexample is really no different from the mathematician's. It's just that the man in the street will be convinced after a few minutes of playing with velcro balls and the mathematician won't.

Timothy Chow
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    Right. Any Jordan curve the "man in the street" is thinking of is basically piecewise linear. – Qiaochu Yuan Jan 09 '11 at 22:32
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    Well, unless they've never seen a circle, piecewise $C^\infty$ ... – Yemon Choi Jan 10 '11 at 01:40
  • Likewise, for invariance of dimension, the "man in the street" (probably more like "vector calculus student" for this question) would be thinking of smooth, or at least piecewise smooth, maps, invariance under which is much easier. – Charles Staats Jan 10 '11 at 05:22
  • @Yemon: "basically" here should be interpreted as "clearly homotopy equivalent to" :) In any case, people don't draw circles, they draw piecewise linear approximations to circles. – Qiaochu Yuan Jan 10 '11 at 14:59
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    I don't draw PL-approximations to circles... – Steven Gubkin Jan 10 '11 at 15:45
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    I don't even draw PL approximations to polygons! – George Lowther Jan 11 '11 at 02:54
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    There is a trick of using your pinky knunckle as the point of a hand compass which allows you to draw "perfect" circles without the aid of a metal compass. I always impress students during office hours with this trick. – Steven Gubkin Jan 11 '11 at 20:55
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    P = NP is the ultimate example of this genre. The exhaustive search is obviously a good strategy for solving an NP problem, and it appears to be best, but one cannot prove this easily! – David Harris Jan 13 '11 at 18:58
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The trefoil knot is knotted.

aorq
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    I don't think this is hard to prove, as long as you don't insist on a "useful" proof. See Tietze's 1908 proof. The moment you have any presentation of the fundamental group (Wirtinger/Dehn/whatever), you find a representation onto a non-abelian group and you finish. – Daniel Moskovich Jan 09 '11 at 18:37
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    @Daniel: that seems like another example of hiding behind machinery to me. You at least need to know that fundamental groups are homotopy invariants and what the fundamental group of the circle is. If you wrote out the complete proofs of all the results you're depending on, would you still consider the resulting proof "easy"? – Qiaochu Yuan Jan 09 '11 at 20:37
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    Everybody knows that a trefoil is knotted, but only in order to state and prove this fact you need lots of technical things! A PL or smooth structure to define the knot itself (otherwise it unknots...), the whole machinery of fundamental groups (which is not obvious), prove that the complement of a knot does not vary when you isotope the knot, the Wirtinger presentation, and some non-trivial surjective homomorphism onto a nontrivial group. Of course one may also use some simple invariants, but they use that two knots are related by Reidemeister moves, which is also non-immediate to prove. – Bruno Martelli Jan 09 '11 at 20:41
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    I strongly disagree with what everyone is saying. There's the formulation, then there's the proof. The formulation says (or at least immediately implies) that a certain 3-manifold (the trefoil complement) is not the solid torus. The proof is about as straightforward as any statement about 3-manifolds could possibly be. Moreover, the proof was found almost immediately when the techniques existed- it was never an "open problem". So yes- if I wrote out all the proofs of everything I use, I'd consider it long but easy. Indeed, I think the fundamental group proof is easiest. – Daniel Moskovich Jan 09 '11 at 20:57
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    Modulo the Reidemeister moves I think an easier proof is noticing that the trefoil is $\mathbb{F}_2$ colorable and the unknot isn't. – solbap Jan 09 '11 at 21:19
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    @Daniel: in the context of the OP's clarification in the comments ("I want theorems for which a plausible argument springs to mind at the level of sophistication required to understand the statement, but for which a proof requires a higher level of sophistication.") I think it is totally reasonable to argue that the statement of the result requires a much lower level of sophistication than any of its proofs. (The plausible argument here is something like "if you try it with a physical trefoil knot it's obviously knotted.") – Qiaochu Yuan Jan 09 '11 at 23:48
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Subgroups of free groups are free. The plausible argument is that any relation satisfied in a subgroup must somehow translate to a relation satisfied in the larger group. Nowadays I guess most people see the proof which proceeds through the fact that the fundamental group of a graph is free, but it's not trivial to set up this machinery (even if one uses a purely combinatorial definition of fundamental group). I don't know how hard the algebraic proof is; perhaps it's easier.

Qiaochu Yuan
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  • I doubt the algebraic proof is easier. – Sean Tilson Jan 09 '11 at 23:03
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    The original algebraic proof (due to Nielsen) is not that bad. The basic idea is that you show that any subgroup has a generating set $S$ such that if $w$ is a reduced word in $S$, then when you plug the free group elements corresponding to elements of $S$ into $w$, at least one letter survives from each "letter" in $w$. The argument that you can do this is (slightly) similar to row reduction in matrices. There's a readable account of this towards the beginning of Lyndon and Schupp's classic book on combinatorial group theory that is well worth reading. – Andy Putman Jan 10 '11 at 04:21
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    For me, the Theorem that subgroups of free groups are free, is not obvious at all. – Martin Brandenburg Jan 12 '11 at 08:03
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    For me neither. The plausible argument above also "shows" that submonoids of a free monoid are free. – Jan Weidner Aug 03 '11 at 16:01
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    Well, I didn't say it was right! – Qiaochu Yuan Aug 03 '11 at 16:50
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The consistency of Peano Arithmetic. This is provably hard to prove, and I think that most would agree that it is obviously true (if not, why are we still doing mathematics?)

