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This is a question regarding Exercise $2.7(b)$ in Baby Rudin (so everything's in a metric space).

Relevant definitions / notation: $\bar E$ is the closure of $E$, i.e., $\bar E = E \cup E'$, where $E'$ is the set of all limit points of $E$.

The Exercise: If $B = \bigcup_{i=1}^{\infty}A_i$, prove that $\bar B \supset \bigcup_{i=1}^{\infty} \bar A_i $.

Where I Am: Well, we have that $$\bar B = B \cup B' = \bigcup_{i=1}^{\infty}A_i \cup \left(\bigcup_{i=1}^{\infty}A_i\right)' = A_1 \cup A_2 \cup \cdots\cup \left( A_1 \cup A_2 \cup \cdots \right)'$$

and

$$\bigcup_{i=1}^{\infty} \bar A_i = \bigcup_{i=1}^{\infty}(A_i \cup A_i ') = (A_1 \cup A_1') \cup (A_2 \cup A_2') \cup \cdots $$

and since $(E_1 \cup E_2 \cup \cdots)' = E_1' \cup E_2' \cdots $ when $E$ is a subset of a metric space, we can write $\bar B$ as

$$ A_1 \cup A_2 \cup \cdots\cup A_1' \cup A_2' \cup \cdots $$

and so (by commutativity and associativity of set operations), we have that

$$ \bar B = \bigcup_{i=1}^{\infty} \bar A_i. $$

BUT, this isn't true, as suggested by the next part of the exercise which states that the inclusion can be proper. Indeed, I believe my confusion lies in the fact that we're dealing with the union of a $countably$ $infinite$ collection of sets, seeing as it $is$ true that

$$ \bar B_n = \bigcup_{i=1}^{n} \bar A_i, \text{for }n = 1,2,3,..., $$

as indicated by part $(a)$ of the exercise. Nevertheless, I'm not sure what would be considered wrong with my approach.

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    $\begingroup$ How do you get $(A_1\cup\cdots)'=A_1'\cup\cdots$? $\endgroup$ – Angina Seng Aug 9 '17 at 20:08
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I don't think it is generally helpful to think of the closure of a set $A$ as the union of $A$ with the set of its limit points. I think of the closure of $A$ as the smallest closed set that contains it.

In your case $\overline B\supseteq B\supseteq A_i$; $\overline B$ is a closed set containing $A_i$. But $\overline{A}_i$ is the smallest closed set containing $A$. Therefore $\overline B\supseteq\overline{A}_i$. As this is true for all $i$, $\overline B\supseteq\bigcup_i\overline{A}_i$.

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  • $\begingroup$ I've always preferred to think of closures that way. It dualizes nicely with the interior being the largest open set below, the closure the smallest closed set above. $\endgroup$ – Randall Aug 9 '17 at 20:19
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Let $A_i =\{\frac{1}{i}\}$ so that $B = \bigcup_i A_i = \{1, \frac{1}{2}, \frac{1}{3}, \ldots\}$. Certainly each $A_i$ is a closed set, so the union of the closures is just $B$ again. However, $\overline{B}$ contains $0$, so the inclusion may be proper.

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  • $\begingroup$ A more drastic example along these lines is $A_i=\{a_i\}$ where the $a_i$ form an enumeration of the rationals. Then $\bigcup_i\overline A_i=\Bbb Q$ and $\overline B=\Bbb R$. $\endgroup$ – Angina Seng Aug 9 '17 at 20:21
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It seems as if you are using that $(\cup_n A_n)'=\cup_n A_n'$. This is false, as the example $A_n:=\{1/n\}$ shows.

It's best to just fix an element in $\cup_n\bar{A_n}$ and show that every neighborhood of this element must intersect $B$.

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