# Basic topology question regarding open covers

Let $$(X,\tau)$$ be a topological space. For a set $$Y\subseteq X$$ let $$\text{int}Y$$ denote the interior of $$Y$$ in $$\tau$$. Suppose we are given a collection of open sets $$(V_{\alpha})_{\alpha\in A}$$ in $$\tau$$ (where $$A$$ is some index set possibly infinite) such that $$\text{int}Y\subseteq \bigcup_{\alpha\in A}V_{\alpha}$$ but $$Y\nsubseteq\bigcup_{\alpha\in A}V_{\alpha}$$.

Does there always exist some collection of open sets $$(W_{\beta})_{\beta\in B}$$ in $$\tau$$ such that $$Y\subseteq \bigcup_{\beta\in B}W_{\beta}$$ and $$W_{\beta}\cap(\bigcup_{\alpha\in A}V_{\alpha})=V_{\beta}$$ for all $$\beta\in B$$?

added: $$B\subseteq A$$.

• The notation $V_\beta$ does not make sense, unless you assume $B \subseteq A$, do you mean that? Commented Oct 7, 2019 at 15:42
• @MarkKamsma yes. Thanks for pointing it out. Commented Oct 7, 2019 at 15:44

The answer is no in general, that is such a collection $$(W_{\beta})_{\beta\in B}$$ need not exist.
Let $$X=\Bbb R^2$$ be the plane, let $$H$$ be the horizontal strip $$H=\{(x,y):0, let $$Q=\{(r,0):r\in\Bbb Q\}=\Bbb Q\times\{0\}$$ where $$\Bbb Q$$ denotes the set of all rational numbers. Let $$Y=H\cup Q$$. Clearly $$\mathrm{int}Y=H$$.
Let $$A=\Bbb R$$, i.e. we are going to use the reals as an index set. For each $$t\in A$$ let $$V_t=B((t,1),1)=$$ the open ball centered at $$(t,1)$$ and of radius $$1$$.
Then $$\mathrm{int}Y=H=\bigcup_{t\in A}V_t$$ and $$Y\nsubseteq\bigcup_{t\in A}V_t$$.
Next we want some index set $$B\subseteq A$$ and some $$W_t$$ open in $$X$$, for each $$t\in B$$, with $$Y\subseteq \bigcup_{t\in B}W_t$$ and $$W_t\cap(\bigcup_{t'\in A}V_{t'})=V_t$$ for all $$t\in B$$. That is, we want that $$W_t\cap H=V_t$$ for all $$t\in B$$.
There is no way to obtain that. Given any rational $$r$$ there ought to be some $$W_t$$ with $$(r,0)\in W_t$$. But then there is some $$\varepsilon>0$$ such that the open ball $$B((r,0),\varepsilon)\subseteq W_t$$. But then $$W_t\cap H\supseteq B((r,0),\varepsilon)\cap H$$, and $$B((r,0),\varepsilon)\cap H\nsubseteq V_t$$ (the latter holds for every $$t\in A$$), so we cannot have $$W_t\cap H=V_t$$.