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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?

Martin Sleziak
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ditlew
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2 Answers2

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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.

Community
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Asaf Karagila
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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 $(A_n)$.

Carl Mummert
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  • clever, thank you. It bothers me that so many books avoid mentioning this in the proof. – ditlew Feb 20 '13 at 18:20
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    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