Fort Space is Scattered

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Theorem

Let $T = \struct {S, \tau_p}$ be a Fort space on an infinite set $S$.


Then $T$ is a scattered space.


Proof 1

Let $H \subseteq T$ such that $H \ne \varnothing$ and $H \ne \left\{{p}\right\}$.

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.

$\blacksquare$


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 $T_1$ 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, $\set q$ is open in $T$.

So $H \cap \set q = \set q$ 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.

$\blacksquare$