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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.

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  • $\begingroup$ Working off a single example can help but it might be misleading. What if we take the union of all the real numbers with $0≤x≤1$? Or those with $0<x≤1$? $\endgroup$
    – lulu
    Jan 25, 2020 at 21:18
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    $\begingroup$ Any set is a disjoint union of singletons. $\endgroup$
    – Andrés E. Caicedo
    Jan 25, 2020 at 21:32
  • $\begingroup$ What @AndresE.Caicedo said is interesting because we can see that a partition of an infinite set doesn't have to contain any infinite set. $\endgroup$
    – PinkyWay
    Feb 24, 2020 at 16:22

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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 $0\notin S$.
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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, $$\bigcup\{S_r\mid r\in(0,1)\}=(0,1)$$

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  • $\begingroup$ Your first result can't be true: here it is proved that a locally compact, connected, Hausdorff space cannot be written as a countable disjoint union of nonempty compact subsets. In your example $(0,1)$ is locally compact, connected and Hausdorff, and each interval in $Q_i$ is compact. See also this MSE question. $\endgroup$
    – Jianing Song
    Mar 19 at 14:41
  • $\begingroup$ You can also see this question that shows that in fact a countable disjoint union of closed subsets of $\mathbb{R}^n$ is never open. $\endgroup$
    – Jianing Song
    Mar 19 at 14:56
  • $\begingroup$ @JianingSong The first example isn't disjoint, and the second isn't countable. There's no contradiction here. $\endgroup$
    – Rushabh Mehta
    Mar 22 at 18:09
  • $\begingroup$ Ah yes, but the question requires the family to be disjoint, otherwise we just note that every open set in $\mathbb{R}^n$ is $F_\sigma$ :) $\endgroup$
    – Jianing Song
    Mar 23 at 9:24

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