Proposition

Show that the only Hausdorff topology on a finite set is the discrete topology.

Solution

Let $X = \{ x_1, \cdots, x_n \}$ be a finite Hausdorff space.

For each $i \in \{ 2, \cdots, n \}$, $x_1, x_i$ have disjoint neighborhoods $U_i, V_i$. Then $U_2 \cap \cdots \cap U_n$ is a neighborhood of $x_1$ that only contains $x_1$. Thus ${ x_1 }$ is open.

This argument can be applied to any point in $X$, so $\{ x_i \}$ is open for each $i$. Therefore, $X$ is discrete.