A basic question on closure of a set in metric space Let $A_1, A_2, A_3,\dots$  be subsets of a metric space. I see in some excercise that closure of infinite union of $A_i$s is a superset (proper) of the infinite union of each of $A_i$s closure whereas for finite union it is equal. I don't understand what is finiteness doing here ?
 A: The finiteness makes a difference because a finite intersection of neighbourhoods of a point $x$ is still a neighbourhood of that point (because finite intersections of open sets are open).
So if you have a finite union of sets $(A_k)_{1\leqslant k \leqslant n}$, and a point $x$ that is not in the closure of any of the $A_k$, then you have neighbourhoods $U_k$ of $x$ with $U_k \cap A_k = \varnothing$. Then $U = \bigcap\limits_{k=1}^n U_k$ is a neighbourhood of $x$ with $U \cap \bigcup_{k=1}^n A_k = \varnothing$, and hence $x \notin \overline{\bigcup_{k=1}^n A_k}$.
If you have infinitely many sets $(A_k)$, and $x \notin \bigcup\limits_{k=1}^\infty \overline{A_k}$, then you still have a neighbourhood $U_k$ of $x$ with $U_k \cap A_k = \varnothing$, but then $\bigcap_{k=1}^\infty U_k$ is in general no longer a neighbourhood of $x$, so you cant deduce $x \notin \overline{\bigcup_{k=1}^n A_k}$.
A simple example is $A_k = \lbrace \frac1m : 1 \leqslant m \leqslant k\} \subset \mathbb{R}$. Each $A_k$ is closed, and none contains $0$, but we have $0 \in \overline{\bigcup_{k=1}^n A_k}$.
