Proof: (a) Suppose $X=K_1∪K_2$ where $K_1,K_2⊆X$ are closed and paracompact. By induction the general case will follow from this one. Suppose $\{U_α\}$ is an open cover of $X$. Then $\{U_α∩K_1\}$ and $\{U_α∩K_2\}$ are open covers of $K_1$ and $K_2$ respectively, so let $\mathcal{B}_1 = \{B_β\}$ and $\mathcal{B}_2=\{B_γ\}$ be locally finite refinements of the two. We claim $\mathcal{C}=\mathcal{B}_1∪\mathcal{B}_2$ is a locally finite refinement of $\{U_α\}$ covering $X$. The refinement and covering conditions are evident, so we proceed to demonstrate local finiteness; assume $x∉K_1$. If $U$ is a neighborhood of $x$ such that $K_2∩U$ intersects only finitely many members of $\mathcal{B}_2$, then $U∩(X-K_1)$ is a neighborhood of $x$ intersecting among the same finite family from $\mathcal{B}_2$, and is disjoint from the members of $\mathcal{B}_1$. So we may assume $x∈K_1$, and similarly $x∈K_2$. As such let $U$ and $V$ be neighborhoods of $x$ such that $K_1∩U$ and $K_2∩V$ intersect only finitely many members of $\mathcal{B}_1$ and $\mathcal{B}_2$ respectively. If $U∩V$ intersects an element of $\mathcal{B}_1$, then since that element is within $K_1$ so too does $U∩V∩K_1⊆U∩K_1$, so that only finitely many elements of $\mathcal{B}_1$ are intersected. Similarly too for $\mathcal{B}_2$, and now $U∩V$ is a neighborhood of $X$ intersecting only finitely many elements of $\mathcal{B}$.
(b) Let $X=∪\text{int }K_i$ where each $K_i$ is paracompact. Again let $\{U_α\}$ be an open cover of $X$. For each $i$, let $\mathcal{B}_i$ be a locally finite refinement of $\{U_α∩K_i\}$ covering $K_i$, and further let $\mathcal{A}_i=\{B∩\text{int }K_i~|~B∈\mathcal{B}_i\}$. Then $\mathcal{A}_i$ is a locally finite open cover of $\text{int }K_i$ for each $i$, so that $\mathcal{A}=∪\mathcal{A}_i$ is a countably locally finite open refinement of $\{U_α\}$ covering $X$, so that $X$ is paracompact by Lemma 41.3.$~\square$
No comments:
Post a Comment