Exercise 3.2.11

Question

\begin{enumerate}[label=(\alph*)] 
  \item Prove that $\overline{A \cup B} = \overline{A} \cup \overline{B}$.
  \item Does this result about closures extend to infinite unions of sets?
   \end{enumerate}

Ulisse Mini · Accepted Answer

\begin{enumerate}[label=(\alph*)] 
  \item Recall that the set of limit points of a set is closed (Exercise 3.2.7). Let $L$ be the set of limit points of $A\cup B$ and let $L_a, L_b$ be the set of limit points for $A$ and $B$ respectively.

Let $x \in L$, thus there exists a sequence $x_n \in A\cup B$ with $x = \lim x_n$, since $(x_n)$ is infinite there exists a subsequence $(x_{n_k})$ where every term is in $A$ or $B$. Thus the limit $\lim (x_{n_k}) = x$ must be a limit point of $A$ or $B$ meaning $x \in L_a \cup L_b$. This shows $\overline{A \cup B} \subseteq \overline{A} \cup \overline{B}$.

Now let $x \in L_a$ ($L_b$ is the same). there exists a sequence $x_n \in A$ with $x=\lim x_n$, now since $x_n \in A \cup B$ as well, $x \in L$. Thus we have shown $\overline{A \cup B}$ completing the proof.
  \item False, take $A_n = \{1/n\}$ as a counterexample
    $$
    \overline{\bigcup_{n=1}^\infty A_n} = \{1/n : n \in \mathbf{N}\} \cup \{0\},
    	ext{ but }
    \bigcup_{n=1}^\infty \overline{A_n} = \{1/n : n \in \mathbf{N}\}
    $$
   \end{enumerate}

Exercise 3.2.11

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Exercise 3.2.11

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