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Here's an exercise I'm trying to solve as I'm studying topology. Please let me know if my proof convinces you and feel free to otherwise criticise it.

Proposition 1. Every finite $T_1$-space is discrete.

To be fully precise, let me fix the definitions here.

Definition. A topological space $X$ is called $T_1$ iff given any two distinct points $x, y \in X$, both have neighbourhoods that do not contain each other i.e. there exist open sets $U$ and $V$ such that $x \in U$, $y \in V$, $y \notin U$, and $x \notin V$.

Definition. A space $X$ is called discrete iff all of its subsets are open.

I will also assume the following lemmata.

Lemma 1. A space $X$ is $T_1$ iff all of its singleton subsets are closed.

Lemma 2. Finite unions of closed sets are closed.

Proof attempt for Prop. 1. Let $X$ be a finite $T_1$-space and let $U \subseteq X$ be an arbitrary subset. We need to show that $U$ is open i.e. that the complement of $U$ (which we denote $U^c$) is closed. As $X$ is finite, $U^c$ must have finitely many elements and must therefore be the union of finitely many singleton sets. By Lemma 1, we know that each of these singleton sets are closed as $X$ is a $T_1$-space. By the closure of closed sets under finite unions (Lemma 2), $U^c$ must be closed.

affibern
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1 Answers1

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Theorem

A topological space is $T_1$ if and only if any singleton is closed.

Corollary

Any finite set of a $T_1$ space is closed.

Now if $X$ is finite then $X:=\{x_1,...,x_n\}$ for some $n\in\Bbb N$ so $X':=\{x_{i_1},...,x_{i_{n-1}}\}$ is closed for any $i_1,...,i_{n-1}\in\{1,...,n-1\}$ and so $X\setminus X'=\{x_{i_n}\}$ is open.

Antonio Maria Di Mauro
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