A point in a First Countable Space is a Limit Point iff it is a Sequential Limit

Last updated Nov 1, 2022

Suppose $(X, \tau)$ is a First Countable Space and let $S \subset X$. Then $\bar{S} = {x : \exists (x_n) \subset S \text{ s.t. } x_{n} \to x}$. That is, $x \in X$ is a Limit Point of $S$ If and Only If it is a sequential limit of a Sequence in $S$.

Denote $T = {x : \exists (x_n) \subset S \text{ s.t. } x_{n} \to x}$.

Because we know Sequential Limits are Limit Points of the Sequence, we know for each $x \in T$, there exists ${x_n}{n=1}^{\infty} \subset S$ so that $x$ is a Limit Point of ${x{n}}$. Since Limit Points of a subset are Limit Points of the original Set, $x$ is also a Limit Point of $S$ and $\bar{S} \supset T$. $\checkmark$

On the other hand, suppose $x \in \bar{S}$. Since $X$ is a First Countable Space, there is some Countable Neighborhood Basis $\mathcal{B}{x} \subset \tau$ of $x$. Enumerate $\mathcal{B}{x} = {B_n}{n=1}^{\infty}$. Construct $(x{n})$ inductively in the following way:

1. Set $A_{1} = B_{1}$. Since $B_{1}$ is Open and $x \in B_{1}$, there exists $x_{1} \in A_{1} \cap S$.
2. Set $A_{2} = B_{2} \cap A_{1}$. Since $B_{2}$ is Open and topological spaces are closed under finite set intersections, $A_{2}$ is Open. Since $x \in A_{1}$ and $x \in B_{2}$, we see $x \in A_{2}$. Thus, as $x$ is a Limit Point, there exists $x_{2} \in A_{2} \cap S$.
3. Continue in this fashion…

Now we show $(x_{n})$ converges to $x$. Suppose $U \subset X$ Open so that $x \in U$. Then there exists $N \in \mathbb{N}$ so that $B_{N} \subset U$ since $\mathcal{B}{x}$ is a Neighborhood Basis of $x$. By construction, $A{n} \subset B_{N}$ for all $n \geq N$. Since $x_{n} \in A_{n}$ by construction, we have that for all $n \geq N$, $$x_{n} \in A_{n} \subset B_{N} \subset U$$ and $(x_{n}) \to x$. Thus $x \in T$ and $\bar{S} \subset T$. $\checkmark$

Therefore $\bar{S} = T$. $\blacksquare$