[Math] Second-countable implies separable/Axiom countable choice

axiom-of-choiceelementary-set-theorygeneral-topologysecond-countableseparable-spaces

Let $(X,\mathscr T)$ be a topological space, and $(B_n)_{n\ge1}$ a
countable basis for X. Under this assumptions, X is separable.

The proof of this assertion is as follows:

We can assume without loss of generality that all the $B_n$ are nonempty, because the empty ones can be discarded. Now, for each $B_n$, pick any element $x_n \in B_n$. Let $D$ be the set of these $x_n$. $D$ is clearly countable. We claim that $D$ is dense in $X$.

To see this, let $U$ be any nonempty open subset of $X$. Then, $U$ contains some $B_n$, and hence, $x_n \in U$. But by construction, $x_n \in D$, so $D$ intersects $U$, proving that $D$ is dense. $\blacksquare$

My question is, can this theorem be proven without the axiom of countable choice?

Best Answer

No. It cannot be proved without the axiom of choice that every second countable space is separable. In fact the following are equivalent:

  1. The axiom of countable choice.
  2. Every second countable space is separable.

For a related topic (with references), Does proving (second countable) $\Rightarrow$ (Lindelöf) require the axiom of choice? Or the following paper:

Horst Herrlich, Choice principles in elementary topology and analysis Comment. Math. Univ. Carolin 38,3 (1997) 545-552.

It is consistent (with the failure of choice) that there is a subset of the real numbers which is infinite Dedekind-finite, that is not finite and does not have any countable infinite subset.

Take $D$ be to such subset, then it is easy to show that $D$ in the relative topology is second-countable, but it clearly not separable.

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