An infinite union of closed sets is a closed set? Question: {$B_n$}$ \in \Bbb R$ is a family of closed sets.
Prove that $\cup _{n=1}^\infty B_n$ is not necessarily a closed set.
What I thought: Using a counterexample: If I say that each $B_i$ is a set of all numbers in range $[i,i+1]$ then I can pick a sequence $a_n \in \cup _{n=1}^\infty B_n$ s.t. $a_n \to \infty$ (because eventually the set includes all positive reals) and since $\infty \notin \Bbb R$ then $\cup _{n=1}^\infty B_n$ is not a closed set.
Is this proof correct?
Thanks

The way I do this is by first proving that the countable intersection of open sets need not be open by this counterexample: $$\bigcap_{\infty}\left (-\frac{1}{n},\frac{1}{n}\right) = \left\{ {0} \right\} $$ Then because a closed set is the complement of an open set we get that the countable union of closed sets need not be closed because $$\left(\bigcap_{i=1}^nA_i\right)^c=\bigcup_{i=1}^nA_i^c\;.$$