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    Peano arithmetic doesn't have to be consistent in order for us to meaningfully do mathematics. It just has to have the property that the shortest contradiction is so long that it is impossible to write down before the end of the universe! – Qiaochu Yuan Jan 09 '11 at 23:00
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    Wait a minute. If there is a contradiction, then everything is provable, right? So you can get nice short contradictions. I guess it could still be that the shortest proof of a contradiction must be very long. – Jeff Strom Jan 09 '11 at 23:13
  • Right, that's what I meant. Apologies for the inaccurate wording. – Qiaochu Yuan Jan 09 '11 at 23:41
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    The consistency of Peano arithmetic is not "hard to prove" in the intended sense of "hard." Logical strength is not the same as psychological difficulty or length of proof or any plausible notion of the hardness of finding a proof. – Timothy Chow Jan 10 '11 at 22:43
  • Tim, I take your point that formal difficulty (logical strength) is not necessarily the same as psychological difficulty. Nonetheless, I think that the formal difficulty of Con(PA) agrees pretty well with psychological difficulty. The proofs of Con(PA) that I am aware of are either quite hard (such as Gentzen's) or else they "cheat" by assuming higher order objects whose consistency is more problematic than that of PA itself. – John Stillwell Jan 11 '11 at 00:59
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    Gentzen's proof cheats as well; if you harbor serious doubts about the consistency of PA then you're likely to harbor doubts about such a strong induction principle. Cheating is inevitable. That doesn't mean that it's hard to cheat. – Timothy Chow Jan 11 '11 at 14:56
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    The proof is extremely easy if done the way a non-logician would do it: point out that $\mathbb N$ is a model. What is hard is a SYNTACTIC proof. – Christian Remling Apr 12 '14 at 18:54
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I agree this question is interesting, but only in a psychological rather than mathematical sense, i.e. the only reason the jordan curve theorem seems obvious is that we do not appreciate the generality of the definition of "continuous", rather taking our simplest intuitive examples as typical. Indeed the proof for smooth functions is pretty easy (cf. Guillemin and Pollack), and how many of us distinguish intuitively between (piecewise) smooth and continuous functions? For instance young students assume the intermediate value theorem is obvious because they do not appreciate the local nature of the definition of continuity, i.e. they intuitively assume that the intermediate value theorem is the definition of continuity, as indeed it was in a less rigorous time. Of course the proof of the IVT is a justification of the reasonableness of the definition of continuity. As Moishezon remarked to us as students: " even if it is obvious, you still have to prove it". Or as Tate said after giving an irresistible pictorial argument in first year honors calc. for the continuity of a composition of continuous functions; :"Of course this is NOT a proof! I have merely rendered it intuitively plausible!" (a statement i did not believe at the time.)

Problems in freshman calculus: 1) Give a characterization of a function g such that g is a primitive of a given Riemann integrable function f. Is it enough to assume that g is continuous and differentiable wherever f is continuous, and that g has derivative equal to f at such points? E.g. is a continuous function which is differentiable with derivative zero a.e. a constant function? If not, what assumptions do you have to add?

2) Give an intrinsic characterization of a function g that is a primitive of some unknown Riemann integrable function on [a,b]. Is it enough to assume that g is Lipschitz continuous?

I guess i would give more credence to this if it concerned say theorems that have physically compelling arguments that are hard to make mathematically rigorous, such as Riemann's arguments for the existence of meromorphic functions of second kind with arbitrary poles.

When someone says it is "obvious" that Euclidean space R^n has dimension n, they are really saying that any definition for which this is false is a bad definition, not that it is easy to give an appropriate definition, nor that it is easy to prove the theorem even for a good definition. So this is just an imprecise use of language.

Let me pose a little fun question: Since everyone knows that if n < m, there can be a continuous surjection, but no homeomorphism from R^n to R^m, what about a continuous injection from R^m to R^n? What is the obvious answer? Is it also the correct answer? How much does your response draw on some non obvious mathematical reasoning?

My best idea in the direction of the original question is: "why is a straight line the shortest smooth curve joining two points?"

roy smith
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I think that the ergodic theorem is a good example of this. In down-to-earth terms it says that if you have a box full of gas then the average velocity of all of the gas particles at a given time (the space average) equals the average velocity of a single given particle over time (the time average). This can be regarded as at least a partial theoretical justification for the fact that gas in a container reaches an equilibrium state over time. And what could be more obvious than that?

Yet the ergodic theorem revealed itself as frustratingly difficult to prove. You might think that the challenge would be to just come up with the right precise formulation of the problem; indeed, I don't think it was until people started to identify the measure theoretic underpinnings of probability theory that this was really possible. But while any student with a semester of measure theory under his/her belt can understand the modern formulation of the pointwise ergodic theorem, I highly doubt that very many could supply a correct proof without a hint. For some reason, the proof simply demands an ingenious combinatorial trick.

Paul Siegel
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Chess is not a forced win for black.

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    @Professor Borcherds: if this is a theorem, could you give some indications of the proof? – Pete L. Clark Jan 09 '11 at 17:40
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    My guess is that Prof. Borcherds said that is should be possible to prove this statement, and that is seems to be true, "obviously", but there is no such proof. As far as I know there is not even an idea of how one could prove such a statement. – Tim van Beek Jan 09 '11 at 18:34
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    All the more surprising, since the game which is just like chess, but where each side gets to make two moves in a row is provably not a forced move for black (proof: exercise) – Igor Rivin Jan 09 '11 at 21:24
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    Along the same lines is, "Chess is a forced win for black if White gives queen odds." – Timothy Chow Jan 09 '11 at 21:49
  • @Igor: Sorry, but is there an obvious way how white can win? I don't see it... – Tim van Beek Jan 09 '11 at 22:29
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    @Tim van Beek: no, but there's an obvious reason white can at least draw. An example of a silly chess variant where white actually has a demonstrable winning strategy is when the first check wins the game. – Chris Eagle Jan 09 '11 at 22:56
  • @Timothy Chow: really? You have a proof? I don't think it is along the same lines at all. – Igor Rivin Jan 10 '11 at 01:27
  • @Igor: I meant that my statement is along the same lines as Richard's statement, not yours. – Timothy Chow Jan 10 '11 at 01:37
  • @Timothy Ah, yes, that is actually a very good example! – Igor Rivin Jan 10 '11 at 01:42
  • @Chris Eagle: Do you have a cite to the proof that white wins such a variant? – I. J. Kennedy Jan 05 '12 at 06:14
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As an undergradued, I remember having precisely this feeling when encountering (a version of) the weak Nullstellensatz, which says that the maximal ideals in $\mathbb C[x_1,\ldots, x_n]$ are the sets of all polynomials vanishing on a fixed point $(x_1,\ldots, x_n)$. This must be pretty obvious, what else a maximal ideal could be?

