# Second-countable implies separable/Axiom countable choice

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?

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.

There is an immediate reversal. Let $(A_n)$ be any countable sequence of nonempty sets. For the purposes of countable choice we may assume the sets are pairwise disjoint. Let $T$ be a space whose points are $\bigcup_n A_n$ and whose topology is generated by the basis $\{A_n : n \in \omega\}$. Let $\{ c_m : m \in \omega\}$ be an enumerated countable dense subset of $T$. For each $n$ let $j(n)$ be minimal such that $c_{j(n)} \in A_n$. Then $\{c_{j(n)} : n \in \omega\}$ is a choice set for the sequence $(A_n)$.

• clever, thank you. It bothers me that so many books avoid mentioning this in the proof. – ditlew Feb 20 '13 at 18:20
• Why would they bother to mention it? The full axiom of choice is a standard axiom in modern mathematics. Should the book also mention which results require the commutativity of natural number addition or the axiom of infinity? – Carl Mummert Feb 20 '13 at 18:21