Finite Complement Space is T1

Theorem

Let $T = \struct {S, \tau}$ be a finite complement topology on any set $S$ which contains at least two points.

Then $T$ is a $T_1$ (Fréchet) space.

Proof

Let $x, y \in S$.

Then:

$S \setminus \set x$ is an open set of $T$ containing $y$ but not $x$.

$S \setminus \set y$ is an open set of $T$ containing $x$ but not $y$.

Hence the result from definition of $T_1$ space.

$\blacksquare$

Sources

• 1978: Lynn Arthur Steen and J. Arthur Seebach, Jr.: Counterexamples in Topology (2nd ed.) ... (previous) ... (next): Part $\text {II}$: Counterexamples: $18 \text { - } 19$. Finite Complement Topology: $5$