Show that the product of a paracompact space and a compact space is paracompact. [Hint: Use the tube lemma.]
Conclude that \(S_\Omega\) is not paracompact.
Solution
1. Part (a): Product of paracompact \(X\) and compact \(Y\) is paracompact.Proof:
1.1. Let \(\mathcal{A}\) be an arbitrary open covering of \(X \times Y\).
1.2. For each \(x \in X\), the slice \(\{x\} \times Y\) is compact, so there exists a finite subcollection \(\mathcal{A}_x \subset \mathcal{A}\) covering \(\{x\} \times Y\).
1.3. Let \(U_x = \bigcup_{A \in \mathcal{A}_x} A\), which is an open neighborhood of \(\{x\} \times Y\) in \(X \times Y\).
1.4. By the Tube Lemma, there exists an open neighborhood \(W_x\) of \(x\) in \(X\) such that \(W_x \times Y \subseteq U_x\).
1.5. The family \(\{W_x\}_{x \in X}\) forms an open covering of \(X\).
1.6. Since \(X\) is paracompact, choose a locally finite open refinement \(\mathcal{V} = \{V_\beta\}_{\beta \in B}\) of \(\{W_x\}_{x \in X}\).
1.7. For each \(\beta \in B\), choose \(x(\beta) \in X\) such that \(V_\beta \subseteq W_{x(\beta)}\).
1.8. Define the collection of open sets in \(X \times Y\):\[\mathcal{C} = \{ (V_\beta \times Y) \cap A \mid \beta \in B, \, A \in \mathcal{A}_{x(\beta)} \}.\]
1.9. Every \((V_\beta \times Y) \cap A\) is open and contained in \(A \in \mathcal{A}\). For any \((x, y) \in X \times Y\), there is \(\beta\) with \(x \in V_\beta \subseteq W_{x(\beta)}\), and since \(\mathcal{A}_{x(\beta)}\) covers \(W_{x(\beta)} \times Y\), \((x, y) \in A\) for some \(A \in \mathcal{A}_{x(\beta)}\), so \((x, y) \in (V_\beta \times Y) \cap A\).
1.10. For any \((x, y) \in X \times Y\), local finiteness of \(\mathcal{V}\) yields a neighborhood \(N_x\) of \(x\) in \(X\) intersecting only finitely many \(V_{\beta_1}, \dots, V_{\beta_k}\). Then \(N_x \times Y\) is an open neighborhood of \((x, y)\) in \(X \times Y\) that intersects only elements of \(\mathcal{C}\) associated with \(\beta_1, \dots, \beta_k\). Since each \(\mathcal{A}_{x(\beta_i)}\) is finite, \(N_x \times Y\) intersects only finitely many members of \(\mathcal{C}\).
1.11. Since \(X\) and \(Y\) are Hausdorff, \(X \times Y\) is Hausdorff. Hence \(X \times Y\) is paracompact.
2. Part (b): \(S_\Omega = [0, \omega_1)\) is not paracompact.Proof:
2.1. Let \(\overline{S}_\Omega = [0, \omega_1]\) be the compact ordinal space.
2.2. If \(S_\Omega\) were paracompact, then by Part (a), the product space \(S_\Omega \times \overline{S}_\Omega\) would be paracompact.
2.3. Every paracompact Hausdorff space is normal.
2.4. \(S_\Omega \times \overline{S}_\Omega\) is not normal: the diagonal \(\Delta = \{(x, x) \mid x \in S_\Omega\}\) and the top edge \(S_\Omega \times \{\Omega\}\) are disjoint closed sets that cannot be separated by disjoint open sets.
2.5. Therefore, \(S_\Omega\) cannot be paracompact. Q.E.D.