Fort Space is Scattered

Theorem
Let $T = \left({S, \tau_p}\right)$ be a Fort space on an infinite set $S$.

Then $T$ is a scattered space.

Proof 1
Let $H \subseteq T$ such that $H \ne \left\{{p}\right\}$.

Then $\exists x \in H: x \ne p$.

From Clopen Points in Fort Space, every point of $T$ apart from $p$ is open in $T$.

So $\left\{{x}\right\}$ is an open set of $T$.

So $H \cap \left\{{x}\right\} = \left\{{x}\right\}$ and so $x$ is isolated in $H$.

Thus $H$ contains at least one point which is isolated in $H$.

On the other hand, suppose $H = \left\{{p}\right\}$.

From Singleton Point is Isolated, $p$ is an isolated point in $\left\{{p}\right\} = H$.

So again $H$ contains at least one point which is isolated in $H$.

So for all $H \subseteq S$ we have that $H$ contains at least one point which is isolated in $H$.

Hence the result, by definition of scattered space.

Proof 2
Suppose that $H \subseteq T$ has no isolated points of $H$.

So, by definition, $H$ is dense in itself.

We have that:
 * a Fort Space is $T_1$
 * a Dense-in-itself Subset of T1 Space is Infinite.

So $H$ is infinite, and so contains more than one point.

So $\exists q \in H: q \ne p$.

But, from Clopen Points in Fort Space, $\left\{{q}\right\}$ is open in $T$.

So $H \cap \left\{{q}\right\} = \left\{{q}\right\}$ and so by definition $q$ is isolated in $H$.

From this contradiction it follows that $H$ is not dense in itself and contains at least one isolated point.

Hence the result, by definition of scattered space.