Finite T1 Space is Discrete
Theorem
Let $S$ be a finite set.
Let $T = \left({S, \tau}\right)$ be a $T_1$ (Fréchet) space.
Then $\tau$ is the discrete topology on $S$.
Proof 1
Let $T = \struct {S, \tau}$ be a $T_1$ space on a finite set $S$.
Let $U \subseteq S$ be any subset of $S$.
Let $H = \relcomp S U$ be the complement of $U$ relative to $S$.
Then by Relative Complement of Relative Complement we have that $U = \relcomp S H$.
We can write $H$ as:
- $\ds H = \bigcup_{x \mathop \in H} \set x$
From Equivalence of Definitions of $T_1$ Space, $\forall x \in H: \set x$ is closed in $T$.
As $S$ is a finite set, it follows that $H$ is a finite union of closed sets of $T$.
By Topology Defined by Closed Sets, $H$ is therefore closed in $T$.
By definition of closed set, $U = \relcomp S H$ is open in $T$.
As $U$ is arbitrary, it follows that:
- $\forall U \subseteq S: S \in \tau$
by definition of a topology.
Hence the result, by definition of the discrete topology.
$\blacksquare$
Proof 2
We have from Finite Complement Topology is Minimal $T_1$ Topology that $\tau$ space is an expansion of a finite complement space.
But from the definition, any such finite complement space on a finite set is discrete.
But as the Discrete Topology is Finest Topology, $\tau$ must itself be the discrete topology on $S$.
$\blacksquare$