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Infinity and an elementary question on continuous manifolds, surjective maps and a dimension formula. The well known "Peano curve" is a continuous surjective map

$$ f(t): I \rightarrow X$$

where $I:=[0,1]$ and $X:=I \times I$ is the unit square in $\mathbb{R}^2$.

There is an equivalence relation on $I$: Two points $s,t$ are equivalent ($s\cong t$) iff $f(s)=f(t)$, and we may pass to the quotient space $I/\cong$. There is the product map

$$\tilde{f}: I \times I \rightarrow X \times X $$

defined by $\tilde{f}(s,t):=(f(s),f(t))$ and the inverse image $R:=\tilde{f}^{-1}(\Delta) \subseteq I\times I$ equals the equivalence relation $\cong$. The diagonal $\Delta \subseteq X \times X$ is a smooth submanifold and the equivalence relation $R$ is "continuous" in the sense that it is the inverse image of a smooth equivalence relation on $X$ under a continuous map $\tilde{f}$. The equivalence relation $R$ gives rise to a continuous bijection

$$b: I/R \cong X$$

of sets. By definition $dim(I)=1, dim(X)=2$. The set $I/R$ is a quotient of $I$ by a continuous equvalence relation on $R$, hence $I/R$ is a "smaller set" than $I$ since we identify points in $I$ when we pass to the equivalence relation. Do you agree this is a "strange" phenomenon? What "happens" to the dimension? Intuitively it should drop. For fractals there is a notion of dimension - the Hausdorff dimension. Is it possible to define the dimension $dim_q(I/R)$ of the quotient $I/R$? If this is possible and $dim_q(I)=1$, we should "intuitively" have the formula $dim_q(I/R) \leq 1$.

Example 1. In general given a surjective map

$$f: M \rightarrow N$$

where $M,N$ are $C^k$-manifolds and $f$ is a surjective map of $C^k$-manifolds - the dimension should drop: Hence $\dim(N) \leq \dim(M)$. This property holds for $C^k$-manifolds when $k$ is "large enough". A continuous manifold is a fundamental object in mathematics and we want the formula to hold in this case as well.

Note. The dimension of a continuous manifold $M$ is a way to measure the "size" of the manifold, and when we take the quotient $M/R$ of $M$ by a "continuous equivalence relation" $R \subseteq M \times M$, we "identify points" and hence the dimension of the quotient should drop. Hence the quotient should not have a continuous bijection with a manifold of higher dimension. The underlying sets of the continuous manifolds $X:=\mathbb{R}$ and $Y:=\mathbb{R}^2$ have the same cardinality, hence there is a set-theoretic bijection $b: X \cong Y$. For the dimension $dim(X),dim(Y)$ to be "well defined", there should be no continuous surjection from $X$ to $Y$.

Question. Is there a way to "change" the definition of the "category of sets" such that the dimension formula holds for the category of continuous manifolds (with corners) and surjective continuous maps? We want to define the category $C:=Cont(\mathbb{R})$ whose objects $Ob(C)$ are finite dimensional continuous manifolds and whose morphisms $Mor(C)$ are continuous maps of continuous manifolds. In this category the following Lemma should hold:

Lemma. If $f: X \rightarrow Y$ is a surjective map in $C$ then $dim(Y) \leq dim(X)$.

Why should this formula hold? If $f$ is surjective then $Y$ is "smaller" than $X$ and hence $dim(Y) \leq dim(X)$ should hold.

This is an "elementary question" on set theory in a sense, and should be interesting for many of the readers of this forum. If you find a solution, please post it here.

A similar question has been asked here

https://mathoverflow.net/questions/383367/infinity-in-mathematics-the-peano-curve-and-other-paradoxes

Example 2. A set is "Dedekind infinite" iff (this is a wikipedia citation):

"In mathematics, a set A is Dedekind-infinite (named after the German mathematician Richard Dedekind) if some proper subset B of A is equinumerous to A. Explicitly, this means that there exists a bijective function from A onto some proper subset B of A. A set is Dedekind-finite if it is not Dedekind-infinite (i.e., no such bijection exists). Proposed by Dedekind in 1888, Dedekind-infiniteness was the first definition of "infinite" that did not rely on the definition of the natural numbers."

