# The Discrete topology on $\mathbb{R}$ is a $T_1$ space and not limit point Compact

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.

• 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". – Pink Panther Mar 10 at 21:42
• 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) – vxnture Mar 10 at 21:43
• we could also claim that R with the discrete topology is countably compact right? @nammie – H_1317 Mar 10 at 21:45
• @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. – vxnture Mar 10 at 21:50
• 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. – vxnture Mar 10 at 22:01