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Reading the comment to this the user makes the argument that one can pick a dense subset of a separable vector spaces without relying the axiom of countable choices. Since the comment got a lot of up votes, I guess it's correct, but to me this seems very strange. My rule of thumb was/is the following: We can get by without a choice function if

  • There is a canonical choice at each iteration.
  • The proof doesn't actually require the iteration process to terminate.

None of these seem to be satisfied in the present case, unless there is a canonical choice for the dense subset, or the set is dense at some finite iteration, so my question is:

When can I tell if an induction arguments relies on the existence of a countable choice function?

TSU
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2 Answers2

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Your misunderstanding seems to come from the fact that you are not iterating making choices -- you are not picking the elements of the countable dense subset one by one, or anything like that. By definition, "$X$ is separable" means that $$ \{Y \in \mathcal P(X) \mid |Y| \leq \aleph_0 \land \overline{Y} = X\} \neq \emptyset. $$ Since it is non-empty, it contains some element. Pick one element from this set -- we can call it $Y$. There is no further iteration to be done; you now have a separating subset $Y$ of $X$.

Mees de Vries
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You are right that a canonical choice function is needed for recursive constructions.

However, if you have an underlying set which is countable—or generally can be well-ordered—then there is a canonical choice function given by enumerating the set, and at each step choosing the element with the least index in the enumeration which "fits".

In those cases, you only need to prove that the recursive construction can actually proceed from $n$ steps to $n+1$ steps. Just like you would in "the usual case where choice is assumed".

Note that to choose one element from a family which is assumed to be non-empty, no choice is needed. So if you assume that a set is countable, for example, you don't need the axiom of choice to enumerate it. You already assume that an enumeration exists.

Also, let me point out this minor issue that "countable choice" only means that you can make a choice once from a countable family of sets. What you want for recursive definitions and/or inductive proofs, is Dependent Choice, which allows you to make a countable sequence of choices, where each choice might depend on the previous ones you made. Just like in a recursive definition.

Asaf Karagila
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  • I'm still a bit confused by the proof here which related to this post. Via recursive construction, we have one $f_n$ defined each step, and can be defined for every $n$. But in the bullet point 4, to claim a function $g$ is defined for infinite dimensional space, with $g=\cup^\infty_{n=1}f_n$, don't I need all $f_n$ at disposal simultaneously? For that, is ordinary 'induction' suffice or the fact that canonical choice exists for all closed intervals has played a role? – lychtalent Mar 13 '19 at 19:59
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    @lychtalent: Since we can canonically and uniformly define the $f_n$'s, then we actually get the sequence of all the $f_n$'s simultaneously. – Asaf Karagila Mar 13 '19 at 22:32
  • Thanks for your reply! Just to check if I get it, defining $f_n$'s cannonically and uniformly as such can never be done by ordinary induction without additional property of $\mathbb{R}$, and the recursive construction is just appearing to be similar to induction, right? For example, if in the original post, $f_n$ is defined on some abstract set, not on $\mathbb{R}$, a function $g$ defined for infinite dimensional space may not exist without choice, right? – lychtalent Mar 13 '19 at 22:44
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    @lychtalent: No, the construction is recursive. It's just that since we have a canonical way of choosing "the next step", we do not need to appeal to choice to ensure that the sequence of functions is also an object. The uniformity and/or canonicity of these things is indeed a structural consequence of the real numbers, yes. – Asaf Karagila Mar 13 '19 at 22:46
  • Thanks! I just realized the poor way I was questioning may have just confused myself. Your 'No' is to state construction is no ordinary induction, therefore my first part of the statement is correct, and your 'yes' in the end is to agree with my second part of the statement. Am I interpreting it correctly? (I add 'right?' in the end of my question out of habit, which clearly is no good) – lychtalent Mar 14 '19 at 09:16
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    @lychtalent: I said "no", because you asked if the construction "only looks like recursion, but it's not really recursion". That's wrong. The construction is by ordinary recursion. – Asaf Karagila Mar 14 '19 at 09:21
  • Oh now I see. I think I got confused by Induction-recursion and Mathematical induction. By 'induction' I actually meant the latter. Hope this time I am correct. Thanks! – lychtalent Mar 14 '19 at 09:30
  • I notice you treat recursion as synonyms of induction here, but what I had in mind the concept of induction is always $P(n-1)\implies P(n)$, the technique used for proof certain property, not use for construction. – lychtalent Mar 14 '19 at 09:47
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    @lychtalent I thought you are constructing the sequence of functions. Not proving that all of its members have some property. – Asaf Karagila Mar 14 '19 at 09:49
  • Then same term induction is used in both 1) constructing something (recursion), and 2) proving some properties (mathematical induction), where it in fact means different things? – lychtalent Mar 14 '19 at 09:53
  • @lychtalent: This will be my last comment on the topic (and on the Hahn–Banach topic in general). The construction is by recursion, not induction. I've been a logician for my entire mathematical career and I know the difference between the two. – Asaf Karagila Mar 14 '19 at 09:56