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So I must prove that "Every infinite set has a countably infinite subset." with following definition of the axiom of choice

Let $\{A_\alpha\}_{\alpha \in \lambda}$ be a collection of non-empty sets. Then there is a function $f:\lambda \rightarrow \cup_{\alpha \in \lambda} A_\alpha$ such that for each $\alpha$ in $\lambda$, $f(\alpha)$ is an element of $A_\alpha$.

Can I state "Let $A$ be an infinite set. Consider the collection of subsets of $A$ $\{A_\alpha\}_{\alpha \in \mathbb{N}}$," without violating the axiom of choice?

ervx
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Colebasaur
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  • It depends how the $A_\alpha$ are defined. You need to know they exist in order to assert the collection has a choice function, and you also need to ensure that your choice function will always give you a new element. – Ian Sep 29 '17 at 23:32
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    What collection of subsets? Also, why is the collection of subsets indexed by $\mathbb N$? – spaceisdarkgreen Sep 29 '17 at 23:33
  • I want it to be indexed by $\mathbb{N}$ in order to select elements to make a bijection to some subset of the naturals (so it is countable). I didn't know if for an infinite set you can assume that this holds. I don't think I can though. – Colebasaur Sep 29 '17 at 23:34
  • Okay I guess my second question was obvious, but my first remains relevant. You must define what collection of subsets (if you mean all of the subsets, that exists without AOC but can't be indexed by the naturals). It's unclear what even you're trying to assume holds. – spaceisdarkgreen Sep 29 '17 at 23:36
  • Okay so I need to prove that there is a collection of subsets of A that can be indexed by $\mathbb{N}$, which I don't see as the best way to tackle this proof given it is an arbitrary infinite set. I know there is an induction method, so I'll use that. Thanks! – Colebasaur Sep 29 '17 at 23:39
  • Since the collection of all subsets of $A$ is not countable, you have to say "a collection of ..." rather than "the collection of." The word "the" implies you are picking the collection of all subsets, which you can't if you index with $\mathbb N$. – Thomas Andrews Sep 29 '17 at 23:55
  • Your question is not proof verification if you don't propose a proof to be verified. – Asaf Karagila Sep 30 '17 at 09:05
  • Also related: 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/1618121/the-smallest-infinity-and-the-axiom-of-choice/ – Asaf Karagila Sep 30 '17 at 09:06

3 Answers3

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It's not a matter of violating the axiom of choice but rather of making sense. It doesn't make sense to say "Consider the collection of subsets of $A$ $\{A_\alpha\}_{\alpha \in \mathbb{N}}$" because you haven't specified which collection of subsets you are talking about. The word "the" implies you have some specific collection in mind, so you need to define what your collection is: given a natural number $\alpha\in\mathbb{N}$, what is the definition of $A_\alpha$?

I think you are getting a bit thrown off by the index set $\lambda$ here. Really, you should think of $\{A_\alpha\}_{\alpha\in\lambda}$ as just any set of sets; you don't have to pick some particular index set. If $S$ is any set of nonempty sets, then you can take $\lambda=S$ and $A_\alpha=\alpha$ for each $\alpha\in S$, and the collection $\{A_\alpha\}_{\alpha\in\lambda}$ is just a really fancy way of saying "$S$". So the axiom of choice says that if you have any set $S$ whose elements are nonempty sets, there exists a function on $S$ which takes each $\alpha\in S$ to an element of $\alpha$.

In this problem, in fact, you probably want to let the set $S$ be the collection of all nonempty subsets of $A$. Think about how you might then use a function $f$ provided by the axiom of choice to recursively pick an infinite sequence of distinct elements of $A$.

Eric Wofsey
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Let A be infinite. Use AxC to well order A. The order cannot
stop at any finite ordinal. Select the first omega_0 elements.

A more formal way is by well ordering A, prove dependent choice
and from dependent choice, prove countable choice and from
countable choice an denumberable subset.

William Elliot
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Let $A$ be infinite. For $n\in \Bbb N$ let $[A]^n$ be the set of all $n$-member subsets of $A.$ (This is standard notation.)

Let $E:\Bbb N\to \cup_{n\in \Bbb N}[A]^n$ such that $\forall n\in \Bbb N\;(E(n)\in [A]^n).$

For $n\in \Bbb N$ let $L(n)$ be the set of all linear orderings of $E(n).$ Let $F:\Bbb N\to \cup_{n\in \Bbb N}L(n)$ such that $\forall n\in \Bbb N\;(F(n)\in L(n)).$

For clarity write $<_n$ for $F(n).$ For brevity let $E'=\cup_{n\in \Bbb N}E(n).$

For $x\in E'$ define $r(x)$ as the least $n\in \Bbb N$ such that $x\in E(n).$

Define $<^*$ on $E'$ by

(i) if $r(x)<r(y)$ then $x<^*y$ and $\neg (y<^*x)$

(ii) if $r(x)=r(y)=n$ then $(x<^*y \iff x<_ny).$

For $x\in E'$ and $r(x)=n$ we have $\{y: y<^*x\}= (\cup_{m<n}E(m))\cup \{y\in E(n):y<_n x\}.$

Let $G(x)$ be the number of members of $\{x\}\cup \{y:y<^*x\}.$ Then $G:E'\to \Bbb N$ is a bijection.

DanielWainfleet
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  • We use Countable Choice for the existence of the functions $E, F$. Countable Choice is a corollary (in fact, a special case) of AC (Axiom of Choice) but does not imply AC. – DanielWainfleet Sep 30 '17 at 08:45