However, the statement now does not look so "obvious" to me anymore... and I don't know if this is a good or a bad thing :-)

Bruno Martelli
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    It's straightforward to see that any maximal ideal with residue field C has this property, so the weak Nullstellensatz is equivalent to the statement that all residue fields are C. I don't think this is "obvious," exactly. Certainly similar obvious statements are false. For example, one might guess that all residue fields of an infinite direct product of copies of R are equal to R, and this is completely wrong. – Qiaochu Yuan Jan 09 '11 at 20:42
  • There are some reasonably simple proofs of the result, my favorite being one that appeared in the Monthly in 2006 by Arrondo http://www.mat.ucm.es/~arrondo/monthly169-171-arrondo.pdf

    Still, they're never quite as obvious as one might think when first encountering the statement.

    – Thierry Zell Jan 10 '11 at 01:37
  • -1. This is not an obvious fact at all. Of course, when you have learnt some classical algebraic geometry, you understand this Theorem and then also its various proofs ... but this does not show that it's obvious "a priori". – Martin Brandenburg Jan 12 '11 at 08:06
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A Theorem is 'obvious' when one does not see an immediate obstruction (for instance a counter-example). Of course it may be true or false, depending on how you are lucky or not. An obvious true theorem whose proof is notoriously difficult is the existence of solutions to linear PDEs $P(i\nabla_x)u=f$ for constant coefficients operators (Malgrange-Ehrenpreis theorem). I don't mean elliptic, hyperbolic, parabolic PDEs, or PDEs of principal type. No, just PDEs. It is not only true but somehow accurate, because it becomes false when the coefficients are non constant, even with analytic coefficients (H. Lewy counter-example).

Dick: At first glance, the Fourier transform reduces the question to the resolution of an algebraic equation $P(\xi)\hat u(\xi)=\hat f(\xi)$. The difficulty is whether $\xi\mapsto\hat f(\xi)/P(\xi)$ is the Fourier transform of a distribution. Because $P$ may vanish, and $P^{-1}(0)$ can be quite singular, this is not a piece of cake. Malgrange had to prove his division theorem to solve it.

Denis Serre
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    Denis, thanks a lot for baiting me. I appreciate very much linquistic lessons I've learned. Of course, there are people to whom "the existence of non-trivial solutions to linear PDEs for constant coefficients operators" is obvious. Enjoy. – Wadim Zudilin Jan 09 '11 at 14:07
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    I believe "H. Levy" should by "H. Lewy", c.f. http://en.wikipedia.org/wiki/Hans_Lewy. – Pete L. Clark Jan 09 '11 at 17:38
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    @Denis: I don't doubt you for a moment, but can you explain a little bit better where the difficulty lies? Doesn't the Fourier transform reduce "analytic" question about linear constant coefficient differential operators to a corresponding "algebraic" question about polynomial multiplication operators on the transform side? That is, doesn't solving of $P(\partial/\partial x^i) f(x) = g(x)$ just reduce to the simple question of solving $P(\xi^i) \hat f(\xi) = \hat g(\xi)$ ? – Dick Palais Jan 09 '11 at 19:57
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    @Dick: Yes, it basically does. But the problem is then to solve this and it is not obvious that there is an invers of this polynomial function in the space of (tempered) distributions. To prove this you need some tricks (a priori estimates to use Hahn-Banach / Bernstein-Sato-polynomials / an explicit formula for the inverse / ...)

    Also the idea to use the Fourier transformation is not obvious for every mathematician.

    – Johannes Hahn Jan 10 '11 at 00:15
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    I don't think the existence of solutions to constant coefficient PDE is so obvious, even though the proof is conceptually straightforward -- my reason being that the statement is false on the torus. Maybe you would consider these obstructions obvious..? – Phil Isett Aug 27 '11 at 03:45
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That every continuous vector field on ${\bf S}^2$ has a zero is pretty "obvious" (when you think about the image of trying to comb the hair on a billiard ball) yet takes considerable machinery to prove.

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    But see Milnor's proof of the Hairy Ball Theorem in the American Mathematical Monthly, July 1978, pp. 521-524. (I wrote about this here: http://topologicalmusings.wordpress.com/2008/07/22/analyzing-the-hairy-ball-theorem/ .) – Todd Trimble Jan 09 '11 at 15:36
  • It is still not obvious, isn't it? Needs work I mean. – Patrick I-Z Jan 09 '11 at 16:25
  • @Patrick: Yes, still takes work; I was really responding to "takes considerable machinery", presumably referring to homology theory. – Todd Trimble Jan 09 '11 at 23:44
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    I misstated Lefschetz' easy argument. If the vector at p is non zero, then on a small disc around p all vectors are roughly parallel to that one. It is possible then to see that on the disc which is the complement of that one, the degree of the map defined by the vectors on the boundary is ±2. (Collapsing the external disc onto the inner disc, by deflating the balloon, reflects the vectors on the boundary circle in the tangent lines to that circle. This fixes the vectors at top and bottom, rotating the rest 180 deg every quarter circle. thus there is a zero outside the original disc.) – roy smith Jan 12 '11 at 18:21
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That the identity map of the circle is not nullhomotopic. [When one thinks about it, it is pretty much equivalent to the Brouwer fixed point theorem, which is not as obvious.]

Lennart Meier
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    This was my introduction to many sophisticated mathematical ideas. When teaching several variable calculus I asked my self how I could convince students that Stokes theorem was useful, and found that it could be used to prove this fact, hence also the fundamental theorem of algebra, Brouwer fix point theorem, etc... Equivalently, why is the one form "dtheta" locally exact but not exact? – roy smith Jan 20 '11 at 06:04
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The Kneser-Poulsen conjecture says that if a finite set of (labeled) unit balls in $\mathbb{R}^n$ is rearranged so that in the new configuration, no pairwise distance is increased, then the volume of the union of the balls does not increase. This was finally proved by Bezdek and Connelly in dimension 2 but remains open in higher dimensions.

There are several other notorious elementary problems in geometry that might qualify, e.g., the equichordal point problem, though this one is not quite as "obvious" as the Kneser-Poulsen conjecture.

Timothy Chow
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  • That's a nice example! – Alec Edgington Jan 11 '11 at 21:13
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    It seems to me far from obvious that that result should be true in all dimensions, especially after the dramatic disproof of the Borsuk conjecture due to Kahn and Kalai or the disproof of the Busemann-Petty conjecture. – gowers Jan 13 '11 at 16:23
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"Global regularity of the Navier-Stokes equation" is not yet in this category, but once a proof is found, I am sure it will be.

