Sets in Modified Fort Space are Disconnected

From ProofWiki
Jump to navigation Jump to search


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

Let $H$ be a subset of $S$ with more than one point.

Then $H$ is disconnected.


By Isolated Points in Subsets of Modified Fort Space:

$\exists x \in H: x$ is isolated

By Point in Topological Space is Open iff Isolated, $\set x$ is open in $T$.

By Modified Fort Space is $T_1$ and definition of $T_1$ space, $\set x$ is closed in $T$.

Therefore $\relcomp S {\set x}$ is open in $T$.

Then we have:

\(\ds H\) \(\subseteq\) \(\ds S\)
\(\ds \) \(=\) \(\ds \set x \cup \relcomp S {\set x}\) Union with Relative Complement
\(\ds H \cap \set x \cap \relcomp S {\set x}\) \(\subseteq\) \(\ds \set x \cap \relcomp S {\set x}\) Intersection is Subset
\(\ds \) \(=\) \(\ds \O\) Intersection with Relative Complement is Empty
\(\ds \leadsto \ \ \) \(\ds H \cap \set x \cap \relcomp S {\set x}\) \(=\) \(\ds \O\) Subset of Empty Set
\(\ds H \cap \set x\) \(=\) \(\ds \set x\)
\(\ds \) \(\ne\) \(\ds \O\)
\(\ds H \cap \relcomp S {\set x}\) \(=\) \(\ds H \setminus \set x\) Set Difference as Intersection with Relative Complement
\(\ds \) \(\ne\) \(\ds \O\) $H$ has more than one point

This shows that $H$ is disconnected.