So in some cases being a strict subset of itself is used to define a "Dedekind infinite set". Has the theory of finite/infinite sets been developed without this definition? I ask for references.

https://en.wikipedia.org/wiki/Dedekind-infinite_set

Noah Schweber
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hm2020
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    There even is a continuous surjection from the zero-dimensional Cantor set $C$ onto $[0,1]^{\Bbb N}$ which is infinite-dimensional. Dimension raising maps are fun! – Henno Brandsma Feb 21 '21 at 10:55
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    Dimemsion $\neq$ size. They behave quite differently. – Henno Brandsma Feb 21 '21 at 11:00
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    You should change the category of manifolds and maps: Consider Lipschitz manifolds and (locally) Lipschitz maps between them. Then everything will work the way you want. – Moishe Kohan Feb 21 '21 at 11:25
  • @HennoBrandsma - A $C^k$-manifold $M$ has dimension $n \geq 1$ ($n$ an integer) iff there is an open cover $U_i$ of $M$ and homeomorphisms $\phi_i: U_i \rightarrow V_i \subseteq \mathbb{R}^n$ ($V_i$ an open set) such that on the intersections $V_i \cap V_j$ the composed maps $\psi_{ij}:=\phi_i^{-1} \circ \phi_j$ is a $C^k$-map. Hence a $C^0$-manifold requires the map $\psi_{ij}$ to be continuous. When you claim dimension $\neq$ size - what does this mean? What is the "size" of the image $f(I)$ in the case of the "Peano map" $f(t)$? – hm2020 Feb 21 '21 at 12:51
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    This question was closed in MO. But the comments there are just as relevant here. https://mathoverflow.net/questions/383367/infinity-in-mathematics-the-peano-curve-and-other-paradoxes – GEdgar Feb 21 '21 at 13:14
  • @GEdgar - the content of the comments in the MO thread is included Example 1 in the post. – hm2020 Feb 21 '21 at 13:18
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    For $\dim(X)$ to be well-defined it only need be preserved by homeomorphisms just like size is by bijections. – Henno Brandsma Feb 21 '21 at 14:06
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    Why do you want to change the category of sets? The obvious thing to do seems to be instead to change the category of spaces (specifically, to restrict attention to particularly "nice" morphisms) you're looking at. I don't see why the category of sets enters the picture at all. – Noah Schweber Feb 21 '21 at 16:13
  • @NoahSchweber - Why not change the "category of sets" to see if this leads to interesting results? To me it seems an infinite set should have some properties in common with a finite set - one property could be that it should not be possible to embed an infinite set as a strict subset of itself - this is a condition on the set of morphisms $Mor_{sets}(S,S)$ of a set $S$. With this definition you cannot compose all maps, though. – hm2020 Feb 22 '21 at 11:08
  • @MoisheKohan - problems related to our current definition of "infinite set" appear everywhere. In algebra there are problems when speaking of vector bundles of infinite rank. The grothendieck group $K_0(C)$ of the category $C$ of projective $A$-modules of countably infinite rank is trivial. One wants a non-trivial grothendieck group. This problem is not neccessarily about the definition of continuity - it is about infinity. – hm2020 Feb 22 '21 at 11:18
  • @MoisheKohan - "Regarding the Question after the Note: As it was pointed out in comments, you should not mess with the category of "sets"." As I mention in the revised post - I'm looking for places where the "category of finite/infinte sets" has been constructed without using the definition of a "Dedekind infinite set". – hm2020 Feb 22 '21 at 12:19
  • @MoisheKohan - I'm asking for an explicit and elementary construction of the "category of sets" without using the definition of "Dedekind infinite sets". – hm2020 Feb 22 '21 at 12:41
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    "To me it seems an infinite set should have some properties in common with a finite set - one property could be that it should not be possible to embed an infinite set as a strict subset of itself" It is difficult for me to adequately express how much I disagree with this. We'd have to somehow outlaw the map $x\mapsto x+1$ as a function from $\mathbb{N}$ to $\mathbb{N}$ - why??? – Noah Schweber Feb 22 '21 at 15:16
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    "problems related to our current definition of "infinite set" appear everywhere." My desire for a nonzero Grothendieck group of a category of projective modules of countably infinite rank is somewhat weaker than my interest in basic properties of the set of natural numbers. I just don't get the motivation here, ultimately. Yes, infinity is somewhat strange - no, it's not so strange that we should utterly do away with it. Besides, strangeness is fun. – Noah Schweber Feb 22 '21 at 15:16
