Clopen Points in Modified Fort Space

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Theorem

Let $T = \left({S, \tau_{a, b}}\right)$ be a modified Fort space.


Then all points in $S \setminus \left\{{a, b}\right\}$ are both open and closed in $T$.


$a$ and $b$ themselves are not open in $T$, but they are closed in $T$.


Proof

Let $p \in S: p \notin \left\{{a, b}\right\}$.

From the definition of modified Fort space, any subset of $S \setminus \left\{{a, b}\right\}$ is open in $T$.

It follows directly that as $p \in S \setminus \left\{{a, b}\right\}$ we have that $\left\{{p}\right\} \subseteq S \setminus \left\{{a, b}\right\}$.

Hence $p$ is open in $T$.

As for $a$ and $b$, we have that $S \setminus \left\{{a}\right\}$ and $S \setminus \left\{{b}\right\}$ are not finite and so $a$ and $b$ are not open in $T$.

$\Box$


For all points $p \in S$ (including $a$ and $b$), we have that $\left\{{p}\right\}$ is (trivially) finite.

It follows that while $S \setminus \left\{{p}\right\}$ contains either $a$ or $b$, it is cofinite.

Thus $S \setminus \left\{{p}\right\}$ is open in $T$.

It follows by definition that $p$ is closed in $T$.

$\blacksquare$


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