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Every open subset $U\subseteq {\mathbb{R}}^N$ is the directed union of finite unions of closed rational cubes contained in $U$. Every compact subset of $U$ lies in one such finite polyhedron, whose homology is finitely generated. The homology of a directed union therefore expresses each $H_i(U;\mathbb{Z})$ as a direct limit over a countable family of countable groups, so it is countable.

List the prime numbers without repetition as $p_0,p_1,p_2,\dots$. Choose mapsand let $T$ be their mapping telescope. A finite initial telescope deformation retracts onto its last sphere. The inclusions of successive finite telescopes induce multiplication by $p_i$ on degree-$n$ reduced homology. The homology of a directed union and the sequential direct limit of multiplication maps on the integers therefore giveEvery finite product $p_0\cdots p_r$ is square-free, and every square-free denominator divides one such product. Henceas in the mapping-telescope realization of the rational group with square-free denominators.

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 $X_b$, soBecause filtered colimits of abelian groups are exact, kernels and images commute with this colimit. Taking homology gives the homology of a directed union:

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 every $H_i(U;\mathbb{Z})$ is countable.

Cohomology behaves differently because $\mathrm{Hom}$ turns a direct sum into a direct product. The connected open sethas one independent loop around each puncture, so $H_1(U;\mathbb{Z})\cong {⨁}_{n\ge 1}\mathbb{Z}$. Since $H_0(U;\mathbb{Z})=\mathbb{Z}$, the universal coefficient theorem for cohomology giveswhich is uncountable. This is the first cohomology of the countably punctured plane.