More generally, there are many PDE which are "obviously" solvable for physical reasons, but for which actually proving existence (particularly in "global", "non-perturbative" situations, and requiring strong (regular) solutions rather than weak ones), is extremely difficult. A typical example is the Boltzmann equation, for which good global regularity results have only become available recently, with the work of Villani and others.

EDIT: Admittedly, many of the global regularity problems become a lot easier if one applies a physically reasonable truncation. For instance, global regularity for Boltzmann is much easier if one can somehow restrict the particle velocities to never exceed some upper bound $c$. But then the non-obvious fact moves elsewhere; rather than global regularity, the issue is whether one has sufficiently quantitative bounds that these thresholds rarely get triggered. Physically, it is intuitively obvious that a Boltzmann gas is not routinely churning out particles travelling at close to the speed of light; but it is remarkably difficult to quantify and then establish this rigorously.

Terry Tao
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The Carpenter's rule: a planar linkage can be straightened without the links running into each other. Although the statement had initially seemed obvious, its truth or falsity was a matter of debate among the experts for several years until Bob Connelly, Eric Demaine, and Günter Rote finally proved it. (The analogous statement in 3 dimensions is actually false.)

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    And the analogous statement in dimensions greater than 3 is true: they can always be straightened! $\mathbb{R}^3$ is the exception. – Joseph O'Rourke Aug 26 '11 at 23:51
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That the surface of a sphere is not homeomorphic to the real plane.

This may be unfair, in that it requires a good understanding of continuous functions. But it is intuitively obvious at a significantly lower level of mathematical sophistication than is required for the proof.

Then again, it is almost equally "obvious" at the same level of sophistication that you can't turn a sphere inside out.

So the notion of "obvious" in this sense is too crude to distinguish between true statements and false ones, and the question shows hides a bias. It would be more balanced also to ask whether there are "obvious" statements which it is hard to prove false.

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    The obviousness of the impossibility of turning a sphere inside out really depends on what intermediate configurations are permissible. – S. Carnahan Jan 09 '11 at 17:45
  • Fair enough Scott. I was aware that a supposedly simple answer was becoming a complex comment. – Mark Bennet Jan 09 '11 at 20:16
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    compactness is a n important concept. – roy smith Jan 10 '11 at 03:58
  • Is it really hard to prove, though? One can say, "A point can go off to infinity on the plane but it can't on the sphere," which can be turned into a rigorous proof about sequential compactness. Right? – Akiva Weinberger Jun 02 '16 at 18:39
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If $T_1, \dots, T_k$ are subtrees of a tree $T$ which pairwise intersect, then $\cap_{i=1}^k T_i$ is non-empty. This is a standard fact in graph theory and almost completely obvious, but it is surprisingly difficult to give a correct proof.

  • Perhaps if you use the word "connected", it becomes easier to give a correct proof. Gerhard "Likes Making Some Proofs Easier" Paseman, 2018.04.10. – Gerhard Paseman Apr 11 '18 at 00:04
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    Is the following wrong? It's enough to do the case $k=3$: once we know this case, then $T_1,\ldots, T_{k-2}, T_{k-1}\cap T_k$ will satisfy the hypotheses of the theorem and we are done by induction. To do the case $k=3$, compute the Euler characteristic of $T_1 \cup T_2 \cup T_3$ using the "inclusion-exclusion" formula $\chi(X \cup Y) = \chi(X)+\chi(Y)-\chi(X \cap Y)$ valid for compact $X$ and $Y$. If $T_1 \cap T_2 \cap T_3 = \emptyset$ then the Euler characteristic is zero, which is impossible. – Dan Petersen Apr 11 '18 at 08:31
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The notorious Dehn's Lemma, and it's generalizations, the Loop Theorem and the Sphere Theorem. It was a bone in the throat of 3-manifold topology for almost half a century, despite being `obvious', until proven by Papakyriakopoulos in 1957.
A comment on why it is obvious: the only singularities one can possibly imagine a disc having are things like "stretch a feeler out and around and around through the disc"- and it's obvious how to re-embed to get rid of those. Dehn's Lemma is a statement in the PL category, not in the topological category, so nothing pathological can occur. There's nothing which could possibly go wrong- no way you could possibly create a singularity in a DISC which you can't kill by re-embedding. But... prove it!

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    -1: All kinds of stuff could go wrong! – Autumn Kent Jan 09 '11 at 16:23
  • Like what? What could possibly go wrong? It's a PL disc, for crying out loud! – Daniel Moskovich Jan 09 '11 at 19:25
  • The simple loop conjecture is also obvious, I take it? – Autumn Kent Jan 09 '11 at 19:59
  • For Dehn's lemma, the disk could have the property that every double curve fails to be simple when you pull it back to the disk, which is not of the "stretch a feeler out" variety. See page 46 of Hempel's "3-manifolds." – Autumn Kent Jan 09 '11 at 20:05
  • The reference is to Johansen (1935), after the problem had already been open for 25 years. I would never have thought of such singularities, and so I contend that it is "obviously true". Figuring out how it could possibly not be blatantly obvious seems a big part of the struggle for the proof- and if a run-of-the-mill undergrad were asked about it, they would surely say it was obviously true. – Daniel Moskovich Jan 10 '11 at 13:47
  • I don't think it's even clear that it should be true at all, much less obvious. I think that if you ask an undergraduate if they think this is true, they would say "no." Especially the Sphere Theorem. Can I give this another -1? :) – Autumn Kent Jan 10 '11 at 16:01
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    I don't know much about 3-manifolds, so perhaps my intuition is undeveloped--that said, when I found out about the sphere theorem, I did a double-take. Were someone to have asked me if I thought this was true before I ran across it, I would have guessed it was wildly optimistic. – Daniel Litt Jan 10 '11 at 18:39
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    @Richard: Almost all undergraduates I know would answer neither "yes" nor "no" but would stare blankly, not understanding the question. – Timothy Chow Aug 26 '11 at 18:26
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I would mention the triangulation and smooth stratification theorems for algebraic varieties and variants thereof (analytic, real analytic etc.) These results are quite useful and I would say they seem obvious, at least from my experience. However, it is tricky to find complete proofs in the literature (especially in the real analytic case, which implies all the rest). I would say this is not because the proof is that difficult; it is not, but it's a bit tedious to spell out all the details.

