# Clopen Sets in Modified Fort Space

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## Theorem

Let $T = \struct {S, \tau_{a, b} }$ be a modified Fort space.

Let $A \subseteq S$ be both closed and open in $T$.

If $a \in A$, then $b \in A$ as well.

That is, any clopen set of $T$ must contain **both** or **neither** of $a$ and $b$.

## Proof

Let $A$ be both closed and open in $T$.

Let $S = A \mid B$ be a separation of $S$ into the open sets $A$ and $B$.

Then by definition $B$ is also both closed and open in $T$.

Let $a \in A$.

Suppose $b \in B$.

As $A$ is open it must contain all but a finite number of points of $S$.

Thus $A$ must itself be infinite.

But then $B$ can not be open, as $b \in B$ and there is more than a finite number of points in $A$.

So if $a \in A$ then $b \in A$ as well.

$\blacksquare$

## Sources

- 1978: Lynn Arthur Steen and J. Arthur Seebach, Jr.:
*Counterexamples in Topology*(2nd ed.) ... (previous) ... (next): Part $\text {II}$: Counterexamples: $27$. Modified Fort Space: $5$