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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

5 Answers 5

up vote 7 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.

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.

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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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