Prove that a fnite topological space is T1 if and only the topology is discrete.

In first part od proof I want to use an equivalent definition of T1: a space is T1 if and only if every singleton is closed.

Proof of my exercise:


Suppose $X$ is $T1$. I have to show that every subset of $X$ is open. Let's take $U \subseteq X$. Let's define $H=X \setminus U$. I can write $H=\bigcup_{x \in H}\{x\}$. By defintion of $T1$ $\quad$ $\forall x \in H: \{ x \} $is closed in $X$. $X$ is finite, it follows that $H$ is the union of a finite number of closed sets of $X$. Sp $H$ is closed in $X$. $U$ is complement of $H$, so is open. We can choose arbitrary $U$, so every subset of of $X$ is open.


Suppose $X$ is topological space with discrete topology. I have to show that for every pair of points $x \neq y $ there is an open set containing $x$ but not $y$. For every $x$ we have an open set $U_x$. Complement of $U_x$ is open and doesn't contain $x$. I don't know what do now.


1 Answer 1


If part: For $x\ne y$, since the topology is discrete, then $\{x\}$ and $\{y\}$ are open sets, and these sets are disjoint.


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