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Axiom of Choice:

Given a collection $A$ of nonempty sets, there exists a function

$$c: A \to \bigcup_{A_{i} \in A}A_{i}$$

such that $c(A_{i})\in A_{i}$ for all $A_{i} \in A$.

Axiom of Dependent Choice:

Given a nonempty set $A$ and a binary relation $\mathcal{R}$ on $A$ such that for all $a\in A$, there exists $b\in A$ such that $a\mathcal{R}b$. There exists a sequence

$$(a_{n})_{n\in \mathbb{N}}$$

such that $a_{n}\mathcal{R}a_{n+1}$ for all $n \in \mathbb{N}$.

Here is my incomplete proof that Axiom of Choice implies Axiom of Dependent Choice:

For $a\in A$, let $R(a)=\{b\in A\mid a\mathcal{R}b\}\implies R(a)\neq\varnothing$ for all $a\in A$.

Using Axiom of Choice for the indexed family of sets $(R(a))_{a\in A}$, there exists a mapping

$$f:A\to A$$

such that $$\forall a\in A:f(a)\in R(a)$$

$\text{That is }\forall a\in A:a\mathcal{R}f(a)$. Let $B=\{(a,f(a))\mid a\in A\}$

I don't know how to proceed to prove the existence of the required sequence $(a_{n})_{n\in \mathbb{N}}$ from set $B$.

Please help me complete my proof! Many thanks for you!

Akira
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3 Answers3

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You're basically done already: pick some $a\in A$ and consider the sequence $$a, f(a),f(f(a)), f(f(f(a))), ...$$ The $f$ that choice provides us is exactly the "strategy" we use to build the desired sequence.

