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In the book Analysis I by Terence Tao, there is the following exercise: 《Let $X $ be a set, show that $X $ is infinite if and only if there exists a proper subset $Y $ of $X $ which has the same cardinality as $X $. (This exercise requires the Axiom of Choice)》

My attempt:

  • If $X $ is finite, then it is trivial (by induction on the number of elements)

  • If $X $ is infinite, then pick $x_0 $ in $X $, we will show a bijection between $X $ and $Y=X\setminus \{x_0\} $. Take a sequence $(x_n ) $ of distinct elements of $Y $: (the definition of the sequence is by induction) If we have already defined $x_i $ for all $0\leq i \leq n $ for some $n $, then take $x_{n+1} $ to be any element in the non-empty set $Y\setminus \{x_1,\ldots , x_n\} $. Then take the bijection $f:X\to Y $ defined by $f (x)=x $ if $x\neq x_i $ for all $i \geq 0$ and $f (x_i)=x_{i+1} $ otherwise.

I don't see where we used the axiom of choice...

Andrés E. Caicedo
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Friedrich
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    Relevant questions: https://math.stackexchange.com/questions/851745/what-is-the-role-of-induction-in-the-proof-of-königs-lemma/, and https://math.stackexchange.com/questions/539665/inductive-definition-with-choice-for-sequence/, and https://math.stackexchange.com/questions/1618121/the-smallest-infinity-and-the-axiom-of-choice/, and https://math.stackexchange.com/questions/300494/is-axiom-of-choice-required-for-there-to-be-an-infinite-linearly-independent-set/, and https://math.stackexchange.com/questions/41590/where-is-the-axiom-of-choice-used/ – Asaf Karagila Oct 04 '17 at 19:28
  • From your second bullet point "...pick $x_0$...take $x_{n+1}$ to be any element in the non-empty set...", and you cannot see where you have used the axiom of choice. Seriously? – Marc van Leeuwen Oct 05 '17 at 08:21
  • @MarcvanLeeuwen The axiom of choice is not necessary to choose an arbitrary element in a non-empty set. See the answers. – Friedrich Oct 05 '17 at 08:27
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    FWIW, the term you're looking for is "Dedekind-infinite set". – Ilmari Karonen Oct 05 '17 at 09:12
  • @Friedrich: My comment was uniquely to point out that it does not require E.S.P. to spot where the proposed proof invokes some choice. It is not and was not intended as an answer. Yes I know the matter is subtle, and one may not be aware that AC (or some weaker form) is used until one tries to formalise the proof. And as to your comment proper: yes I am aware that it does not require AC to show for an arbitrary nonempty set that it admits an element (choosing is never a mathematical goal in itself, it always serves as a means to establish some existence) – Marc van Leeuwen Oct 05 '17 at 10:08

3 Answers3

20

You use the axiom of choice while constructing the sequence $(x_n \mid n \in \mathbb N)$. To make this more apparent, we need to be a bit more formal in how you obtain such a sequence:

Consider the set $S$ of all finite sequences $\vec{y} = (y_n \mid n \le m)$ such that $y_0 = x_0$ as fixed above and such that for all $n < m$ $y_{n+1} \in X \setminus \{ y_0, \ldots, y_n \}$. Now consider the relation $R$ on $S \times S$ given by $$ (y_0, \ldots, y_m) R (z_0, \ldots, z_l) : \iff l > m \wedge \forall n \le m \colon y_n = z_n, $$

in simpler words: $\vec{y}R\vec{z}$ iff $\vec{z}$ is an end-extension of $\vec{y}$.

Your induction proves that for all $\vec{y} \in S$ there is some $\vec{z} \in S$ such that $\vec{y} R \vec{z}$. The axiom of choice (it suffices to have the axiom of dependent choice) now allows you to pick an infinite sequence $( \vec{y}_n \mid n \in \mathbb N)$ such that for all $n \in \mathbb N$ $$ \vec{y}_n R \vec{y}_{n+1}. $$

If you now let $x_n$ be the $n$-th element of $\vec{y}_n$, the resulting sequence $(x_n \mid n \in \mathbb N)$ is as desired.

Note that this doesn't prove that some form of choice is necessary. It only highlights where you implicitly use the axiom of (dependent) choice. To see that, in general, we need some form of choice here is much harder and (for the moment) you really shouldn't worry about that.

