1
$\begingroup$

Under $\mathsf{ZF+AC_\omega}$, the space $\Bbb{R}$ with finite complement topology is not second-countable, since a countable union of countable subsets of $\Bbb{R}$. However, such argument doesn't work if we does not assume any form of choice.

My question is, $\Bbb{R}$ with finite complement topology is not second-countable under ZF (Zermelo-Fraenkel set theory without choice), or it is consistently second-countable? I would appreciate your answer.


Proof of "$\Bbb{R}$ with finite complement topology is not second-countable" under choice: Let assume $\{V_n\}_{n<\omega}$ be a countable basis of the space. We know that $\Bbb{R}-\{x\}$ is open for each $x\in \Bbb{R}$ so for each $x$ there is $n<\omega$ s.t. $V_n\subset \Bbb{R}-\{x\}$ so $$\bigcap_{n<\omega} V_n\subseteq \bigcap_{x\in\Bbb{R}} \Bbb{R}-\{x\} = \varnothing.$$ However, $\bigcap_{n<\omega} V_n = \Bbb{R} - \bigcup_{n<\omega} (\Bbb{R}-V_n)$ and $(\Bbb{R}-V_n)$ is finite set. Since countable union of at most countable sets is at most countable, $\bigcup_{n<\omega} (\Bbb{R}-V_n)$ is countable so $\bigcap_{n<\omega} V_n$ is not empty, a contradiction.

$\endgroup$
1
$\begingroup$

Here's the thing about $\Bbb R$. It's linearly ordered. And this means that finite subsets are well-ordered canonically by that linear ordering.

In particular, this means that the countable union of finite sets of real numbers is always countable. So the usual proof works, with the minor modification that now we need to appeal to the properties of $\Bbb R$ to claim that $\bigcup(\Bbb R-V_n)$ is countable.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.