A curiosity on covers of compact

Question

Let $(\mathbb{R}^N,\tau)$ a topological space, where $\tau$ is the usual topology. Let $A\subset\mathbb{R}^N$ a compact. If $(A_n)_n$ is a family of open such that \begin{equation} \bigcup_nA_n\supset A, \end{equation} then, from compact definition \begin{equation} \bigcup_{i=1}^{k}A_i\supset A \end{equation} Now, if I find a family of open $(B_n)_n$ such that \begin{equation} cl\bigg(\bigcup_n B_n\bigg)\supset A, \end{equation} can I say that \begin{equation} cl\bigg(\bigcup_{i=1}^kB_i\bigg)\supset A\quad? \end{equation} Thanks for the attention!

Answer 1

I am afraid the answer is no.

For example let $A=[0,1]$ which is a compact subset of $\mathbb{R}$. Now let $B_n = \left(-1,1-\frac{1}{n}\right)$. We then have that

$$\bigcup_{n=1}^\infty B_n = (-1,1)$$

Hence $A \subseteq cl\left(\bigcup_{n=1}^\infty B_n\right) = [-1,1]$. However for every $k \in \mathbb{N}$,

$$\bigcup_{n=1}^k B_n = \left(-1,1-\frac{1}{k}\right)$$

and hence $cl\left(\bigcup_{n=1}^k B_n\right) = \left[-1,1-\frac{1}{k}\right]$ and $A$ is not a subset of $\left[-1,1-\frac{1}{k}\right]$

Answer 2

No. Take $A = [0,1] \subseteq \mathbb{R}$ and $B_n = (\frac{1}{n}, 1)$ for all positive $n$. Then $\bigcup B_n = (0,1)$, so its closure is $A$; but any finite union of the $B_n$ is $(\frac{1}{m}, 1)$ for some $m$, so its closure is $[\frac{1}{m}, 1]$, which doesn't contain $A$.