Construct compact space with some homology group not finitely generated I would like to construct a compact space $X$ that has a non finitely generated singular homology group $H_n (X)$ for some $n$. I thought about taking a countable wedge sum of 1-spheres, but this space is not compact. Another idea would be the Hawaiian earring which should be compact and have an infinitely generated homology group, but this is quite difficult to calculate.
 A: The Cantor set $C \subset [0,1]$ has uncountably many path components (in fact, each single point subset is a path component of $C$). Hence $H_0(C)$ is a free abelian group with uncountably many generators.
Taking the suspension $\Sigma$ and noting that $\tilde{H}_{n+1}(\Sigma X) \approx \tilde{H}_n(X)$, where $\tilde{H}_*$ denotes reduced homology, you can construct examples for all $H_i$ with $i \ge 0$. Recall that $\tilde{H}_i = H_i$ for $i > 0$.
Edited: Instead of $C$ you can take any space $X$ with infinitely many path components. If you take $X = \{ 0 \} \cup \{ 1/n \mid n \in \mathbb{N} \}$ as in Eric Wofsey's answer, you get $\Sigma X$ = Hawaiian earring.
A: The Hawaiian earring works just fine.  You don't have to explicitly compute its homology to show it is not finitely generated.  Specifically let $X$ be the Hawaiian earring space.  Note that for any $n\in\mathbb{N}$, $X$ retracts onto a wedge of $n$ circles (just take $n$ of the circles that make up $X$, and map all the rest of the circles to the point where the circles meet).  This implies that $H_1(X)$ has $\mathbb{Z}^n$ as a direct summand for all $n\in\mathbb{Z}$.  This implies $H_1(X)$ is not finitely generated.
For an even easier example, you could take $X$ to be any infinite compact totally disconnected space (say, $\{0\}\cup\{1/n:n\in\mathbb{Z}_+\}$, or a Cantor set).  Then $H_0(X)$ is not finitely generated, since it is freely generated by the path-components of $X$ and there are infinitely many path-components.