While I'm at it, let me also mention that proofs of these theorems can be found in the references given in the answers to this MO question: Embeddings and triangulations of real analytic varieties (many thanks again to Mohan and Benoit).

algori
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Inspired by the trefoil knot example: "If two knots are smoothly* isotopic, then their complements are homeomorphic." I'm not sure exactly how hard the proof is, but it certainly seems obvious, and I don't think there is a one line proof.

*Thanks to Richard Kent for pointing out that I need this adverb.

David E Speyer
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    You need to put "smoothly" in front of "isotopic," otherwise the statement is false: all knots are isotopic (by shrinking all the "knottedness" to a point, a.k.a. bachelor's unknotting). Usually, one fixes this by defining knots to be equivalent if they are ambient isotopic, in which case you get the homeomorphism for free. (In the smooth case, the proof amounts to proving that the isotopy extends to the 3-sphere.) – Autumn Kent Jan 10 '11 at 00:36
  • It's the definition of ambient isotopy! See e.g. Page 2 of Burde-Zieschang. – Daniel Moskovich Jan 10 '11 at 14:30
  • Right, which is why I said isotopy, not ambient isotopy. Let $\phi : S^1 \times [0,1] \to S^3$ be a continuous map which is injective on $S^1 \times \{ t \}$ for every $t$. Why is it clear that this can be extended to a family of automorphisms of $S^3$? (Possibly I should have said "smooth" rather than "continuous", I haven't absorbed Richard Kent's comment yet.) – David E Speyer Jan 10 '11 at 15:01
  • For example, http://en.wikipedia.org/wiki/Homotopy#Isotopy gives the defn: "When should two knots be considered the same? We take two knots, K1 and K2, in three-dimensional space. A knot is an embedding of a one-dimensional space, the "loop of string", into this space, and an embedding is simply a homeomorphism. The intuitive idea of deforming one to the other should correspond to a path of embeddings: a continuous function starting at t=0 with the K1 embedding, ending at t=1 with the K2 embedding, with all intermediate values being embeddings; this corresponds to the definition of isotopy." – David E Speyer Jan 10 '11 at 15:05
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    That was a misleading passage in Wikipedia. I tried to clarify it. – Douglas Zare Jan 11 '11 at 01:30
  • Ah, thanks, that is helpful. And now I see why you need smoothness. But I still maintain that the fact that smooth isotopy implies ambient isotopy is "obvious", but not easy to prove. – David E Speyer Jan 11 '11 at 02:45
  • @David: The proof follows immediately from the isotopy extension theorem, which, using the basic theorem of ODEs amounts to extending a time dependent vector field. Perhaps you meant that the converse is obvious but hard: two knots with homeomorphic complements are isotopic, ie.the Gordon-Lueke theorem? – Paul Jan 11 '11 at 03:53
  • @Paul : I can't imagine anyone who would call the Gordon-Luecke theorem "obvious". It's hard and subtle (for instance, it is false for links). – Andy Putman Jan 11 '11 at 04:08
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    Glad to help, David. I agree with the assessment that the extension is obvious but not easy. – Autumn Kent Jan 11 '11 at 04:22
  • I don't understand this, because a knot is an embedded object- it lives in the ambient manifold. So "smooth isotopy" isn't the most natural relation to consider. You make your life hard by choosing this formalism, and then claim it's obvious but hard that the smooth isotopy extends- but you created that difficulty yourself by your choice of formalism. And after all that, the argument for the isotopy to extend is precisely the standard one- it's no harder, and no easier, than one would expect. – Daniel Moskovich Jan 11 '11 at 13:31
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    @Daniel I disagree about which is the natural formalism. My physical intuition for why you can't untie a trefoil is that it would have to pass through itself, which I capture with the injectivity hypothesis. The fact that I would have to move space out of the way is completely irrelevant. After all, if I put a knotted rope in a vacuum, it is still knotted! – David E Speyer Jan 11 '11 at 14:50
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Here's a "trick" answer:

$[0, 1]$ is connected.

But I consider this a trick answer because the real difficulty is turning this into a completely rigorous statement. Understanding the rigorous definition of connectedness (and understanding the point of making definitions like this) can be a substantial hurdle, but once this hurdle is crossed, the proof is not difficult.

Frank Thorne
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In the same vein, chating: My advisor used to say "An interesting theorem is a theorem true which looks false". Well, tastes and colors... ;-)

Patrick I-Z
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Speaking of thesis advisers, mine said, "I think something should be called obvious only if it is obvious in the logical sense of if A implies B and if B implies C then A implies C". All else is subjective and hence capable of misuse. I have tried, but not necessarily succeeded, to follow this. I am constantly amazed/amused at how people coming at a problem from different points of views will find certain facts obscure or well known.

meh
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The independence of the Parallel Postulate (especially since the proof, which consists of demonstrating that elliptic geometry satisfy the other axioms, is not hard, yet too 2500 years to find).

Igor Rivin
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Inspired by ``the trefoil knot is knotted" answer, how about the fact that Reidemeister moves generate isotopy of PL knots? This is pretty obvious but a full proof requires a lot of machinery. Indeed, the proof was not known to Reidemeister, who took the fact that his eponymous moves generate isotopy as an unproven axiom. (See Daniel Moskovich's comment.)

Jim Conant
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    Are you sure? I think Reidemeister did prove it in Knottentheorie (1932), entirely combinatorially. There is a "general position" issue, which is an issue for all of PL topology- but it's no harder for the Reidemeister Theorem than anywhere else. See http://mathoverflow.net/questions/15217/proof-of-the-reidemeister-theorem – Daniel Moskovich Jan 10 '11 at 00:52
  • Actually, Reidemeister's "Knotentheorie" is my source for this, but it is somewhat dimly recalled. – Jim Conant Jan 10 '11 at 01:20
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    I stand corrected. It didn't make it into the original edition of Knotentheorie, but was proved 6 years earlier by Reidemeister (1926), and independently by Alexander and Briggs (1927), as I found by nosing around in the math library this morning. All details are there, including general position arguments. Wikipedia, to my surprise, is entirely accurate: http://en.wikipedia.org/wiki/Reidemeister_move – Daniel Moskovich Jan 10 '11 at 13:40
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All isometries of the plane are affine linear.

