Limit Points in T1 Space

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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 $H$ if and only if every neighborhood of $x$ contains infinitely many points of $H$.


Proof

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, by Definition of Limit Point of Set, $x$ is a limit point of $H$.

$\Box$


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

This is a finite intersection of open sets.

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

By Definition of Set Intersection, 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:

\(\displaystyle \) \(\) \(\displaystyle z \in H \land z \in U \setminus \set x\) Definition of Set Intersection
\(\displaystyle \) \(\leadsto\) \(\displaystyle z \in H \land z \notin \set x \land z \in U\) Definition of Set Complement
\(\displaystyle \) \(\leadsto\) \(\displaystyle z \in H \land z \notin \set x \land z \in N \land \paren {\forall y} \paren {y \in H \cap \paren {N \setminus \set x} \implies z \in U_y}\) Definition of Set Intersection
\(\displaystyle \) \(\leadsto\) \(\displaystyle z \in H \land z \in N \setminus \set x \land \paren {\forall y} \paren {y \in H \cap \paren {N \setminus \set x} \implies z \in U_y}\) Definition of Set Complement
\(\displaystyle \) \(\leadsto\) \(\displaystyle z \in H \cap \paren {N \setminus \set x} \land \paren {\forall y} \paren {y \in H \cap \paren {N \setminus \set x} \implies z \in U_y}\) Definition of Set Intersection
\(\displaystyle \) \(\leadsto\) \(\displaystyle z \in H \cap \paren {N \setminus \set x} \land z \in U_z\) Universal Instantiation

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 Definition of Limit Point of Set, $x$ is not a limit point of $H$.

Thus the contrapositive is proved.

$\blacksquare$