Finite T1 Space is Discrete

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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 $T = \left({S, \tau}\right)$ be a $T_1$ space on a finite set $S$.


Let $U \subseteq S$ be any subset of $S$.

Let $H = \complement_S \left({U}\right)$ be the complement of $U$ relative to $S$.

Then by Relative Complement of Relative Complement we have that $U = \complement_S \left({H}\right)$.


We can write $H$ as:

$\displaystyle H = \bigcup_{x \mathop \in H} \left\{{x}\right\}$


From Equivalence of Definitions of $T_1$ Space, $\forall x \in H: \left\{{x}\right\}$ 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, $U = \complement_S \left({H}\right)$ 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 $S$.

$\blacksquare$