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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3No, countable unions of open sets are open. Finite unions of closed sets are closed. – njguliyev Sep 04 '13 at 06:47
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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. – Andrés E. Caicedo Sep 04 '13 at 06:47
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2@njguliyev, Citizen: Arbitrary unions of open sets are open, no countability needed. – Jonas Meyer Sep 04 '13 at 06:49
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@JonasMeyer: I know, but I just copied and paste the OP's expression. :-) – njguliyev Sep 04 '13 at 06:52
5 Answers
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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 $A$.
Dan Brumleve
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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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Infinite unions of closed sets are not in general closed. For example, consider
$$\bigcup_{n=1}^\infty \left[-\left(1-\frac1n\right), \left(1-\frac1n\right)\right] = (-1, 1)$$
MJD
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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.
Pocho la pantera
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