Finite Complement Topology is Minimal T1 Topology

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Theorem

Let $T = \left({S, \tau}\right)$ be a finite complement space.

Let $\tau'$ be a topology on $S$ such that $T' = \left({S, \tau'}\right)$ is a $T_1$ (Fréchet) space.


Then $\tau$ is comparable with $\tau'$ such that $\tau$ is coarser than $\tau'$.

That is, of all the topologies on $S$ fulfilling the $T_1$ separation axiom, the finite complement space is the smallest.


Thus the finite complement topology is known as the minimal $T_1$ topology on any given set.


Proof

Let $T = \left({S, \tau}\right)$ be the finite complement space on $S$.


Let $U \in \tau$ be any open set of $T$.

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

By definition of finite complement topology, $H \subseteq S$ is a finite subset of $S$.


Let $T' = \left({S, \tau'}\right)$ be any arbitrary $T_1$ (Fréchet) space on $S$.

From Equivalence of Definitions of $T_1$ Space:

$\forall x \in S: \left\{{x}\right\}$ is closed in $T'$.

We can write $H$ as:

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

As $H$ is finite, 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 Relative Complement of Relative Complement we have that $U = \complement_S \left({H}\right)$.

So, by definition of closed set, we have that $U$ is open in $T'$.

So we have shown that for any arbitrary $U \in \tau$ it follows that $U \in \tau'$.

So $\tau \subseteq \tau'$ and so by definition $\tau$ is coarser than $\tau'$.

$\blacksquare$


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