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Proof of sequence lemma

Here is a proof of the sequence lemma I lifted from this post. It seems to me naively that the axiom of choice is being used to selected the $x_n$'s. Is the axiom of choice being invoked in this argument? Is it necessary?

The top answer in this post says that we can choose some element from each set in a finite collection of non-empty subsets without choice, though I'm not sure how to do that rigorously. Could some inductive argument be used in the case above?

Physical Mathematics
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    it's being used because $A$ is a generic set. If it is finite or countable you don't need it (because $\mathbb{N}$ is well ordered) – Marcelo Jan 15 '19 at 15:49
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    @J.G. If $A$ is well-ordered and $S\Subset A$ you can "choose" an element of $S$ without AC, by selecting the smallest element. – David C. Ullrich Jan 15 '19 at 15:54
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    like @DavidC.Ullrich said... and if $A$ is a generic finite/countable set with $f:B\subseteq\Bbb{N}\to A$ bijection you can chose $f(s)$ where $s$ is the least element of $B$ – Marcelo Jan 15 '19 at 15:57
  • What everyone has said so far makes sense. Can someone explain why this kind of argument is flawed? For each $n \in \mathbb{N}$ we know $B_d(x,\frac{1}{n}) \cap A$ is non-empty, i.e. $\forall n \in \mathbb{N} \exists c : c \in B_d(x,\frac{1}{n}) \cap A$. So by existential instantiation (https://en.wikipedia.org/wiki/Existential_instantiation), $\forall n \in \mathbb{N} : c_n \in B_d(x,\frac{1}{n}) \cap A$, and then we just consider the sequence ${c_n}$. – Physical Mathematics Jan 15 '19 at 16:06
  • That sort of argument only uses the definition of non-empty and the rules of first-order logic. Though it seems that we could use it to get around choice in all cases. – Physical Mathematics Jan 15 '19 at 16:08
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    @KeeferRowan, there is a subtlety in your notation. You move from considering $c$ to $c_n$ -- once you adopt the latter notation, you are already suggesting that $c_-$ is a function from $\mathbb N$ to $A$. For each $n$ you can use existential quantification, but you cannot "gather these together" into a collection or function. In a sense, the axiom of choice is a trick that lets you do lots of instances of existential instantiation simultaneously. – Mees de Vries Jan 15 '19 at 16:09
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    @KeeferRowan, you might be interested in this question (with answer and comments). – Mees de Vries Jan 15 '19 at 16:11
  • @MeesdeVries Ok so we can use existential instantiation a finite number of times to get a choice function on a finite collection of sets, but there is no reason, without a new axiom to allow it, that we can do it infinitely many times, e.g. for each $n \in \mathbb{N}$? – Physical Mathematics Jan 15 '19 at 16:25
  • Finally, to clarify, the proof above heavily relies on Choice and no one knows of a non-choice version of the proof? Though clearly countable choice would be sufficient. – Physical Mathematics Jan 15 '19 at 16:26
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    The statement is equivalent to countable choice. – Asaf Karagila Jan 15 '19 at 16:28

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