Theorem   A point of a topological space is in the closure of a subspace if and only if there is a net of points of the subspace converging to the point.

Proof. Let $X$ be a topological space, $x$ a point of $X$ , and $A$ a subspace of $X$ . Suppose first that $x\in\bar{A}$ , and let $\mathcal{U}$ be the collection of neighborhoods of $x$ , partially ordered by reverse inclusion. For each $U\in\mathcal{U}$ , select a point $x_U\in U\cap A$ (such a point is guaranteed to exist because $x\in\bar{A}$ ); then $(x_U)_{U\in\mathcal{U}}$ is a net of points in $A$ , and we claim that $x_U\rightarrow x$ . To see this, let $V$ be a neighborhood of $x$ in $X$ , and note that, by construction, $x_V\in V$ ; furthermore, if $U\in\mathcal{U}$ satisfies $V\supset U$ , then because $x_U\in U$ , $x_U\in V$ . It follows that $x_U\rightarrow x$ . Conversely, suppose there exists a net $(x_\alpha)_{\alpha\in J}$ of points of $A$ converging to $x$ , and let $U\subset X$ be a neighborhood of $x$ . Since $x_\alpha\rightarrow x$ , there exists $\beta\in J$ such that $x_\alpha\in U$ whenever $\beta\preceq\alpha$ . Because $x_\alpha\in A$ for each $\alpha\in J$ by hypothesis, we may conclude that $U\cap A\neq\emptyset$ , hence that $x\in\bar{A}$ .