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To provide some context, I was trying to prove $(X, d)$ separable $\implies$ X second countable, specifically without using any form of choice.

Let $U \subseteq X$ open and $D$ our countable dense subset, and consider $\mathscr B := \{B_q(d) \mid q\in \mathbb Q_{\geq 0}, d\in D\}$, which should be our basis.

In particular, I will verify that $$ U = \bigcup \underbrace{\{B_q(d) \mid q\in \mathbb Q_{\geq 0}, d\in D, B_q(d)\subseteq U\}}_{\mathscr S}. $$

Proving $\supseteq$ is trivial. I now proceeded to prove that for every $u\in U$, I can find some ball $B_\delta(u)\subseteq U$. To avoid any choice, set $\delta := \sup\{\delta^\prime\mid B_{\delta^\prime}(u)\subseteq U\}$. However, now we have to choose a point $x\in D\cup B_\delta(u)$ close enough to $u$ – here, the restriction $d(x,u)<\delta/2$ suffices – and for such an $x$, a rational number $q>0$ such that $d(x,u)<q<\delta/2$, which works because the rationals are dense in the reals. It is not too hard to verify that this is enough for $u\in B_q(x)\subseteq B_{\delta}(u)\subseteq U$.

I concluded that $U\subseteq \bigcup \mathscr \{\ldots\}$ as above. Did I accidentally use the axiom of choice here?

All I found was Remark 1.3 in this paper, which states that the theorem holds in ZF. However it gives neither proof nor reference for that.

Lukas Juhrich
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  • You should give more details about $U$, $\mathscr S$ and how you find $q(u) \in \mathbb Q$. – Paul Frost Oct 20 '19 at 23:18
  • You might not need to use choice, because you can endow $\Bbb Q$ with an explicit well-order $\preceq$ induced by a bijection with $\Bbb N$. Then, for any property $P$ you can pick the $\preceq$-least rational number which satisfies it. –  Oct 20 '19 at 23:31
  • I added the specific scenario. I noticed that I effectively reproduced [https://math.stackexchange.com/a/926236/91103](this proof on MSE), but it makes no mention about whether it relies on choice. – Lukas Juhrich Oct 20 '19 at 23:33
  • @Gae.S. indeed, I assume an enumeration of $\mathbb Q$ and $D$ is known. – Lukas Juhrich Oct 20 '19 at 23:33
  • Note you don't have to worry about choice in choosing $\delta$ since the topology induced by a metric is defined as the topology generated by open sets of the form $B_\delta(x)$ for $x\in X, \delta > 0$. Thus for $U$ open, $U = \bigcup_{i \in I} B_{\delta_i}(x_i)$, so $x \in U \implies \exists i \in I: x \in B_{\delta_i}(x_i)$. – Physical Mathematics Oct 20 '19 at 23:39
  • Where do you think AC is being used? – Physical Mathematics Oct 20 '19 at 23:48

1 Answers1

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Let $x \in U$. Then for some $x \in X, \delta > 0$, $x \in B_\delta(x)$ (see comment above). Then consider $B_{\delta/2}(x) \neq \emptyset$. Taking our definition of dense to be "A set $D$ is dense if every nonempty open set contains a point of $D$", then $\exists p \in D$ s.t. $p \in B_{\delta/2}(x)$. Then there exists $q \in \mathbb{Q}$ s.t. $d(x,p) < q < \delta/2$ since $(d(x,p),\delta/2)$ is a nonempty open set in $\mathbb{R}$ and $\mathbb{Q}$ is dense in $\mathbb{R}$. Then you can easily verify $B_q(p) \subseteq B_\delta(x)$ and $x \in B_q(p)$. Then $B_q(p) \in \mathcal{S}$. Thus $\forall x \in U \exists V \in \mathcal{S} : s \in V \in \mathcal{S}$. Then $\cup \mathcal{S}$ is defined as $\{x \mid \exists V \in \mathcal{S}: x \in V\}$. From this immediately follows $\forall x \in U: x \in \cup \mathcal{S}$, or equivalently $U \subseteq \cup \mathcal{S}$.

I hope this more explicit proof shows AC is nowhere needed.

Physical Mathematics
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