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    Is this considered hard? – Todd Trimble Jan 10 '11 at 15:20
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    Todd, I think something closely resembling David's statement might count as hard: the Mazur-Ulam Theorem, that every bijective isometry between real normed spaces is affine. An indication that it's not obvious is that it's false without the word "bijective": e.g. you can cook up a non-affine isometry from R to (R^2 with the infinity-norm). Reference: http://www.helsinki.fi/~jvaisala/mazurulam.pdf . I don't know whether it's appreciably easier if you stick to two dimensions. – Tom Leinster Jan 13 '11 at 18:21
  • But maybe you know all that, Todd... – Tom Leinster Jan 13 '11 at 18:22
  • Sorry Tom, didn't see your comment until now. Thanks for your comment. I had assumed "isometry" meant with respect to the usual Euclidean distance (2-norm). Here is a sketch of the proof I had in mind. First, by translation, assume WLOG that the origin is taken to itself. Then |f(v)|^2 = |v|^2 for all v. Since d(fv, fw) = d(v, w) for all v, w, it follows that |fv - fw|^2 = |v - w|^2. Since the inner product of v and w can be defined in terms of |v|^2, |w|^2, and |v-w|^2, it follows that f preserves inner products. So it takes an orthonormal basis to an orthonormal basis. Finally, [cont'd] – Todd Trimble Mar 05 '11 at 15:34
  • Since the inner product is linear in each argument, it quickly follows that if e_i is orthonormal, then the inner product of f(av + bw) against all the f(e_i) matches the inner product of af(v) + bf(w) against all the f(e_i). Since f(e_i) is an orthonormal basis, this shows f preserves linear combinations av + bw. – Todd Trimble Mar 05 '11 at 15:39
  • @TomLeinster The link to the pdf seems to be broken. Could you take a look at it, please? –  May 20 '19 at 15:00
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    @Brahadeesh The citation is Jussi Väisälä, A proof of the Mazur-Ulam theorem, American Mathematical Monthly 110 (2003), 633-635, https://doi.org/10.1080/00029890.2003.11920004. I couldn't quickly find a free copy online, but maybe you'll have better luck. – Tom Leinster Aug 04 '19 at 13:27
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The first of the Tait Conjectures seems intuitively obvious:

Any reduced diagram of an alternating link has the fewest possible crossings.

This 19th century conjecture is difficult to prove, with the proof coming only in 1987 by Kauffman, Murasugi, and Thistlethwaite, using the Jones Polynomial. The discovery of this proof was a huge coup for quantum topology; a quantum invariant was used to prove a difficult classical open problem.
While this is certainly hard to prove, I don't think it's unexpectedly hard to prove. Knot diagrams modulo Reidemeister moves form a rather complicated algebraic structure; and there's no reason to expect that any statement about knot diagrams should be easy to prove.

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On page 33 of Stable Mappings and Their Singularities by Golubitsky and Guillemin (GTM 014; 1974), the following proposition is characterized as an "obvious, but surprisingly complicated result":

Proposition 1.10: Let S be a nonempty open subset of $\mathbb{R}^n$. Then S is not of measure zero.

Edit: As pointed out in the comments, Gowers already mentioned an equivalent result whose difficulty is more clear. Please vote this answer down (I can't vote down my own answers).

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    I must be completely missing the point here. Why is this hard? Can't you just show that there is a small cube inside $S$? – Deane Yang Feb 28 '11 at 02:14
  • I'm also missing the point. I always thought this was a trivial proposition. – Jim Conant Feb 28 '11 at 02:50
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    I suppose that this is trivial if one constructs Lebesgue measure by first assigning measure to rectangles and in the end declares a set to be of measure zero if...well, if it has Lebesgue measure zero. If one merely has the classical definition of a zero measure set (a set such that $\forall \varepsilon > 0$ can be covered by countably many rectangles so that the sum of their volumes is at most $\varepsilon$), this is nontrivial. Actually Gowers already mentioned an equivalent result, in which the difficulty is more clear. – Mark Feb 28 '11 at 03:34
  • Aha. That makes sense. – Jim Conant Feb 28 '11 at 15:39
  • Yes, thanks, Mark, for the explanation. – Deane Yang Feb 28 '11 at 15:52
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    Does this mean that it is hard to show that a rectangle does not have measure zero? – Deane Yang Feb 28 '11 at 15:54
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Dynamic programing principle (DPP) is one of the 'obvious' and also intuitive one, in the control problem. Many papers proves its validity in various setup, and all proofs are very complicated. But, there is rarely a counter example of DPP. I wonder, if there is general framework on it. See, Dynamic programming principle (DPP)

kenneth
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P is not equal to NP. This is "obvious", is a straightforward arithmetic proposition doesn't involve any fancy set theory or spacefilling curves, and yet it's so hard that there have whole workshops ("Barriers") and important papers (BGS, natural proofs etc.) devoted to the question of what makes it so hard. Scott Aaronson describes "a would-be P≠NP prover who hasn’t yet grasped the sheer number of mangled skeletons and severed heads that line his path." P≠PSPACE is even more obvious and yet there is a comparable lack of progress.

none
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I think a very good example is Kepler Conjecture:

http://en.wikipedia.org/wiki/Kepler%27s_conjecture

This conjecture stated that the most "tight" stack of same balls have only two kinds of arrangement with a fixed density.

Every physicist knows that it's true, no mathmaticans ever proved it.

Fortunately, Hales used computers to step forward a little little bit.

Henry.L
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The Hodge decomposition theorem

It is obvious that there is a unique point in any given affine plane in a finite-dimensional euclidean vector space which is closest to the origin.

Therefore it would seem similarly obvious that every de Rham cohomology class on a compact oriented riemannian manifold should have a unique representative with minimal $L^2$ norm: namely, its harmonic representative.

Yet it takes some effort (elliptic regularity,...) to prove that the harmonic representative does in fact exist, i.e., that it is a smooth differential form.

