TITLE: Confusion regarding the fact that $\bar B \supset \bigcup_{i=1}^{\infty} \bar A_i $.
QUESTION [4 upvotes]: 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.

REPLY [6 votes]: 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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