Limit Points in T1 Space

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Theorem

Let $T = \left({S, \tau}\right)$ 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

Let $x$ be a limit point of $H$.

Let $N_x$ be a neighborhood of $x$.

Then by definition $\exists U \in \tau: x \in U \subseteq N_x$.


Suppose $N_x$ were finite, of cardinality $n$.

Then $N_x = \left\{{x, y_1, y_2, \ldots, y_{n - 1}}\right\}$.

From Equivalence of Definitions of $T_1$ Space, each of the elements of $N_x$ are closed points.

Now as $N_x$ is finite, from Topology Defined by Closed Sets, it follows that $\displaystyle \bigcup_{s \mathop \in N_x} \left\{{s}\right\}$ is also closed.

Hence, so is any subset of $N_x$.

This includes $\displaystyle V := \left({\bigcup_{s \mathop \in N_x} \left\{{s}\right\}}\right) \setminus \left\{{x}\right\}$.

As $V$ is closed, $S \setminus V$ is open, and it follows that $\left({S \setminus V}\right) \cap N_x = \left\{{x}\right\}$.

That is, $x$ is an isolated point of $N_x$.