Noah Schweber
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  • I think your "strategy" looks like defining a sequence recursively. Is it true that your "strategy" implicitly appeals to Theorem of Recursion (https://www.wikiwand.com/en/Recursion#/The_recursion_theorem)? – Akira Feb 23 '18 at 01:27
  • @leanhdung No, actually! It follows just from the Separation axiom: we pick $a\in A$ and let $B$ be the set of $b\in A$ such that there is a finite chain of $f$-applications linking $a$ to $b$. This doesn't need the recursion theorem at all. (And incidentally, the recursion theorem doesn't need choice, so even if it was necessary that would be fine.) – Noah Schweber Feb 23 '18 at 01:34
  • The key point that lets us avoid the recursion theorem (that is, that makes this argument work in the axiom system Z, which is ZF minus Replacement and within which the recursion theorem is not provable) is that everything is taking place inside a fixed set $A$. – Noah Schweber Feb 23 '18 at 01:35
  • Yes, we only need Axiom of Separation to form such sets $R(a)={b\in A\mid a\mathcal{R}b}$ for all $a\in A$, but what I meant is that you must use Recursion Theorem at the end to form $(a_{n})_{n\in \mathbb{N}}$ from mapping $f:A\to A$. – Akira Feb 23 '18 at 07:58
  • @leanhdung No, you don't - you can just use Separation. The application of Separation is a bit more complicated, as my comment says above. The recursion theorem is not needed here. – Noah Schweber Feb 23 '18 at 11:39
  • The version of the recursion theorem at the link the OP gave also takes place within a predetermined set $X$ - so it seems like that version also should be provable without replacement. Namely, the function is defined to be the union of all subsets of $\mathbb{N} \times X$ such that it forms a function $n \to X$ satisfying the recursive formula for some natural number $n$. (i.e. the version there is not the metatheorem which uses a formula defining a relation on $V$ to represent $f$.) – Daniel Schepler Feb 23 '18 at 17:42
  • @DanielSchepler, Yes, the version, which I gave here, is proved without Axiom of Replacement and is defined to be the union of all subsets of $\mathbb{N} \times X$ such that it forms a function $n \to X$ satisfying the recursive formula. I think you get my confusion. I have two questions: 1. Can you point out the differences between my version and "the version there"? 2. Is it true that the answer above must appeal to Recursion Theorem (the version in the link I gave) to prove the existence of the required sequence $(a_{n})_{n\in \mathbb{N}}$? – Akira Feb 24 '18 at 04:20
  • @NoahSchweber, I meant your answer must appeal to Recursion Theorem (the version in the link I gave) to prove the EXISTENCE of the required sequence $(a_{n})_{n\in \mathbb{N}}$. If you think i have not got what you meant, please explain more slowly and in more detail the part "The application of Separation" and "a finite chain of $f$-applications linking $a$ to $b$". I'm just exposed to set theory ^^ – Akira Feb 24 '18 at 04:25
  • @leanhdung Yes, I'm aware - and I'm saying it's not needed there. I understand what you're claiming, and I'm saying it's wrong. We can build that sequence all at once, as follows. (0) Fix $a\in A$. (1) Let $X=\mathbb{N}\times A$. (2) By separation, let $Y$ be the subset of $A$ consisting of all pairs $(n, x)$ such that there is a sequence of length $n$ whose first term is $a$, whose last term is $x$, and such that each term after the first is $f$ of the previous term. (cont'd) – Noah Schweber Feb 24 '18 at 04:30
  • It is now easy by straightforward induction to show that for each $n\in\mathbb{N}$, there is exactly one $y\in A$ with $(n, y)\in Y$. This $Y$ literally is the sequence $\langle a, f(a), f(f(a)), ...\rangle$, and we never needed to use the recursion theorem at all. The only "recursion-like" step was the argument saying that for each $n$ there is a unique $y$ with $(n, y)\in Y$; but this just uses standard induction, with no need to appeal to the recursion theorem. The recursion theorem is a useful hammer, but not everything is a nail. – Noah Schweber Feb 24 '18 at 04:33
  • (That said, you might recognize this style of argument from the proof of the recursion theorem itself! The point is that in this case we only need the "germ" of that idea. In particular, the real meat of the recursion theorem happens at the "limit stages" of the construction, and there aren't any of those in this argument. So the recursion theorem isn't irrelevant here, but saying that it's needed seems ... off.) – Noah Schweber Feb 24 '18 at 04:41
  • Well, the version of the theorem given in the link also isn't the full generality version which allows you to recursively define functions on any well-founded domain. It's just the special case with domain $\mathbb{N}$ there. (And even then, if you're given a bounded range as a set, I think you don't need replacement there. The full power metatheorem requiring replacement would only be needed for cases like defining the rank of a set.) – Daniel Schepler Feb 24 '18 at 04:56
  • And yes, technically you never have to use a theorem, if you just repeat its proof... – Daniel Schepler Feb 24 '18 at 04:57
  • @DanielSchepler But I didn't just repeat its proof: the argument I used is simpler than the recursion theorem in a very important way (no limit steps). – Noah Schweber Feb 24 '18 at 04:58
  • @NoahSchweber, I think you meant "let $Y$ be the subset of $X$" rather than "let $Y$ be the subset of $A$". – Akira Feb 24 '18 at 05:05
  • @leanhdung Whoops, yes, good catch. – Noah Schweber Feb 24 '18 at 05:24
  • @NoahSchweber. Although i have a proof of Recusion Theorem at https://math.stackexchange.com/questions/2613220/possible-typos-in-the-proof-of-recursion-theorem, i don't understand what "limit stages" is in the proof I gave in the link. I think i can have a simpler and shorter proof of Recusion Theorem with your approach. In the above proof, proof Garling did not use Axiom of Separation. Instead, he defined the required function to be the intersection of all subsets of $N×X$ such that it forms a function $n→X$ satisfying the recursive formula. – Akira Feb 24 '18 at 06:32
  • @DanielSchepler, i think you meant "Namely, the function is defined to be the INTERSECTION of all subset..." rather than "Namely, the function is defined to be the UNION of all subset...". – Akira Feb 24 '18 at 06:40
  • @leanhdung Well, that would be another way to construct it, as the intersection of all subsets such that $(n, x) \in S \Rightarrow (n + 1, f(x)) \in S$. I was giving a different construction, as the union from below of partial functions ${ m \mid m < n } \to X$ satisfying the recurrence relation for $m$ where $m + 1 < n$. – Daniel Schepler Feb 24 '18 at 17:38
  • @DanielSchepler, It seems that my approach is top-down, while yours is bottom-up! – Akira Feb 26 '18 at 03:53
  • @DanielSchepler, I has a feeling that your bottom-up approach is like k-approximation by Brian M. Scott at https://math.stackexchange.com/questions/402266/prove-the-recursion-theorem and like one by spaceisdarkgreen at https://math.stackexchange.com/questions/2511846/confusion-with-recursion-theorem?rq=1. – Akira Feb 26 '18 at 08:28
  • @leanhdung One technical advantage of the bottom-up approach is that it still works in ZFC (with some modifications) for the cases where the domain or possible range are proper classes too large to be a set, whereas the top-down approach would require something along the lines of Morse-Kelley set theory. – Daniel Schepler Feb 26 '18 at 17:59
  • Oh, and looking back at my previous comment, I left something out: it would need to be the intersection of sets $S$ such that $(0, x_0) \in S$ and $(n, x) \in S \Rightarrow (n+1, f(x))$. (Without the requirement $(0, x_0) \in S$, the intersection would of course be $\emptyset$.) – Daniel Schepler Feb 26 '18 at 18:02
  • @DanielSchepler, I have just presented the detail proof using the bottom-up approach at https://math.stackexchange.com/questions/2667272/bottom-up-approach-for-proving-recursion-theorem-proof-verification, please have a look and check my proof! – Akira Feb 27 '18 at 00:50
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Once you have $f$, use the assumption that $A$ is nonempty to choose $x \in A$. Now, you can define a sequence recursively by $a_0 = x$, $a_{n+1} = f(a_n)$. (So for example, $a_1 = f(x)$, $a_2 = f(f(x))$, $a_3 = f(f(f(x)))$, etc.)

Daniel Schepler
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  • Since your answer mentions "define a sequence recursively", is it true that you implicitly appeal to Recursion Theorem (https://www.wikiwand.com/en/Recursion#/The_recursion_theorem)? at the end to form $(a_{n})_{n\in \mathbb{N}}$? – Akira Feb 23 '18 at 08:02
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    Yes, it uses the existence part of that theorem. – Daniel Schepler Feb 23 '18 at 17:32
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I think you're almost there: now define $$ a_0 := a, a_n := f^n(a)$$ for some $a \in A$.

57Jimmy
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  • The way you acquire $(a_{n})_{n\in \mathbb{N}}$ makes me feel like you implicitly appeal to Recursion Theorem (https://www.wikiwand.com/en/Recursion#/The_recursion_theorem). Is my understanding true? – Akira Feb 23 '18 at 08:13
  • I don't think so. $f^n$ is a well-defined function for all $n \in \mathbb{N}$, so I am defining $a_n$ "explicitly". But I might be wrong, I am not an expert in set theory. – 57Jimmy Feb 23 '18 at 08:42
  • I am not sure, but isn't the theorem more focused on uniqueness than on existence? Here we don't need uniqueness, and existence is clear from the definition of $f^n$ (I think) – 57Jimmy Feb 23 '18 at 08:48