Recall that \(X\) is said to be \(\sigma\)-compact if there is a countable collection of compact subspaces of \(X\) whose interiors cover \(X\).
Theorem. Let \(X\) be a \(\sigma\)-compact Hausdorff space. If every compact subspace of \(X\) has topological dimension at most \(m\), then so does \(X\).
Let \(\mathcal{A}\) be an open cover of \(X\). Find an open cover \(\mathcal{B}\) of \(X\) refining \(\mathcal{A}\) that has order at most \(m + 1\), as follows:
Show that \(X = \bigcup X_n\), where \(X_n\) is compact and \(X_n \subset \operatorname{Int} X_{n+1}\) for each \(n\). Let \(X_0 = \varnothing\).
Find an open covering \(\mathcal{B}_0\) of \(X\) refining \(\mathcal{A}\) such that for each \(n\), each element of \(\mathcal{B}_0\) that intersects \(X_n\) lies in \(X_{n+1}\).
Suppose \(n \geq 0\) and \(\mathcal{B}_n\) is an open covering of \(X\) refining \(\mathcal{B}_0\) such that \(\mathcal{B}_n\) has order at most \(m + 1\) at points of \(X_n\). Choose an open covering \(\mathcal{C}\) of \(X\) refining \(\mathcal{B}_n\) that has order at most \(m + 1\) at points of \(X_{n+1}\). Choose \(f: \mathcal{C} \to \mathcal{B}_n\) so that \(C \subset f(C)\). For \(B \in \mathcal{B}_n\), let \(D(B)\) be the union of those \(C\) for which \(f(C) = B\). Let \(\mathcal{B}_{n+1}\) consist of all sets \(B \in \mathcal{B}_n\) for which \(B \cap X_{n-1} \neq \varnothing\), along with all sets \(D(B)\) for which \(B \in \mathcal{B}_n\) and \(B \cap X_{n-1} = \varnothing\). Show that \(\mathcal{B}_{n+1}\) is an open covering of \(X\) that refines \(\mathcal{B}_n\) and has order at most \(m + 1\) at points of \(X_{n+1}\).
Define \(\mathcal{B}\) as follows. Given a set \(B\), it belongs to \(\mathcal{B}\) if there is an \(N\) such that \(B \in \mathcal{B}_n\) for all \(n \geq N\).
Solution
Let \(\mathcal A\) be an open cover of \(X\).
By \(\sigma\)-compactness choose compact sets \(K_1,K_2,\dots\) whose interiors cover \(X\). We construct integers \[ 1\le r_1<r_2<\cdots \] so that \[ X_n=K_1\cup\cdots\cup K_{r_n} \] satisfies \(X_n\subset\operatorname{Int}X_{n+1}\). Once \(r_n\) is chosen, the compact set \(X_n\) is covered by the family \(\{\operatorname{Int}K_j\}\); a finite subfamily suffices, and taking \(r_{n+1}\) larger than all its indices gives the desired inclusion. Since \(r_n\ge n\), \(X_n\supseteq K_n\), so the \(X_n\)’s cover \(X\). Put \(X_0=\varnothing\).
For each \(x\in X\), let \(r\ge1\) be least such that \(x\in\operatorname{Int}X_r\). Choose \(A_x\in\mathcal A\) with \(x\in A_x\), and choose an open neighborhood \[ x\in B_x\subset A_x\cap\operatorname{Int}X_r \] that is disjoint from \(X_{r-2}\) (with \(X_{-1}=X_0=\varnothing\)). Let \(\mathcal B_0=\{B_x:x\in X\}\). If \(B_x\cap X_n\ne\varnothing\), then \(n\ge r-1\), hence \[ B_x\subset X_r\subset X_{n+1}. \] Thus \(\mathcal B_0\) has the required property and refines \(\mathcal A\).
Assume \(\mathcal B_n\) has order at most \(m+1\) at points of \(X_n\). Since \(X_{n+1}\) is compact and has dimension at most \(m\), the restriction of \(\mathcal B_n\) to \(X_{n+1}\) has an open refinement of order at most \(m+1\). Extending these sets to \(X\) and adjoining \(X-X_{n+1}\) if necessary gives an open cover \(\mathcal C\) of \(X\) refining \(\mathcal B_n\) and having order at most \(m+1\) at points of \(X_{n+1}\).
Choose \(f:\mathcal C\to\mathcal B_n\) with \(C\subset f(C)\), and set \[ D(B)=\bigcup\{C\in\mathcal C:f(C)=B\}. \] Each \(D(B)\) is open and contained in \(B\). Define \(\mathcal B_{n+1}\) as in the exercise. It is an open refinement of \(\mathcal B_n\). It covers \(X\): points lying in a retained \(B\) remain covered; otherwise a point lies in some \(C\in\mathcal C\), hence in \(D(f(C))\).
For the order bound, let \(x\in X_{n+1}\). If \(x\in X_n\), every member of \(\mathcal B_{n+1}\) containing \(x\) is contained in a distinct member of \(\mathcal B_n\) containing \(x\), so there are at most \(m+1\). If \(x\in X_{n+1}-X_n\), no retained \(B\) can contain \(x\): a retained \(B\) meets \(X_{n-1}\), hence by the property inherited from \(\mathcal B_0\) it lies in \(X_n\). Thus only sets \(D(B)\) occur at \(x\). Distinct such sets yield distinct elements \(C\in\mathcal C\) through \(x\), so their number is at most \(m+1\). Hence \(\mathcal B_{n+1}\) has order at most \(m+1\) on \(X_{n+1}\).
- Let \(\mathcal B\) consist of the sets that eventually remain unchanged in all \(\mathcal B_n\). Every such set refines \(\mathcal B_0\), hence \(\mathcal A\). To see that \(\mathcal B\) covers \(X\), take \(x\in X_n\) and choose \(B\in\mathcal B_{n+1}\) containing \(x\). At every later stage this \(B\) meets the corresponding retention set \(X_{k-1}\), so it is retained forever; hence \(B\in\mathcal B\).
Finally fix \(x\in X_n\). Any member of \(\mathcal B\) containing \(x\) must already occur by stage \(n+1\): after that stage, a set containing \(x\) can never be newly created as a \(D(B)\), because its predecessor meets \(X_n\) and is therefore retained. Thus the members of \(\mathcal B\) through \(x\) are among those of \(\mathcal B_{n+1}\) through \(x\), at most \(m+1\) of them. Therefore \(\mathcal B\) has order at most \(m+1\) everywhere.
Hence every open cover of \(X\) has an open refinement of order at most \(m+1\), so \(\dim X\le m\).