nets and closures of subspaces

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 $x\in \overline{A}$, and let $𝒰$ be the collection of neighborhoods of $x$, http://planetmath.org/node/123partially ordered by reverse . For each $U\in 𝒰$, select a point $x_U\in U\cap A$ (such a point is guaranteed to exist because $x\in \overline{A}$); then ${(x_U)}_{U\in 𝒰}$ is a net of points in $A$, and we claim that $x_U\to x$. To see this, let $V$ be a neighborhood of $x$ in $X$, and note that, by construction, $x_V\in V$; furthermore, if $U\in 𝒰$ satisfies $V\supset U$, then because $x_U\in U$, $x_U\in V$. It follows that $x_U\to x$.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 $x_{\alpha }\to x$, there exists $\beta \in J$ such that $x_{\alpha }\in U$ whenever $\beta ⪯\alpha$. Because $x_{\alpha }\in A$ for each $\alpha \in J$ by hypothesis, we may conclude that $U\cap A\ne \mathrm{\varnothing }$, hence that $x\in \overline{A}$.∎

The forward implication of the preceding is a generalization of the result that a point of a topological space is in the closure of a subspace if there is a sequence of points of the subspace converging to the point, as a sequence is just a net with the positive integers as its http://planetmath.org/node/360domain; however, the converse (if a point is in the closure of a subspace then there exists a sequence of points of the subspace converging to the point) requires the additional condition that the ambient topological space be first countable.