Limit Points in Fort Space

Theorem

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

Let $x \in S$ such that $x \ne p$.

Then $p$ is the only limit point of $x$.

Proof

From Definition of Fort Space, we have $\relcomp S {\set x} \in \tau_p$.

For any $y \ne x$, $y \in \relcomp S {\set x}$.

Therefore $\relcomp s {\set x}$ is an open neighborhood of $y$.

From Definition of Relative Complement we also have $x \notin \relcomp S {\set x}$.

Hence $y$ is not a limit point of $x$.

By Definition of Limit Point of Point, $x$ cannot be a limit point of $x$.

Therefore $x$ has no limit points in $S$.

Sources

• 1978: Lynn Arthur Steen and J. Arthur Seebach, Jr.: Counterexamples in Topology (2nd ed.) ... (previous) ... (next): Part $\text {II}$: Counterexamples: $23 \text { - } 24$. Fort Space: $7$