  • @NoahSchweber - "Yes, infinity is somewhat strange - no, it's not so strange that we should utterly do away with it. Besides, strangeness is fun." I'm not asking people to "do away with infinity" or "outlaw" things. I'm asking if an explicit and elementary construction of "a category of sets" where we make this assumption has been written down in the litterature. – hm2020 Feb 22 '21 at 15:59
  • @NoahSchweber - "Besides, strangeness is fun." If there is a problem with inconsistency of the ZF system we are all wasting our time writing down complicated proofs - it is meaningless to do this. This is why you should care. – hm2020 Feb 22 '21 at 16:01
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    "I'm asking if an explicit and elementary construction of "a category of sets" where we make this assumption has been written down in the litterature." How about simply the category of finite sets? Presumably this doesn't satisfy your goal - ok, what do you want your category of sets to do exactly? I really don't understand what you're looking for here. – Noah Schweber Feb 22 '21 at 16:56
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    Also keep in mind that we can rephrase the $\mathsf{ZF}$ axioms to avoid anything resembling a Dedekind-infinite set in their statements. Specifically, only the axiom of infinity directly refers to something like Dedekind-infiniteness, but it turns out that it can be equivalently replaced with an axiom asserting the existence of some "appropriately unbounded" object. For example, "There is a nonempty linear order with no greatest element" does the job - nontrivially, since it's consistent with $\mathsf{ZF}$ that there are Dedekind-finite linear orders without greatest elements! – Noah Schweber Feb 22 '21 at 16:58
  • (Proof hint: look at the suborders of such a linear order which are both well- and co-well-ordered.) Essentially, the existence of Dedekind-infinite sets is basically a direct consequence of any set theory which can (i) perform basic set operations and (ii) develop any sort of "unbounded" objects; it doesn't have to be built in from the get-go. This is why I see this whole approach as throwing the baby out with the bathwater: why not simply be more careful about the category you work in than drastically (and counterintuitively in its own right) reshape the entire set-theoretic framework? – Noah Schweber Feb 22 '21 at 17:02
  • (Unrelatedly, I've fixed the MO link in the question.) – Noah Schweber Feb 22 '21 at 17:07
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    "If there is a problem with inconsistency of the ZF system we are all wasting our time writing down complicated proofs" First of all that's not really true - $\mathsf{ZF}$ is massive overkill for basically everything and has many well-understood drastic weakenings, so this actually wouldn't necessarily be that big a deal - but even ignoring that what does that have to do with your question? At no point have you indicated anything suggesting a contradiction in ZF. Counterintuitive phenomena ≠ inconsistencies. – Noah Schweber Feb 22 '21 at 17:21
  • @NoahSchweber - "How about simply the category of finite sets?" In the post I'm asking for references - could you provide one? – hm2020 Feb 22 '21 at 18:24
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    The category of finite sets is treated in a number of places - Goldblatt's book Topoi goes into a fair amount of detail. But it's a very limited object, and it's hard to see how to even talk about (say) topological manifolds within it in any meaningful way. – Noah Schweber Feb 22 '21 at 20:25
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    The question about dimension and surjective maps is an interesting one, and Moishe Kohan gave a great answer. All the stuff about changing the category of sets and injective but not surjective maps is completely unrelated and distracting. If you wanted to ask for references about alternative set theories or the category of finite sets, you should have done it in a separate question, and without suggesting that mathematical phenomena that you happen to find counterintuitive should also be a problem for everyone else. – Alex Kruckman Feb 24 '21 at 14:17
  • My intuition was - based on the Peano map - that "there are too many maps of infinite sets" and that an infinite set should have fewer "endomorphisms". I agree there are complications since you cannot compose maps when you make this restriction. – hm2020 Feb 24 '21 at 14:35

