Countable closed sets

There is a theorem that states that the finite union of closed sets is closed but I was wondering if we have a set that consists of countable many subsets that are all closed if that set is closed. I really want to believe that the set is closed but I've been wrong in past so if anyone can supply me with an answer I would be very grateful.

Thank you.

Solution 1:

No. Consider the following two collections:

1. For $n \in \mathbb{N} = \{ 1,2,\ldots \}$, let $A_n = \{ n \}$. Clearly each $A_n$ is closed (all singletons are closed) and their union $\bigcup_{n \in \mathbb{N}} A_n = \mathbb{N}$ is also a closed subset of $\mathbb{R}$.
2. For $n \in \mathbb{N}$, let $B_n = \{ \frac{1}{n} \}$. Again, each $B_n$ is closed, but their union $\bigcup_{n \in \mathbb{N}} B_n = \{ \frac{1}{n} : n \in \mathbb{N} \}$ is not closed, because $0$ is a limit point of that set.

Added:

The examples presented here might almost lead you to believe that the countable union of closed sets can be almost anything. This is not exactly true, and we call a countable union of closed sets an $\text{F}_\sigma$-set. There are many sets do not belong to this class; the set $\mathbb{R} \setminus \mathbb{Q}$ of all irrational numbers is but one example.

However, there are conditions on a family $\{ F_n \}_{n \in \mathbb{N}}$ of closed sets which imply that their union is also closed. One example is the following: If for each $x \in \mathbb{R}$ there is a $\delta > 0$ such that $F_n \cap ( x-\delta , x+\delta) = \emptyset$ for all but finitely many $n$, then the union $\bigcup_{n\in\mathbb{N}} F_n$ is closed.

Solution 2:

no. Countable unions of closed sets need not to be closed, for example $$ (0,1) = \bigcup_{n\ge 2} \left[\frac 1n, 1-\frac 1n\right] $$ is not closed in $\mathbb R$.