2

The stability of Minkowski Spacetime

An asymptotically flat initial data set for the Einstein equations that is sufficiently "close" to the initial data for Minkowski spacetime generates a solution to the Einstein equations that approaches Minkowski spacetime asymptotically. (try saying that fast 3 times)

It is "obvious" because of our physical experience and intuition with gravity, and it is hard to prove because Einstein's equations are quite subtle and complicated.

There are other theorems in mathematical relativity that fall into this category, but this one is especially striking since it is particularly difficult to prove, while it "feels" blatantly true.

  • The problem is if the near-flat initial data consists of a ton of gravitational waves all going inward, and making a black hole. This result is only obvious once you bound the waves in wavelength and energy to prevent this scenerio, and then it is just as complicated looking as the proof. – Ron Maimon Aug 27 '11 at 05:13
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No exatly what asked..

But historically: The fifth postulate of Euclid (to a point outside a straight line passes exactly one line parallel to this line). At first it seems an obvious fact, and tried to prove it....

Buschi Sergio
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Very good candidates for this question are theorems that amount to saying that some sequence behaves randomly in a way. Both the fact that the statements are obvious and the fact that they are usually hard to prove, are explained by the fact that there is just no reason for the sequence to nót behave randomly. The primes are of course notorious for this. Easy example that springs to mind: there are about as many primes whose last (decimal) digit is $1$ as there are primes whose last digit is $3$.

Woett
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In the same genre, if not the same type: The Fundamental Theorem of Algebra. Easily understood by high schoolers, plausible, beautifully simple to state. As far as I know, there are no nice proofs understandable by a good (not brilliant) high school student.

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    I do not know why this is plausible. What would be an "obvious" reason to expect that a degree 6 polynomial with real coefficients has a complex root? – Andrés E. Caicedo Jan 10 '11 at 23:30
  • The division theorem implies a non constant polynomial defines an open map from the sphere to itself. Since any such map is also continuous, hence closed, it is surjective. that's the best I can do. It seems conceivable to make that seem plausible to someone with a little intuition, if not obvious. – roy smith Jan 11 '11 at 00:43
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    Here is a proof whose main idea is understandable by many high school students. The winding number of the image of the circle of radius $r$ changes from $0$ at $r=0$ to the degree of the polynomial for $r$ large, and it can only change when there is a $0$ of the polynomial. – Douglas Zare Jan 11 '11 at 01:11
  • Doug's argument is made completely explicit in the book of Steenrod and Chinn, aimed precisely at high school students. – roy smith Jan 11 '11 at 04:52
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    If I'm following this correctly, some comments say the example is unsuitable because proving the theorem is actually easy, while the oldest comment says it's unsuitable because it's not obvious. What a mess! Keeping in mind how long it took between the result being conjectured and the first actual proof, I think we are too far removed in time from the result to truly appreciate it from a historical perspective, and the FTA is too fundamentally rooted in students educations to imagine how hard it would be for someone who was a blank slate. – Thierry Zell Jan 11 '11 at 14:43
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    @Thierry: agreed. Anything Euler attempted unsuccessfully to prove can't be all that easy. I think your comments apply to comments on several other answers as well in relation to how rooted algebraic topology is in many mathematicians' educations these days. – Qiaochu Yuan Jan 12 '11 at 14:34
  • Have you seen the proof which looks at Re(f) and Im(f) separately and studies the combinatorics of how they cross? I think a high school student could understand this. I like this exposition http://arxiv.org/abs/math/0511248 – David E Speyer Feb 28 '11 at 19:32
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On an elementary level, the intermediate value theorem is surprisingly deep.

On a less elementary level, the prime number theorem is "obvious" from $\sum_{p\leq x}1/p\sim \log\log x$ (that was noticed by Euler) and Dirichlet's theorem on primes in arithmetic progressions is "obvious" if you use the sieve of Erathostenes.

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A differentiable manifold M that is homeomorphic to the n-sphere is also diffeomorphic to the n-sphere . Obvious, but wrong ! (But right for 1-, 2-, 3-, 5-, 6- and 12-spheres).

jjcale
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Since this was resurrected, here is the statement that at this time seems to me to have the greatest gap between obviousness of truth and obviousness of proof:

  • There exists a natural set theoretic universe in which every subset of [0,1] is Lebesgue measurable, so that the reals admit no well order and do not inject into aleph-1.

Here are a different class of obvious theorems, these are only obvious in the sense of physical intuition. They took a long time to prove:

  • The existence of solid matter occupying space (in the lowest energy state, the electron-nucleus system occupies a volume proportional to the number of nuclei)
  • The positive energy theorem--- every asymptotically flat solution of GR obeying the appropriate energy condition has a positive mass at infinity, with zero mass only for Minkowski space.
  • Hard sphere collisions on a negatively curved space are ergodic.

Here is a physically obviously true statement, which can be seen from physical intuition, but which is not proven (as far as I know):

  • The asymptotic fate of GR with a positive cosmological constant is within any causal patch, and except for a set of initial conditions of measure zero, a deSitter space.

The reason this is obvious is because the deSitter space maximizes the horizon area, which is a measure of the entropy.

Ron Maimon
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I think Godel's completeness theorem is very intuitive. For example, can you imagine a first order theorem that would be true for all groups, that you wouldn't be able to prove (by Godel's definition of `prove'). Of course not! But the proof of the completeness theorem is hard.

  • among people trained on the incompleteness theorem, this is not obvious. – Will Sawin Sep 30 '11 at 03:59
  • The completeness theorem is not really hard to prove. The propositional part may be a bit messy (which can be alleviated by a wise choice of a proof system), but the main Henkin construction is quite straightforward. – Emil Jeřábek Sep 30 '11 at 11:36
  • Will, I disagree. The incompleteness theorem talks about something quite different. About whether there is a small set of axioms that imply every true statement about the integers (or other models). I dislike the comparison people make between Godel's completeness theorem and his incompleteness theorem. They talk about very different things. Emil, you are correct. The proof is not very hard. But it is much easier to state the theorem and believe it than to actually prove it. – James D. Taylor Oct 01 '11 at 15:24
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    I agree that the completeness theorem sounds very intuitive, but I think this is misleading. It takes some serious thought to convince yourself that a particular definition of formal proofs captures mathematical practice (even if you believe intuitively that some definition should work, it's much less obvious that a given deductive system really is complete). Furthermore, I'd bet that many mathematicians would find it equally intuitive that there should be a complete proof system for second-order logic, and of course incompleteness tells us there isn't. So completeness is pretty subtle... – Henry Cohn Sep 08 '12 at 18:17
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Fermat's Last Theorem. (Should I state it?)

