Limit Points in T1 Space

Theorem
Let $T = \struct {S, \tau}$ be a topological space which satisfies the $T_1$ (Fréchet) axiom.

Let $H \subset S$ be any subset of $S$.

Let $x \in H$.

Then $x$ is a limit point of $H$ every neighborhood of $x$ contains infinitely many points of $H$.

Necessary Condition
Suppose every neighborhood of $x$ contains infinitely many points of $H$.

Let $N$ be an open neighborhood of $x$.

Then $H \cap N$ is infinite, and so is $H \cap \paren {N \setminus \set x}$.

Since $N$ is arbitrary, $x$ is a limit point of $H$ by definition.

Sufficient Condition
We prove the contrapositive:

Suppose there exists some neighborhood $V$ of $x$ containing only finitely many points of $H$.

Then there exists some open set $N \subseteq V$ such that $x \in N$.

By Set Intersection Preserves Subsets, $N \cap H \subseteq V \cap H$.

By Subset of Finite Set is Finite, $N \cap H$ is finite.

For each $y \in H \cap N$ and $y \ne x$, since $T$ is $T_1$:


 * $\exists U_y \in \tau: x \in U_y, y \notin U_y$

Consider the set $\ds U = N \cap \bigcap_{y \mathop \in H \cap \paren {N \setminus \set x} } U_y$.

This is a finite intersection of open sets.

By General Intersection Property of Topological Space, $U \in \tau$.

By, since $x \in N$ and $x \in U_y$ for all $y \in H \cap \paren {N \setminus \set x}$, we have $x \in U$.

Therefore $U$ is an open neighborhood of $x$.

Let $z \in H \cap \paren {U \setminus \set x}$.

Then:

But from definition of $U_z$:


 * $z \in H \cap \paren {N \setminus \set x} \implies z \notin U_z$

Which is a contradiction.

This shows that $H \cap \paren {U \setminus \set x} = \O$.

By, $x$ is not a limit point of $H$.

Thus the contrapositive is proved.