A point is a Limit Point iff it is a Net Limit

Last updated Nov 1, 2022

Let $X$ be a Topological Space and let $S \subset X$. Then $x \in X$ is a Limit Point of $S$ If and Only If there exists a Net ${x_{\alpha}}{\alpha \in A} \subset S$ such that $x{\alpha}\to x$.

$(\Rightarrow)$ Suppose $x \in X$ is a Limit Point of $S$. Consider the collection $A := {U \subset X: U \text{ is open}, x \in U}$. Then observe $A$ is a Directed Partial Ordering when ordered by reverse Subset Relation since If $U, V \in A$, then $U \cap V \ni x$ and is Open, so $U \cap V \in A$. By definition of Set Intersection, $U \cap V \subset U, V$, so $U \cap V \geq U, V$. Since $x$ is a Limit Point, for each $U \in A$, there exists $x_{U} \in U \cap S$. We will show that the Net ${x_{U}}{U \in A} \subset S$ converges to $x$. Observe that for any Open $V \subset X$ so that $V \ni x$, every $U \geq V$ (that is $U \subset V$) has $x{U} \in U \subset V$. Thus $\lim\limits_{U \in A}x_{U} = x$ and we have a Net that converges to the Limit Point $x$ $\checkmark$.

$(\Leftarrow)$: Suppose $x \in X$ is a limit of Net ${x_{\alpha}}{\alpha \in A} \subset S$. Let $U \subset X$ so that $x \in U$. Then there exists some $\alpha{0} \in A$ so that $x_{\alpha_{0}} \in U$. Since $x_{\alpha_{0}} \in S$ already, we have $x_{\alpha_{0}} \in U \cap S$. Thus $x$ is a Limit Point of $S$ $\checkmark$. $\blacksquare$