Wadim Zudilin
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    Why is Fermat's last theorem 'obvious' ? – Denis Serre Jan 09 '11 at 13:40
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    It was already obvious to Fermat! ;-) There are different ways to interpret the word "obvious". Not so many people doubt it's true. And just a few understand its proof. – Wadim Zudilin Jan 09 '11 at 13:48
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    Citing the dictionary, "Obvious: easily perceived or understood; clear, self-evident." Citing the OP, "I'm interested in examples of theorems that are 'obvious', and known to be true." – Wadim Zudilin Jan 09 '11 at 13:52
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    Wadim, sorry for the downvote; but I think you're misunderstanding the English language here. Although the meaning of the statement is obvious, its truth is definitely not. Suppose you'd never heard of it before, and someone asked you to guess the truth or falsity of the statement, in less than an hour, say for a bet of 100 dollars or similar. I cannot believe that any human being could possibly guess "true" and have enough "reasonable" confidence to want to make the bet. – Zen Harper Jan 09 '11 at 14:06
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    Zen, I guess you are not the one who tells me who could tell me that the Jordan curve theorem is obvious (how many students in your class would agree?). The word "obvious" is too subjective. The things which are obvious to you may be not obvious to others. Therefore, we have a perfect voting system. Everybody feel free to vote down! I have to go to bed to get rid of this nightmare. English language lessons are too much for Sunday. G'night. – Wadim Zudilin Jan 09 '11 at 14:16
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    Wadim, for what it's worth I also disagree that the Jordan curve theorem is obvious. It's certainly not any more obvious than the nonexistence of space-filling curves. – Qiaochu Yuan Jan 09 '11 at 15:04
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    I also don't see any reasonable sense in which FLT is obvious. Let me just restrict attention to FLT(3). This problem is equivalent to the following: you are given a rational elliptic curve which has three obvious (!!) torsion points on it. You look by trial and error for a while and don't find any further rational points. What is the chance that there are no further rational points on this curve? Current wisdom suggests: precisely $\frac{1}{2}$. So that there are no further rational points even on this one curve is far from obvious: one should not even bet on it at unequal odds. – Pete L. Clark Jan 10 '11 at 04:44
  • Some obviousness for the FLT. If you look at the statement of $a^q - b^q = c^q$ then it is firstly obvious that all primefactors of $c$ must occur with the exponent $q$. But from the little theorem of fermat and some checking around it becomes "obvious", that the form of the lhs $ \frac{a^q - b^q}{a-b} $ in the most cases allows primefactors only with low exponent: the number of occurences of high values of a so-called fermat-quotient is somehow reciprocal to its height, but it is required, that it must be $q$. So, few simple (well: not trivial) heuristics make FLT (a bit more) "obvious"... – Gottfried Helms Jan 10 '11 at 15:55
  • Couldn't correct the previous comment. Surely it should be written: "primefactors of $c^q$ must occur..." . Later I meant: "it is required, that its exponent must be $q$ ". Sorry... – Gottfried Helms Jan 10 '11 at 15:59
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Maybe on the boundary of what's allowed, but I would say most basic geometric things like Pythagoras' Theorem, trigonometry with sine/cosine, the area of a circle, etc. etc. etc.; here of course the difficulty is in defining what we mean by length, area, angle, etc. - in which case some of these become axiomatic, but then the difficulty is shifted onto proving that things do work correctly.

Zen Harper
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    -1 (sorry for the downvote): The difficulty is in statement of the theorems, not the proof. – Daniel Moskovich Jan 09 '11 at 14:31
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    Same kind: the surface of the sphere = 4π (Archimedes). Well, not really obvious and not really easy to prove. I'm down :( – Patrick I-Z Jan 09 '11 at 16:29
  • for the surface of a sphere, if you assume the volume of the sphere is known, and the earlier result on the volume of a pyramid, then archimedes' comment that the sphere is a pyramid with base equal to its surface, you are done. – roy smith Oct 03 '18 at 02:22
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I think this answers your question in a perverse way: All statements in the theory of Natural Numbers provable from the ZFC axioms of set theory. They are obviously true by definition.


EDIT: Looking at this objectively, it probably sounds like I'm saying if a statement is true, then it's obviously true. However, that was not my intent, and I apologize for what may have sounded like a thoughtless response. This is how I see it:

All statements expressible in the language of arithmetic can be represented by formulas in the language of set theory that are only $\Delta_1$ in the Levy hierarchy. In particular, all transitive models will agree on whether they are true. If we further restrict ourselves to only consider the true statements in $\mathbb{N}$ that are ZFC theorems, then all ZFC models will agree that these statements must be true so they are about as obvious to ZFC models as possible. Now if you are an oracle having knowledge of all such true statements, then you will probably develop an intuition that makes them all seem "intuitively obvious." This reflects the answers suggesting that a theorem is obvious after you prove it.

To add one more related point here, when addressing G$\ddot{\textrm{o}}$del's Incompleteness Theorem, one can naively ask about completing PA in the "obvious" way, i.e., by extending it to be the theory consisting of all true statements in $\mathbb{N}$. But of course such a completion is not computable.

Jason
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    What is this definition of "obviously true" that makes these obviously true? – Chris Eagle Jan 10 '11 at 04:00
  • By definition, the theory of the Natural numbers is the collection of all statements in the language of arithmetic that are true in $\mathbb{N}$. When we say that a statement in the language of arithmetic is true, we mean that is true in $\mathbb{N}$. Consequently, a statement in the language of arithmetic is true if and only if it is in the theory of the Natural numbers. The argument of whether the statement is in the theory of the Natural numbers takes place in the set-theoretic universe $V$, which models all of the theorems of ZFC. – Jason Jan 10 '11 at 04:29
  • My point was that given you know a statement is in the theory of the Natural numbers, if you were asked whether it were true or not, you'd be able to respond without hesitation with an affirmative answer. – Jason Jan 10 '11 at 05:03