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Let $\Omega$ be a region (nonempty connected open subset of the complex plane). Let $K_n$ be a sequence of compact sets whose union is $\Omega$, such that $K_n \subset \mathring{K_{n+1}}$ (the interior of $K_{n+1}$)

Now the book says: "Choose a countable set $E \subset \Omega$ such that $E \cap K_n$ is dense in $K_n$ for all $n$"

It's not clear to me why such $E$ exists. I know that $K_n$ are compact hence separable, so it exists $E_n$ countable and dense in each $K_n$. So of course $E = \bigcup E_n$ is such that $E \cap K_n$ is dense in every $K_n$, but we need the axiom of choice to show that $E$ is countable, don't we?

Is there a more immediate way to show that such $E$ exists? Maybe I am overlooking something obvious

Thanks!

Camilo Arosemena-Serrato
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Ant
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  • @AlexR why? $E_1$ is dense in $K_1$, which is the smallest of them. I don't think this implies it's dense in $K_n$ for $n > 1$.. But I may be wrong, this is a new topic for me – Ant Apr 27 '15 at 13:53
  • Ooops, Sorry I somehow thought $K_1$ was the biggest. Still I think we only need countable choice. But Asaf may be able to provide full insight. – AlexR Apr 27 '15 at 13:55
  • I think if you would try to keep track of every instance of a possible hidden use of countable choice in your analysis classes, you would lose sight of the main goal (learning analysis). I would have, in addition, also gone insane :O. – hot_queen Apr 27 '15 at 21:45

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"Choosing" dense subsets of compact sets in plane (or $\mathbb{R}^n$) can be done uniformly as follows: Given $K$ compact, let $\{r_n : n < \omega \}$ list all rational points in plane. For each $n$, let $d_n$ be the distance of $r_n$ from $K$. If $d_n = 0$, let $x_n = r_n$, otherwise choose $x_n \in K$ to be the unique point in $K$ whose distance from $r_n$ is $x_n$ and which makes the smallest angle (in $[0, 2\pi)$) with the horizontal axis. Now check that $\{x_n : n < \omega\}$ is dense in $K$.

hot_queen
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You don't need the full axiom of choice to show that a countable union of countable sets is countable. Let $\varphi_n : E_n \to \mathbb N$ be a bijection (you can pick such bijections using countable choice, that is, a product of countably many non-empty sets is non-empty, which allows you to pick a bijection for each $n \in \mathbb N$), and let $$ \coprod_{n \ge 1} E_n \to \mathbb N^2, \qquad (n, x) \mapsto (n, \varphi_n(x)). $$ (the symbol $\coprod$ denotes disjoint union). This is a bijection, and you can easily compute a bijection of $\mathbb N^2$ to $\mathbb N$. The inclusions $E_n \to \bigcup_{n \ge 1} E$ can be put together to give a map $\coprod_{n \ge 1} E_n \to E$ which is surjective, showing that $|E| \le \left| \coprod_{n \ge 1} E_n \right| = |\mathbb N|$. (If you want to define $\le$ using only injections, you can use the Cantor-Bernstein-Schröder Theorem, which does not rely on the axiom of choice.)

Hope that helps,

Patrick Da Silva
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  • to be fair, I am having trouble understanding what you've written (I am a little confused by the notation). While I try to figure it out, I read just earlier today this post (http://math.stackexchange.com/questions/603456/prove-that-the-union-of-countably-many-countable-sets-is-countable/604420#604420) which says the exact opposite! :) – Ant Apr 27 '15 at 14:01
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    Hrm, how do you choose $\varphi_n$? – Asaf Karagila Apr 27 '15 at 15:45
  • @AsafKaragila : Right. I guess I need countable choice there? since I want a bijection $\varphi_n$ for each $n \in \mathbb N$. I never did set theory formally, I should get down to it someday.. – Patrick Da Silva Apr 27 '15 at 21:55
  • Yes, but there are infinitely many injections. And you choose one for infinitely many sets. If that is not using the axiom of choice, what is? – Asaf Karagila Apr 28 '15 at 03:25
  • I mean choice is not about picking an element in an infinite set, but rather about picking infinitely many elements simultaneously (because that means picking an element in a product of sets, which without choice cannot be assumed non-empty). If a set is non-empty I don't need choice to pick an element inside of it – Patrick Da Silva Apr 28 '15 at 20:15