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This question already has an answer here:

Say the set $A$ is countable, therefore we can write $A = \{x_0,x_1,x_2,......\}$ $\subseteq \mathbb{R}^2$. But, we can also write, $A = \bigcup \{x_i\} $. And we know a single point in $\mathbb{R}^2$ is a closed, therefore $A$ is closed since it is a countable union. Is this true?

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marked as duplicate by Jonas Meyer, MJD, Asaf Karagila, Julian Kuelshammer, user61527 Sep 4 '13 at 7:15

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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    $\begingroup$ No, countable unions of open sets are open. Finite unions of closed sets are closed. $\endgroup$ – njguliyev Sep 4 '13 at 6:47
  • $\begingroup$ To avoid ambiguities, rather than $\bigcup\{x_i\}$, write $\bigcup_{i=0}^\infty\{x_i\}$. The latter is what you mean. The former can be misinterpreted as the union of the set $\{x_i\}$, which is just $x_i$ if $x_i$ happens to be a set, and nonsense otherwise. $\endgroup$ – Andrés E. Caicedo Sep 4 '13 at 6:47
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    $\begingroup$ @njguliyev, Citizen: Arbitrary unions of open sets are open, no countability needed. $\endgroup$ – Jonas Meyer Sep 4 '13 at 6:49
  • $\begingroup$ @JonasMeyer: I know, but I just copied and paste the OP's expression. :-) $\endgroup$ – njguliyev Sep 4 '13 at 6:52
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Infinite unions of closed sets are not in general closed. For example, consider

$$\bigcup_{n=1}^\infty \left[-\left(1-\frac1n\right), \left(1-\frac1n\right)\right] = (-1, 1)$$

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No, a counterexample (using $\mathbb{R}$ instead of $\mathbb{R^2}$) is $x_n = \frac{1}{n+1}$, in that case $A$ is open because $0$ is a limit point of $A$ not contained in $A$.

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  • $\begingroup$ That set is not open. You meant "non-closed". $\endgroup$ – MJD Sep 5 '13 at 6:42
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No. A countable union of closed sets need not be closed; consider

$$\bigcup\limits_{n = 1}^{\infty} [-1 + \frac 1 n, 1 - \frac 1 n]$$

In the case that we're restricting to a countable set, then consider

$$x_n := \frac{1}{n}$$

Then $\{x_n\}$ has $0$ as a limit point not in the set.

For another counterexample, let $A = \Bbb{Q} \times \Bbb{Q}$, which has every ordered pair of real numbers as a limit point.

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Definitely not. A $\it{finite}$ union of closed sets is closed.

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Not necessarily. For example if $x_n=(\frac{1}{n},0)$, then $(0,0)\in \overline{A}\setminus A$. So, A is not closed.

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