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$\overline{A}$ is the closure of the set A.

I think the set $A_n = (-\infty, -\frac{1}{n}] \cup [\frac{1}{n}, +\infty)$ is a correct example. Because $A_n$ is closed, so its closure $\overline{A_n} = A_n$. $\cup{\overline{A_n}}$ does not contain 0. But $\overline{\cup{A_a}}$ contains $0$.

Is my example correct? If it is, could you give another example?


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This looks fine. I think $A_n = \{1/n\}$ is a good example as well (it shares the important of the features of yours, but it looks simpler). – Dylan Moreland Aug 27 '11 at 5:41
up vote 6 down vote accepted

Let $A_q=\{q\}$ for all rational numbers $q\in\mathbb Q$.

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And generalize to dense subsets (in reasonable spaces) – gary Aug 27 '11 at 5:43

Consider the topological space $\mathbb{R}$ with usual topology. For each $n\in \mathbb{N}$, consider $A_n=(1/n).\mathbb{Z}$. Since $\mathbb{Z}$ is closed, so is each $A_n$. So RHS in your question becomes $\bigcup A_n=\mathbb{Q}$, while LHS is $\bar{\mathbb{Q}}=\mathbb{R}$.

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I’ll generalize Dylan’s example. If $\sigma = \langle x_n:n\in\omega\rangle$ is a convergent sequence of distinct points in a $T_1$-space $X$, the sets $A_n = \{x_n\}$ are an example. Because $X$ is $T_1$, singleton sets are closed, and the limit of the sequence clearly belongs to $\left(\operatorname{cl}\bigcup\limits_{n\in\omega}A_n\right)\setminus\bigcup\limits_{n\in\omega}A_n$.

In fact, all you need is a space $X$ with a non-isolated point $x$ of countable pseudocharacter, meaning one that is the intersection of countably many open sets: if $\{x\} = \bigcap\limits_{n\in\omega}V_n$, where each $V_n$ is open, just let $A_n = X \setminus V_n$ for each $n\in\omega$. Clearly the $A_n$ are closed and their union is $X\setminus \{x\}$; but $x$ isn’t isolated, so is must be a limit point of $\bigcup\limits_{n\in\omega}A_n$.

If the family of sets needn’t be countable, any non-discrete $T_1$-space works. Let $X$ be such a space, and let $x$ be a non-isolated point of $X$. For each $y\in X \setminus \{x\}$ let $A_y = \{y\}$. As in my first example, each $A_y$ is closed, but the union of the $A_y$ is $X\setminus \{x\}$, which clearly has the non-isolated point $x$ in its closure.

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