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Here is an answer, of sorts, to your questions:

  1. (i) Do you agree this is a "strange" phenomenon? (ii) What "happens" to the dimension? (iii) Intuitively it should drop.

Answer: (i) No, I do not find it strange at all. I also find it completely normal that there are continuous nowhere differentiable functions and that there is a continuous function mapping the Cantor set onto the unit interval (Cantor function). Such functions even appear "naturally" in real analysis and if one teaches RA one usually gives these as examples of functions which are continuous but not absolutely continuous (a monotonic nonconstant continuous function whose derivative vanishes almost everywhere).

(ii) Dimension increased, that's all.

(iii) The intuition that dimension should drop is based on a "naive" understanding of continuous functions which suffices for undergraduate calculus but which one typically looses after taking a real analysis class. Incidentally, there are some interesting inequalities regarding the behavior of dimension under continuous maps; the earliest ones are called Hurewicz formulae. Here is one of these inequalities.

Definition. Let $f: X\to Y$ be a map of topological spaces. Then $\dim(f)$ is defined as the supremum of dimensions of the subsets $f^{-1}(y), y\in Y$. In particular, if $f$ is a finite-to-one map, $\dim(f)=0$.

Theorem. Suppose that $X, Y$ are separable metrizable spaces (say, $R^n$ and $R^m$ respectively) and $f: X\to Y$ is a closed map (i.e. a map sending closed subsets to closed subsets). Then $$ \dim(X)\le \dim(Y) + \dim(f). $$ For instance, if $f$ is finite-to-one (as in the case of the classical Peano curve or the Cantor function mapping the Cantor set onto $[0,1]$), then $\dim(X)\le \dim(Y)$, hence, one gets the opposite of the inequality that you expected.

One can find a proof of this result for instance in

Hurewicz, Witold; Wallman, Henry, Dimension theory, Princeton: Princeton University Press. 165 p. (1948). ZBL0036.12501.

  1. Not a question but a sentiment expressed in the Note: "For the dimension (),() to be "well defined", there should be no continuous surjection from to ."

No, (as it was pointed out in comments) for a topological dimension to be well-defined it should be invariant under homeomorphisms and it is. There are other notions of dimension in topology, ones which make sense for spaces more general than manifolds. The earlier one is the Lebesgue covering dimension. It is defined in such a way that it is manifestly invariant under homeomorphisms; then one proves that it equals $n$ for (nonempty) $n$-dimensional manifolds. Ditto for the cohomological dimension, etc.

  1. Regarding the Question after the Note: As it was pointed out in comments, you should not mess with the category of "sets" but rather modify the category of "spaces" to exclude examples of morphisms which increase the dimension (whatever the definition of the dimension is). Such a request came earlier from Alexander Grothendieck who asked for something similar in his Esquisse d'un Programme; he was also troubled by Peano curves. See for instance, the discussion here.

(a) One solution comes from logic, the o-minimal theory. Somebody more knowledgeable in logic should provide the details. Otherwise, just read one of the references given in the Wikipedia article.

(b) As a geometer, I would argue for the "Lipschitz category." You can find many more details in my answer here, especially on how to define this category without metrics and how to relate it to Banach star-algebras. The thing is that Lipschitz maps of (compact, or, more generally, complete 2nd countable, locally compact) metric spaces cannot increase the Hausdorff dimension because if the $\alpha$-Hausdorff measure of a metric space $X$ is zero then for every Lipachitz map $f: X\to Y$, the $\alpha$-Hausdorff measure of $f(X)$ is also zero.

(c) As a topologist, I also like the categories of PL and DIFF (piecewise-linear and smooth manifolds respectively), where morphisms are, respectively, PL and smooth. Again, morphisms cannot increase the dimension. PL category naturally extends to the one of simplicial complexes which still have the same property (plus fibers of simplicial maps are simplicial subcomplexes, which is something else that Grothendieck did not like about the category of manifolds). Still, if one is interested in topological manifolds, the Lipschitz viewpoint is the best because of the following theorem due to Dennis Sullivan:

Theorem. Every topological manifold of dimension $n\ne 4$ admits a Lipchitz structure, i.e. an atlas with Lipschitz transition maps. Moreover, this structure is unique, i.e. every homeomorphism is isotopic to a bilipschitz homeomorphism.

See:

Sullivan, Dennis, Hyperbolic geometry and homeomorphisms, Geometric topology, Proc. Conf., Athens/Ga. 1977, 543-555 (1979). ZBL0478.57007.

In other words, in order to study topological manifolds (of dimension $\ne 4$) one looses nothing by working in the Lipschitz category. It was later proven by Donaldson and Sullivan that dimension 4 is a genuine exception, for instance, the E8-manifold does not admit a Lipschitz structure:

Donaldson, S. K.; Sullivan, D. P., Quasiconformal 4-manifolds, Acta Math. 163, No. 3-4, 181-252 (1989). ZBL0704.57008.

Which (if any) of these categories you should use, is up to you, and depends on what your further objective is.

Moishe Kohan
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