Question:
I need to show that
$A = \bigcup_{n=1}^{\infty} [2n, 2n+1]$ is closed.
I thought that in order for the union of closed sets to be closed it must be a finite union. Since I can take a sequence
$x_k = 2k, $ then $x_k \to \infty $ as $ k \to \infty$ Since $\{x_k\} \subseteq A$ $ \forall$ $ k \in \Bbb{N} $ so $x_k $ does not converge, then $A$ is open.
Apparently something is wrong with my understanding.
Best Answer
There are several misunderstandings here. Let's go through them one by one.
First, in a topological space, any finite union of closed sets is closed. It is not necessarily the case that an infinite union of closed sets is closed. However, it is entirely possible that in some particular examples, an infinite union of closed sets might be closed. As an example, if $B\subset\mathbb{R}$ is any infinite closed set, then $B$ is the union of all the singleton sets $\{x\}$ where $x\in B$. These singletons are closed, so $B$ is an infinite union of closed sets which is closed.
Second, for your set $A$ to be closed, you need to know that whenever $(x_k)$ is a sequence of points in $A$ which converges to some limit $x\in\mathbb{R}$, $x\in A$ as well. Your example sequence does not converge to any $x\in\mathbb{R}$, so that's OK. You only need the limit of a sequence to be in $A$ when the limit exists in $\mathbb{R}$ to begin with.
Third, "open" and "not closed" do not mean the same thing. So even if you found that $A$ is not closed, that would not mean that it is open. A set is open iff its complement is closed, and a set is closed iff its complement is open.
In fact, in this case, the easiest way to prove your set $A$ is closed may be to prove that its complement is open. Think about what $\mathbb{R}\setminus A$ looks like and see if you can write it as a union of open intervals.