Every <singular simplex> has compact image, and a <singular chain> is a finite sum of simplices. The image of any chain is therefore compact and lies in some $X_a$. Directedness puts any finite collection of chains into one common $X_b$, so
$$
C_*(X;\mathbb Z)=\varinjlim_aC_*(X_a;\mathbb Z).
$$
Because <filtered colimits> of <abelian groups> are exact, kernels and images commute with this colimit. Taking homology gives the <homology of a directed union>:
$$
\boxed{H_i(X;\mathbb Z)\cong\varinjlim_aH_i(X_a;\mathbb Z).}
$$

For an open $U\subseteq\mathbb R^N$, use the directed family of finite unions of closed rational cubes contained in $U$. Every compact subset of $U$ lies in one such finite polyhedron, and each polyhedron has finitely generated <cellular homology>. There are only countably many of them, so their direct limit is countable. Thus \b[every $H_i(U;\mathbb Z)$ is countable].

Cohomology behaves differently because $\operatorname{Hom}$ turns a direct sum into a direct product. The connected open set
$$
U=\mathbb R^2\setminus\{(n,0):n\geq1\}
$$
has one independent loop around each puncture, so $H_1(U;\mathbb Z)\cong\bigoplus_{n\geq1}\mathbb Z$. Since $H_0(U;\mathbb Z)=\mathbb Z$, the <universal coefficient theorem for cohomology> gives
$$
\boxed{H^1(U;\mathbb Z)\cong\operatorname{Hom}\left(\bigoplus_{n\geq1}\mathbb Z,\mathbb Z\right)
\cong\prod_{n\geq1}\mathbb Z,}
$$
which is uncountable. This is the <first cohomology of the countably punctured plane>.

Solution

Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 114 / 2 / 2 / Solution

{title2=$H^1(\mathbb R^2\setminus\{(n,0):n\geq1\};\mathbb Z)\cong\prod_{n\geq1}\mathbb Z$}

The connected open set $X=\mathbb R^2\setminus\{(n,0):n\geq1\}$ retracts onto a locally finite graph having one independent loop around each puncture. Hence $H_1(X;\mathbb Z)\cong\bigoplus_{n\geq1}\mathbb Z$, while the <universal coefficient theorem for cohomology> gives
$$
H^1(X;\mathbb Z)\cong\operatorname{Hom}\left(\bigoplus_{n\geq1}\mathbb Z,\mathbb Z\right)\cong\prod_{n\geq1}\mathbb Z,
$$
an uncountable group.

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