Sets in Modified Fort Space are Disconnected

Theorem

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.

Proof

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.

$\blacksquare$

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: $4$