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
- 1978: Lynn Arthur Steen and J. Arthur Seebach, Jr.: Counterexamples in Topology (2nd ed.) ... (previous) ... (next): Part $\text {II}$: Counterexamples: $27$. Modified Fort Space: $6$