Is the infinite (countable or uncountable) union of disjoint closed sets closed?

Question

Is the infinite (countable or uncountable) union of disjoint closed sets closed?

I think infinite (countable or uncountable) union of disjoint closed singleton sets should be open because if there are the union of infinitely many disjoint singletons, then we have the real numbers which is an open set.

What about the case of non-singleton uncountable disjoint union?

I don't known about the countable union of disjoint closed sets.

Answer 1

An infinite union of closed sets doesn't have to be closed, or open. For instance $S=\bigcup_{n\in\mathbb N}\left\{\frac1n\right\}\left(=\left\{\frac1n\,\middle|\,n\in\mathbb N\right\}\right)$ is neither closed nor open:

• it is not open, because $1\in S$, but $S$ contains no interval centered at $1$;
• it is not closed because the sequence $\left(\frac1n\right)_{n\in\mathbb N}$ is a sequence of elements of $S$ which converges to $0$ and $0\notin S$.

Answer 2

Not only does an infinite union of closed sets not have to be closed, it can also be open. Let $Q_i$ be a set of closed sets defined as below: $$Q_i=\left\{\left[\frac{3k-2}{3i},\frac{3k-1}{3i}\right]\text{ for }0<k\leq i\right\}$$

Let $$Q=\bigcup\limits_{i\in\mathbb N}Q_i$$Note that $$\bigcup Q=(0,1)$$But all of the elements of $Q$ are disjoint closed sets.

Even more trivially, for $r\in\mathbb R$, let $S_r=\{r\}$. Then, note that $S_r$ is closed for all $r$ since $\mathbb R$ is Hausdorff, and no two $S_r$ intersect. Moreover, $$\bigcup\{S_r\mid r\in(0,1)\}=(0,1)$$