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

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?

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 $$i_*(b') = [\sum_i n_i\phi_i]_U = b$$.