Theorem. Let \(A \subset X\). Then \(x \in \overline{A}\) if and only if there is a net of points of \(A\) converging to \(x\).
[Hint: To prove the implication \(\Rightarrow\), take as index set the collection of all neighborhoods of \(x\), partially ordered by reverse inclusion.]
Solution
If a net \((a_\alpha)\) in \(A\) converges to \(x\), then every neighborhood of \(x\) contains some tail point \(a_\alpha\in A\); hence every neighborhood of \(x\) meets \(A\), so \(x\in\overline A\).
Conversely, assume \(x\in\overline A\). Let \(J\) be the set of neighborhoods of \(x\), ordered by reverse inclusion: \[ U\preceq V\iff U\supset V. \] This is directed because \(U\cap V\) is a neighborhood of \(x\) and satisfies \(U\preceq U\cap V\) and \(V\preceq U\cap V\). For each \(U\in J\), choose \(a_U\in A\cap U\), possible because \(x\in\overline A\). Then the net \((a_U)_{U\in J}\) converges to \(x\): if \(W\) is a neighborhood of \(x\), then for every \(U\succeq W\) we have \(U\subset W\), so \(a_U\in W\).