Stefan Mesken
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    You're very welcome! Sometimes the usage of the axiom of choice seems so natural that you end up overlooking how and where you actually relied on it... That certainly has happened to me and to many famous mathematicians ;-) – Stefan Mesken Oct 04 '17 at 19:01
  • Doesn't "pick $x_0$ in $X$" require the Axiom of Choice all by itself? – Spencer Oct 05 '17 at 02:50
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    @Spencer No, it doesn't. $X$ is nonempty, hence there is some $x_0 \in X$. Fix that. (The axiom of choice is needed, roughly speaking, if we want to simultaneously pick infinitely many elements without being able to give an explicit procedure producing such a sequence. This is demonstrated by the fact that the axiom of choice is equivalent to: For any collection ${ X_i \mid i \in I }$ of nonempty sets, the set $\prod_{i \in I} X_i$ is nonempty.) – Stefan Mesken Oct 05 '17 at 06:52
  • @StefanMesken I may have to ask why it doesn't scale down to a single element with no isolating predicate as a followup question, then, because I was pretty sure it did. – Spencer Oct 05 '17 at 10:52
  • @Spencer Not sure that I understand your question. If you ask whether you could apply the axiom of choice to select an element of a nonempty set $X$, the answer is yes. However, the axiom of choice is absolutely not required to do that. (If you break it down to a formal proof, what happens is that we end up with a formula that looks something like this: $\forall x_0 \in X \exists \vec{x} \phi(x_0, \vec{x})$ where $\phi$ says that $\vec{x}$ is a sequence $\vec{x} = (x_n \mid n \in \mathbb N)$ with first element $x_0$ of elements of $X$ such that for all $n \in \mathbb N \colon x_{n+1} \in x_n$) – Stefan Mesken Oct 05 '17 at 10:56
  • @Spencer: Let me recommend an old answer of mine to that question. – hmakholm left over Monica Oct 05 '17 at 12:35
  • Wouldn't this proof only be valid for infinite countable sets? I don's see how we can build a sequence for uncontable sets. – Pablo Oct 06 '17 at 10:23
  • @Pablo If you mean that $X$ must be countable - no. We didn't use that anywhere. If you meant that we only produce a countable sequence $(x_n \mid n < \omega)$ - sure. That's the point of the proof. – Stefan Mesken Oct 06 '17 at 10:25
  • Sorry, I misunderstood the proof, you are right. – Pablo Oct 06 '17 at 10:29
  • @Pablo No worries. I'm glad that I could clear things up. (-; – Stefan Mesken Oct 06 '17 at 10:30
  • I think it's correct to say you need the axiom of choice to prove it and also correct to say you need the axiom of dependent choice to prove it. That's because saying you need the axiom of choice to prove it is another way of saying it can be proven in ZFC but not in ZF and saying you need the axiom of dependent choice to prove it is another way of saying that in ZF, you can't prove it but can deduce it from the axiom of dependent choice. – Timothy Apr 06 '18 at 19:14
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the definition of the sequence is by induction

Well, no, not really - you haven't shown how to define the sequence at all, since the choice of $x_n$ given $x_0, ..., x_{n-1}$ isn't unique. What you can prove by induction is that for each $n$, there is a sequence of $n$-many distinct elements of $X$, but this doesn't give you a way to get an infinite sequence of elements of $X$; this is exactly the sort of thing, in fact, that choice is needed for.

Noah Schweber
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The other answers have correctly identified the issue. Let me highlight the difficulty: it is relatively consistent with the axioms of set theory except for the axiom of choice that there are infinite sets which do not contain a copy of the natural numbers (that is, there are infinite sets $X$ such that there is no injection $f\!:\mathbb N\to X$).

This means that the suggested strategy fails for these sets, but it is even worse, since it is not true that these sets have a proper subset of the same size. Perhaps this feels too abstract, so let's drive the point home:

Suppose $X$ is an infinite closed or open set of reals. We can explicitly find a countably infinite subset of $X$ in this case (without invoking the axiom of choice); this is a decent exercise you may want to attempt. The same is true if $X$ is slightly more complicated, for instance if $X$ is a countable union of closed sets or even a countable union of countable intersections of open sets (this is already challenging).

However, one cannot even prove without using the axiom of choice that if $X$ is a countable intersection of countable unions of closed sets of reals then $X$ has a proper subset of the same size. These are still fairly concrete sets, and already the axiom of choice is needed here.

Andrés E. Caicedo
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