Why does a compact set contain open with compact closure with extra condition on homology as in the book by Vick?

Question

I am having trouble understanding a step in the proof of theorem 6.3 in the book by Vick on Homology Theory: An Introduction to Algebraic Topology.

Let $U\subset \mathbb{R}^n$ be open. For $n\geq 1$, take $b\in H_{n-1}(U)$. Let $b$ be represented by the cycle $\sum_i n_i\phi_i$, then we can take $X=\bigcup_i\phi_i(\sigma^{n-1})$ a compact subset of $U$.

Vick claims that there exists an open set $V$ with compact closure $\overline{V}\subset U$ such that there exists a $b'\in H_{n-1}(V)$ which under the induced morphism from the inclusion $i:V\to U$, gets mapped to $i_*(b')=b$.

The first part is true, since in a Hausdorff space $X$ is closed and taking the interior will be an open set for which the closure is either $X$ itself or the empty set. But I do not know why such an element exists. It seems this is not even true when the interior of $X$ is the emptyset, since then $H_{n-1}(V)$ is empty.

Can anyone tell me what Vick intended with this or where I maybe made an error in my reasoning?

Answer 1

If $W = \text{int} X$, then you know that $\overline W \subset X$. But in general we have $\emptyset \subsetneqq \overline W \subsetneqq X$, though $\overline W = \emptyset$ or $\overline W = X$ is possible.

To find $V$, note that $X$ is a compact subset of the open set $U$. Since $\mathbb R^n$ is locally compact, there exists an open $V \subset \mathbb R^n$ such that $\overline V$ is compact and $X \subset V \subset \overline V \subset U$. But now the cycle $\sum_i n_i\phi_i$ in $U$ can be regarded as a cycle in $V$ (or even in $X$ if you want). Taking $b' = [\sum_i n_i\phi_i]_V$ we get $i_*(b') = [\sum_i n_i\phi_i]_U = b$.