A space is locally compact if it can be covered by open sets each of which is contained in a compact subspace of \(X\). It is said to be \(\sigma\)-compact if it can be covered by countably many such open sets.
Show that if \(X\) is locally compact and second-countable, it is \(\sigma\)-compact.
Let \((Y, d)\) be a metric space. Show that if \(X\) is \(\sigma\)-compact, there is a metric for the topology of compact convergence on \(Y^X\) such that if \((Y, d)\) is complete, \(Y^X\) is complete in this metric. [Hint: Let \(A_1, A_2, \ldots\) be a countable collection of compact subspaces of \(X\) whose interiors cover \(X\). Let \(Y_i\) denote the set of all functions from \(A_i\) to \(Y\), in the uniform topology. Define a homeomorphism of \(Y^X\) with a closed subspace of the product space \(Y_1 \times Y_2 \times \cdots\).]
Solution
Let \(\mathcal B\) be a countable basis for the locally compact space \(X\). For each \(x\), choose an open \(U_x\) and compact \(K_x\) with \(x\in U_x\subset K_x\). Choose a basis element \(B_x\) with \(x\in B_x\subset U_x\). Only countably many distinct \(B_x\) occur; choosing one compact \(K_x\) for each such basis element gives countably many compact sets whose interiors cover \(X\). Thus \(X\) is \(\sigma\)-compact.
Choose compact \(A_1,A_2,\dots\) whose interiors cover \(X\), and replace them by finite unions so that \(A_i\subset A_{i+1}\). Put on \(Y_i=Y^{A_i}\) the bounded uniform metric \[ \rho_i(u,v)=\min\{1,\sup_{A_i}d(u,v)\}. \] Define \[ \rho(f,g)=\sum_{i=1}^\infty2^{-i}\rho_i(f|_{A_i},g|_{A_i}). \] This induces exactly compact convergence: every compact \(C\subset X\) is covered by finitely many \(\operatorname{Int}A_i\), hence lies in some \(A_N\), while each \(A_i\) itself is compact.
The restriction map \[ R:Y^X\to\prod_iY_i,\qquad f\mapsto(f|_{A_i}) \] identifies \(Y^X\) with the closed compatibility subspace \(u_{i+1}|_{A_i}=u_i\). If \(Y\) is complete, each uniform function space \(Y_i\) is complete (pointwise limits of uniform Cauchy families exist and convergence is uniform), so the metric \(\sum_i 2^{-i}\rho_i\) on \(\prod_iY_i\) is complete, and \(R\) is an isometry from \((Y^X,\rho)\) onto the compatibility subspace. A closed subspace is complete, so \((Y^X,\rho)\) is complete.