$V \subset \mathbb{R}^n$ is closed if and only if $V$ is a union of closed balls I've proved that $V\subset \mathbb{R}^n$ is open if and only if $V=\bigcup\limits_{\alpha\in A}B_\alpha$, where $B_\alpha$ is an open ball. Now I'm asked the following question:

What happens to this result when open is replaced by closed?

My answer is that this result will hold for closed sets as well, but in order to prove it, we must take points $\vec{x}\in V-∂V$, that is, not include any points on the boundary. We take these points and show that a closed ball $B_{r_{\vec{x}}}(\vec{x})$ is in the union of closed balls, etc.
But is my point valid?
 A: The result becomes false when "open" is replaced by "closed". Take $n=1$, and let $B_k$ be the closed ball around $0$ of radius $k/(k+1)$, so $B_k = [-\frac k{k+1}, \frac k {k+1}]$.  Then $\bigcup_k B_k = (-1,1)$ is not closed.
The $\Rightarrow$ direction is true, in a somewhat trivial way: for a closed set $C$, $C = \bigcup_{x\in C} \{x\} = \bigcup_{x\in C}\mathcal{B}(x; 0)$, where $\mathcal{B}(x; \delta)$ is the closed ball of radius $\delta$ around $x$, because $\mathcal{B}(x; 0) = \{x\}$. (Some infinite unions of closed sets are closed.)
The example above shows that the $\Leftarrow$ direction is in general false (though it's true for finite unions).
A: You could also use the fact that $\cap_{n\in\mathbb{N}}U_n$ need not be open for a collection of open sets $\{U_n\}$. So if $\{U_n\}$ is a collection of open sets such that $\cap_{n\in\mathbb{N}}U_n$ = $C$ for some closed set $C$, what happens when we look at the complements? You can take the complement of the sets $B_k = [-\frac k{k+1}, \frac k {k+1}]$ mentioned above if you want a concrete example of a suitable collection $\{U_n\}$.
