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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.

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Try $\bigcup_{n=1}^\infty \{1/n\}$ in $\mathbb{R}$. – t.b. Mar 26 '12 at 9:24
"...but I was wondering if we have a set that consists of countable many subsets that are all closed if that set is closed..." -- That formulation does not seem to correspond to your actual question. – TMM Mar 26 '12 at 9:27
up vote 11 down vote accepted

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.


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.

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OK I understand. Thanks! – docjay Mar 26 '12 at 9:28

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 $\mathbb R$.

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No. $(0,2)=\bigcup_{n=1}^\infty [1/n,2-1/n]$. Actually, one can show that every open set in a metric space is a countable union of closed sets.

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Take the real line endowed with the usual topology, and $S_n:=\{n^{-1}\}$ for each integer $n$. $0$ is in the closure of the union of $S_n$ but not in this union, so this one cannot be closed.

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The are many cases for example $\bigcup [-n,n]=\mathbb{R}$ closed (not compact) and open and $\bigcup [-n^{-1},n^{-1}]=[-1,1]$ (compact) . And there are cases like $\bigcup [-1+n^{-1},1-n^{-1}]=(-1,1)$ (open not closed), $\bigcup [-1,1-n^{-1}]=[-1,1)$ (neither open nor closed).

However theses sets are important (when the union is countable) because they are "near" (measure sense) to closed set and they are called in Measure Theory $F_\sigma$ it comes French Fermé somme what means literally "closed sum".

Correction: $\bigcup [-1,1+n^{-1}]=[-1,2]$

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Actually, $F_{\sigma}$ sets behave more like open sets, while $G_{\delta}$ sets behave more like closed sets. – Dave L. Renfro Mar 26 '12 at 19:11

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