Show that a direct limit of countable abelian groups over a countable indexing set is countable. Apply this to show that if \(X\) is an open set in \(\mathbb{R}^n\) then \(H_i(X; \mathbb{Z})\) is countable for all \(i\).
Solution
If the indexing set is countable and every \(G_\alpha\) is countable, then the disjoint union \[ \coprod_\alpha G_\alpha \] is countable. The direct limit is a quotient of this set, hence is countable.
Now let \(X\subset\mathbb R^n\) be open. Choose an exhaustion \[ K_1\subset\operatorname{int}K_2\subset\operatorname{int}K_3\subset\cdots, \qquad X=\bigcup_j K_j, \] where each \(K_j\) is a finite cubical polyhedron contained in \(X\). Such an exhaustion is obtained from a rational cubulation by taking finitely many cubes whose distance from \(\mathbb R^n-X\) and from infinity is bounded below.
Every compact subset of \(X\) lies in some \(K_j\). In particular every singular simplex, and every finite singular chain or chain homotopy, is contained in some stage. Hence \[ C_*(X)=\varinjlim_j C_*(K_j) \] and Exercise 17 gives \[ H_i(X;\mathbb Z)\cong\varinjlim_j H_i(K_j;\mathbb Z). \] Each \(K_j\) is a finite CW complex, so \(H_i(K_j;\mathbb Z)\) is finitely generated and therefore countable. The indexing set is countable, so the first part implies \[ \boxed{H_i(X;\mathbb Z)\text{ is countable for every }i.} \]