Closed Set in Particular Point Space has no Limit Points

From ProofWiki
Jump to navigation Jump to search

Theorem

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

Let $H \subsetneq S$ be closed in $T$.


Then $H$ has no limit points.


Proof

Let $H$ be closed in $T$.

Then by definition $p \notin H$.


Let $x \in H$.

By definition, $x$ is a limit point of $H$ if every open set $U \in \tau$ such that $x \in U$ contains some point of $H$ other than $x$.


Consider the set $U_x := \set {x, p} \subseteq S$.

As $p \in U_x$ we have that $U_x$ is open in $T$.

But there is no $y \in U_x: y \in H, y \ne x$ and so $x$ is not a limit point of $H$.


So $H$ has no limit points.

$\blacksquare$


Sources