Finite Complement Space is T1

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Theorem

Let $T = \left({S, \tau}\right)$ 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:

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

Hence the result from definition of $T_1$ space.

$\blacksquare$


Sources