Show that if \(X\) is Hausdorff, a net in \(X\) converges to at most one point.
Solution
Goal: Prove that in a Hausdorff topological space \(X\), limits of nets are unique (i.e. every net converges to at most one point).
1. Setting and hypothesis:Let \((x_\alpha)_{\alpha \in J}\) be a net indexed by a directed set \((J, \succeq)\) in a Hausdorff space \(X\).Suppose that \((x_\alpha) \to x\) and \((x_\alpha) \to y\) for points \(x, y \in X\).We must show that \(x = y\).
2. Separation by disjoint open neighborhoods:Proof:
2.1. Suppose for contradiction that \(x \neq y\).
2.2. Because \(X\) is Hausdorff (\(T_2\)), there exist open sets \(U, V \subseteq X\) such that \(x \in U\), \(y \in V\), and \(U \cap V = \varnothing\).
3. Eventual containment in both neighborhoods:Proof:
3.1. Since \((x_\alpha) \to x\) and \(U\) is an open neighborhood of \(x\), there exists \(\alpha_1 \in J\) such that:\[\alpha \succeq \alpha_1 \implies x_\alpha \in U.\]
3.2. Since \((x_\alpha) \to y\) and \(V\) is an open neighborhood of \(y\), there exists \(\alpha_2 \in J\) such that:\[\alpha \succeq \alpha_2 \implies x_\alpha \in V.\]
4. Derivation of contradiction:Proof:
4.1. Because \((J, \succeq)\) is a directed set, there exists an upper bound \(\alpha_0 \in J\) satisfying \(\alpha_0 \succeq \alpha_1\) and \(\alpha_0 \succeq \alpha_2\).
4.2. By 3, \(\alpha_0 \succeq \alpha_1\) implies \(x_{\alpha_0} \in U\), and \(\alpha_0 \succeq \alpha_2\) implies \(x_{\alpha_0} \in V\).
4.3. Therefore \(x_{\alpha_0} \in U \cap V = \varnothing\), which is impossible.
5. Conclusion:The assumption \(x \neq y\) is false, so \(x = y\). Thus every net in \(X\) has at most one limit. Q.E.D.