Let $(\mathbb{R}, \tau)$ be the countable complement topology on the real numbers. I want to know if: (i) $(\mathbb{R}, \tau)$ is metrizable, and if (ii) $(\mathbb{R}, \tau)$ is compact.

I think I have (ii) completed. Let $K_q=\mathbb{R}\setminus\mathbb{Q}\cup\{q\}$ where $q \in \mathbb{Q}$. Then $\bigcup\limits_{q \in \mathbb{Q}} K_q$ is an open cover of $(\mathbb{R}, \tau)$ that does not have a finite subcover.

Does (i) follow from the fact that $(\mathbb{R}, \tau)$ does not have a countable base?

  • $\begingroup$ is $(\Bbb R, \tau)$ first countable ...? $\endgroup$ – Camilo Arosemena-Serrato May 1 '13 at 1:03
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    $\begingroup$ every two non-empty open sets intersect in $\tau$ - could this happen if $\tau$ were metrizable? $\endgroup$ – yoyo May 1 '13 at 1:08

If $(\mathbb R, \tau)$ is metrizable, then it is must be Hausdorff, since every metrizable space is Hausdorff. However the space is not even Hausdorff.

Proof: for any distinct points $x, y \in \mathbb R$, and any open nbhd $U$ of $x$, and $V$ of $y$. Then $\mathbb R \setminus U$ and $\mathbb R \setminus V$ are both countable. Then $\mathbb R \setminus U \cup \mathbb R \setminus V$ is countable. However $\mathbb R$ is uncountable, so $\mathbb R \setminus (\mathbb R \setminus U \cup \mathbb R \setminus V)\not= \emptyset$, i.e., $U \cap V \not=\emptyset$, which witnesses that $(\mathbb R, \tau)$ is not Hausdorff, hence not metrizable.

It may be helpful for you.

  • $\begingroup$ I'm not sure it follows that $\mathbb{R} \setminus U$ is countable for open $U$. For example, take $U = \mathbb{R} \setminus [0,1]$, which is open since it's complement is the uncountable, compact set $[0,1]$. $\endgroup$ – Austin Mohr May 12 '15 at 17:19
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    $\begingroup$ U is defined to be open if its complement is countable! $\endgroup$ – user202542 May 24 '17 at 18:45

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