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Consider the following definition: We say that $O \subseteq \mathbb{R}$ is an open set if for every $x \in O$ there exists an open interval $(a,b)$ such that $x \in (a,b)$ and $(a,b) \subseteq O$.

Now one can prove that every open set in $\mathbb{R}$ is a union of open intervals. Suppose $O$ is open in $\mathbb{R}$, then for every $x$ in $O$ there is an open interval $(a(x),b(x))$ such that $x \in (a(x),b(x))$ and $(a(x),b(x)) \subseteq O$. But then $O \subseteq \bigcup_{x \in O} (a(x),b(x))$, which gives the equality.

My question is: does one need the axiom of choice for this kind of argument? I think that the notation "$(a(x),b(x))$" implicitly uses choice. This interval clearly doesn't have to be unique, so I don't see how one would formally write down the union of these intervals without using the axiom of choice. With AOC one simply gets a choice function $f$ and could call these intervals $f(x)$ based on $x$, but I don't see why this would be possible without it.

One could also try to get rid of the notation and simply infer that for every $x$ there exists an open interval $(a,b)$ such that the properties hold. But then I don't know how one would write down the union of the sets (again, I think one might need choice here).

user103178981
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    Instead of choosing a single open interval per point, why not choose all intervals at once? – Akiva Weinberger Nov 15 '22 at 15:43
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    As written, you are making infinitely many choices without an indication of how you are doing them, so yes, you are invoking choice. But in this particular setting you don't need to use choice, because you can use intervals with rational endpoints, which can be explicitly well-ordered, so you can just pick "the first interval" containing your point and contained in $O$. – Arturo Magidin Nov 15 '22 at 15:44
  • @FShrike That's essentially "finite choice", which is, as you say, not independent (it doesn't require "actual" choice). But if we want to do this for all infinitely many $x$ at once… – Akiva Weinberger Nov 15 '22 at 15:46
  • @FShrike: No, $(a,b)$ is not given. That paragraph is the definition of "open" being used. $O$ is the given set. – Arturo Magidin Nov 15 '22 at 15:47
  • That said, if we let $U_x$ be the set of all open intervals containing $x$, then it's not hard to show that $O=\bigcup_{x\in O}U_x$. This doesn't require choice. – Akiva Weinberger Nov 15 '22 at 15:47
  • @ArturoMagidin I stand corrected then. I said that because I saw someone trustworthy say exactly that in a similar situation, I expected it applied here. – FShrike Nov 15 '22 at 15:48
  • @AkivaWeinberger: You mean all open intervals containing $x$ contained in $O$. – Arturo Magidin Nov 15 '22 at 15:49
  • @Fshrike: It is true that existential instantiation does not require AC. The thing is that I think you misread the situation here, because $(a,b)$ is not "given". The problem is: given an open set $O$, show that $O$ is a union of open intervals. – Arturo Magidin Nov 15 '22 at 15:50
  • I'm not sure I'm following the argument in the comments that it requires choice. By definition of $O$ being open for every $x\in O$ there is an open interval $I_x$ so that $x\in I_x; I_x \subset X$. We are not choosing the interval nor are we choosing its endpoints. All we are doing is acknowledging that an Interval exists by definition... I think the notation $a(x),b(x)$ is misleading as it implies you have two functions in mind of mapping a specific pair of endpoints for $x$ and that is choice, but that's just notation. As long as we don't claim w know anything about which they are – fleablood Nov 15 '22 at 16:05
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    @Fleablood: The object isn't to show the existence of $I_x$ for each $x$; the objective is to show that $O$ is a union of intervals. In order to claim that $O=\cup I_x$, you need to have an $I_x$ for each $x$. That's where you are using AC, to guarantee that the family of intervals indexed by the $x\in O$ exists. It's not a problem of describing them (quite the opposite: it's the fact that we don't describe them). – Arturo Magidin Nov 15 '22 at 17:05
  • Ah, I see. That is subtle. Kind of strange that Akiva's solution that seems way overkill and intutively a worse abuse is actually a solution. Axiom of choice is sooooooo weird. – fleablood Nov 15 '22 at 18:09

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Your proof requires choice as stated. However, it can be easily repaired, as follows:

If we let $U_x$ be the set of all open intervals containing $x$ contained in $O$, then it's not hard to show that $O$ is the union of all elements of $\bigcup_{x\in O}U_x$. This doesn't require choice.

(Another option is Arturo's idea. In my opinion it's less elegant, but the idea is to use the fact that $\Bbb R$ has a countable basis: every point in an open set $O$ contains an interval with rational endpoints contained in $O$. Thus we may pick the interval with the smallest denominator or some such.)

EDIT: The following may be simpler. Let $U$ equal the set of all open intervals contained in $O$. Clearly $\bigcup U\subseteq O$ (regardless of if $O$ is open or not). Since (because $O$ is open) every element $x\in O$ has at least one element of $U$ containing it, $x\in\bigcup U$. Thus, $O=\bigcup U$.

Akiva Weinberger
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