Exercise 17.7

Question

\def\a{\alpha}
\def\b{\beta}
Criticize the following ``proof'' that $\overline{\bigcup A_\a} \subset \bigcup \overline{A}_\a$: if $\left\{A_\a\right\}$ is a collection of sets in $X$ and if $x \in \overline{\bigcup A_\a}$, then every neighborhood $U$ of $x$ intersects $\bigcup A_\a$.
Thus $U$ must intersect some $A_\a$, so that $x$ must belong to the closure of some $A_\a$.
Therefore, $x \in \bigcup \overline{A}_\a$.

Dan Whitman · Accepted Answer

\def\a{\alpha}
\def\b{\beta}

The problem with this ``proof'' is that just because every neighborhood $U$ intersects \emph{some} $A_\a$, it does not mean that \emph{every} $U$ intersects a single $A_\a$, which is what is required for $x$ to be in $\overline{A}_\a$.
This is illustrated in the counterexample above at the end of Exercise~17.6 part~(c).
There, every neighborhood of $0$ clearly intersects \emph{some} set $A_n = (1/n, 2]$, but, for any given $n \in {{\mathbb{Z}}_+}$, not every neighborhood of $0$ intersects $A_n$, for example the neighborhood $(-1, 1/n)$ does not.

Exercise 17.7

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Exercise 17.7

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