Exercise 2.7

Question

Exercise 7:  Let $A_1,A_2,A_3,\ldots$ be subsets of a metric space.
\begin{enumerate}
\item[(a)] If $B_n = \bigcup_{i=1}^n A_i$, prove that $\bar{B_n} = \bigcup_{i=1}^n \bar{A_i}$, for $n = 1,2,3,\ldots$.

\item[(b)] If $B =\bigcup_{i=1}^\infty A_i$, prove that $\bar{B} \supset \bigcup_{i=1}^\infty \bar{A_i}$.

Show, by an example, that this inclusion can be proper.
\end{enumerate}

ghostofgarborg · Accepted Answer

\def\eps{\varepsilon}

(a) Assume $1 \leq i \leq n$. Since $A_i \subseteq B_n$, a limit point of $A_i$ is a limit point of $B_n$. 
    Let $x$  be a limit point of $ B_n $, and consider the neighborhoods $N_k = N_{\frac 1 k}(x)$
    for $k \in \mathbf N$. There must be a $j$ for which there is no $K$ such that 
    $ k > K$ implies that $ N_k $ does not intersect $A_j$, as the negation of that statement
    would imply that $x$ is not a limit point of $B_n$. Since $N_\ell \subset N_k$ whenever 
    $\ell > k$, this implies that $N_k$ intersects $A_j$ for all $k \in \mathbb N$. 
    Given any neighborhood $N_\epsilon(x)$ of $x$, we can find a $k$ such that $N_k \subseteq N_\epsilon(x)$,
    so all neighborhoods of $x$ intersect $A_j$, and $x \in \bar A_j$.

(b) It is clear that a limit point of $A_i$ is a limit point of $B$, so that the inclusion holds. 
    To see that the inclusion is proper, let $ A_i = (\frac 1 i, 1) $. Then $\bar A_i = [\frac 1 i, 1]$,
    so that $\bigcup_i \bar A_i = (0,1]$.  However, $ B = \bigcup_i (\frac 1 i, 1) = (0,1)$, 
    so that $\bar B = [0,1]$, and $\bar B \supsetneq \bigcup \bar A_i$.

Richard Ganaye · Answer

\begin{proof} 
Note that
\begin{align}
U \subset V \Rightarrow \overline{U} \subset \overline{V}.
\end{align}
Indeed, every limit point $u$ of  $U$ is a limit point of $V$ (see the solution of Ex. 2.6), so $\overline{U} = U \cup U' \subset V \cup V'= \overline{V}$.

\begin{enumerate}
\item[(a)] Since $A_i \subset B_n$ for all $i$, $\bar{A_i} \subset \bar{B_n}$ by (1), thus
$$\bigcup_{i=1}^n \bar{A_i} \subset \bar{B_n}.$$
Conversely, 
$$B_n = \bigcup_{i=1}^n A_i \subset \bigcup_{i=1}^n \bar{A_i}.$$
Moreover,   $F = \bigcup\limits_{i=1}^n \overline{A_i}$ is closed by Theorem 2.24 (d) and Theorem 2.27(a). Since $B_n \subset F$, where $F$ is closed, Theorem 2.27 (c) shows that
$$\overline{B_n} \subset F    = \bigcup\limits_{i=1}^n \bar{A_i}.$$
This shows that
$$\overline{B_n}  = \bigcup\limits_{i=1}^n \overline{A_i}.$$

\item[(b)]
$B = \bigcup_{i=1}^\infty A_i \supset B_n = \bigcup_{i=1}^n A_i$ for all $n$. By part (a), and (1),
$$\overline{B} \supset \bar{B_n} = \bigcup_{i=1}^n \bar{A_i}\qquad (n = 1,2,3,\ldots).$$
Since 
$$\bigcup\limits_{i=1}^\infty \bar{A_i} = \bigcup_{n=1}^\infty \bigcup\limits_{i=1}^n \bar{A_i},$$
we obtain
$$\overline{B} \supset \bigcup\limits_{i=1}^\infty \bar{A_i} .$$

Now consider the example
$$A_i = \{1,1/2,1/3,\ldots,1/i\}.$$
Since $A_i$ is a finite set, $\overline{A_i} = A_i$. Moreover $B = \{1/n \mid n \in \mathbb{N}^*\}$, thus
$$\bar{B} = \{0\} \cup B 
e B = \bigcup\limits_{i=1}^\infty A_i =  \bigcup\limits_{i=1}^\infty \bar{A_i} ,$$
so
$$\bar{B} 
e \bigcup\limits_{i=1}^\infty \bar{A_i}.$$
\end{enumerate}
\end{proof}

Exercise 2.7

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Exercise 2.7

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