I was hoping someone could review my proofs below. I'm not totally sure both statements are true (or even if one of them is true). Thanks!

Statement 1:

$\mathbb{R}$ with the discrete topology is a $T_1$ space

Statement 2:

The discrete topology on $\mathbb{R}$ is not limit point compact.

Proof of statement 1:

We wish to show that finite sets are closed. It suffices to show that single point sets are closed, since then the finite union of closed singletons will produce a closed finite set.

Take any set $\{a\}$. Then $X - \{a\} = (-\infty, a) \cup (a,\infty)$. Since each set in the above union is open as it can be written as the infinite union of the open singleton points inside each set, we have that $X - \{a\}$ is open. Hence $\{a\}$ is closed. Hence any finite set is closed.

Proof of statement 2:

Let $X$ be the discrete topology on $\mathbb{R}$. Take the interval $[0,1]$. This interval is infinite. The interval also contains no limit points since for any proposed limit point $x$, we can take the open set $\{x\}$ as an open set that does not intersect $[0,1]$ in any place other than itself. Hence $X$ is not limit point compact.

  • $\begingroup$ just a note, "the discrete topology on R" itself is not a space. One should rather write "$\mathbb R$ with the discrete topology is a T1 space". $\endgroup$ – Pink Panther Mar 10 at 21:42
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    $\begingroup$ Both statements are true and your proofs are correct! You can simplify them by generalising this to any space with the discrete topology (hint: every subset of a space with the discrete topology is both closed and open) $\endgroup$ – vxnture Mar 10 at 21:43
  • $\begingroup$ we could also claim that R with the discrete topology is countably compact right? @nammie $\endgroup$ – H_1317 Mar 10 at 21:45
  • $\begingroup$ @H_1317 any (infinite) space with the discrete topology is not countably compact. You can prove this directly (as every set is open), or use that in $T1$ spaces, countably compact and limit point compact are equivalent. $\endgroup$ – vxnture Mar 10 at 21:50
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    $\begingroup$ Yes! The intervals $(n, n+2)$ form an open cover of $\mathbb{R}$ with no finite subcover in both the discrete topology and the usual Euclidean one. $\endgroup$ – vxnture Mar 10 at 22:01

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