Isolated Points in Subsets of Modified Fort Space

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Theorem

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

Let $H \subseteq S$ contain more than two points.


Then $H$ contains an isolated point.


Proof

Let $H \subseteq S$ contain more than two points.

From Definition of Modified Fort Space, we can write $S = N \cup \set a \cup \set b$, where $N$ is an infinite set.


Suppose $H \cap N \ne \O$.

Let $x \in H \cap N$.

Then:

$\set x \subset N$

Therefore:

$\set x \in \tau_{a, b}$

This shows that $x$ is isolated.


Suppose $H \cap N = \O$.

By Empty Intersection iff Subset of Complement:

$H \subseteq \relcomp S N = \set {a, b}$

Because $H$ contains at least $2$ elements:

$H = \set {a, b}$

Because $a \in \relcomp S {\set b}$ and $\relcomp S {\relcomp S {\set b} } = \set b$ is a finite set:

$\relcomp S {\set b} \in \tau_{a, b}$

Thus:

$H \cap \relcomp S {\set b} = \set a$

This shows that $a$ is isolated.


Therefore $H$ must contain an isolated point.

$\blacksquare$


Sources