Show that if \(X\) has the discrete topology, then \(X\) is paracompact.
Show that if \(f: X \to Y\) is continuous and \(X\) is paracompact, the subspace \(f(X)\) of \(Y\) need not be paracompact.
Solution
1. Part (a): Discrete spaces are paracompact.Proof:
1.1. In the discrete topology on \(X\), every singleton \(\{x\}\) is open and closed, so \(X\) is Hausdorff.
1.2. Let \(\mathcal{U} = \{U_\alpha\}_{\alpha \in J}\) be an arbitrary open cover of \(X\).
1.3. Consider the singleton collection \(\mathcal{V} = \{\{x\} \mid x \in X\}\).
1.4. In the discrete topology, each \(\{x\}\) is open, and \(\bigcup_{x \in X} \{x\} = X\), so \(\mathcal{V}\) is an open cover of \(X\).
1.5. For each \(x \in X\), since \(\mathcal{U}\) covers \(X\), there exists an index \(\alpha \in J\) such that \(x \in U_\alpha\). Thus \(\{x\} \subseteq U_\alpha\), so \(\mathcal{V}\) refines \(\mathcal{U}\).
1.6. For any point \(p \in X\), the open neighborhood \(W = \{p\}\) intersects only one member of \(\mathcal{V}\) (namely \(\{p\}\) itself).
1.7. Thus \(\mathcal{V}\) is a locally finite open refinement of \(\mathcal{U}\), proving \(X\) is paracompact.
2. Part (b): Continuous images of paracompact spaces need not be paracompact.Proof:
2.1. Let \(Y = [0, \omega_1)\) be the space of countable ordinals equipped with the order topology.
2.2. The space \(Y\) is Hausdorff and countably compact, but not compact (the open cover \(\{[0, \alpha)\}_{\alpha < \omega_1}\) has no finite subcover).
2.3. Since every paracompact and countably compact Hausdorff space is compact, \(Y\) is not paracompact.
2.4. Let \(X = Y_d\) denote the set \([0, \omega_1)\) equipped with the discrete topology.
2.5. By 1, \(X\) is paracompact.
2.6. The identity map \(f: X \to Y\) defined by \(f(x) = x\) is continuous because the domain \(X\) is discrete, and \(f(X) = Y\) is surjective.
2.7. Thus \(f(X) = Y\) is a continuous image of the paracompact space \(X\) that fails to be paracompact. Q.E.D.