Let $A$, $B$, and $A_\alpha$ denote subsets of a space $X$. Prove the following:
(a) If$A\subseteq B$,then$\overline{A} \subseteq \overline{B}$. (b) $\overline{A\cup B}=\overline{A}\cup \overline{B}$.
(c) $\overline{ \bigcup A_{\alpha}} \supset \bigcup \overline{A_\alpha }$ ; give an example where equality fails.
My attempt: (a) $A\subseteq B \subseteq \overline{B}$. Since $\overline{B}$ is closed and $A\subseteq \overline{B}$, we have $\overline{A} \subseteq \overline{B}$. Is this proof correct? If you think last step is bit of a jump, you can look at https://courses-archive.maths.ox.ac.uk/node/view_material/50743 lecture notes, page no. 6.
(b) Approach(1): We first show $(A\cup B)’ = A’ \cup B’$. Let $x\in (A\cup B)’ \Rightarrow \forall U\in \mathcal{N}_x, (U-\{x\})\cap (A\cup B)\neq \phi$. By De Morgan's laws, $[(U-\{x\}) \cap A] \cup [(U-\{x\})\cap B]\neq \phi$. Which means $(U-\{x\}) \cap A$ or $(U-\{x\})\cap B$ or both is non empty. By definition of limit point, $x\in A’$ or $x\in B’$. So $x\in A’ \cup B’$. Thus $(A\cup B)’\subseteq A’\cup B’$.
Conversely, $x\in A’ \cup B’$. If $x\in A’$, then $\forall U\in \mathcal{N}_x, (U-\{x\})\cap A\neq \phi$. So $(U-\{x\})\cap (A\cup B)= [(U-\{x\}) \cap A] \cup [(U-\{x\})\cap B]\neq \phi$. Thus $x\in (A\cup B)’$. Similarly, if $x\in B’$, then $x\in (A\cup B)’$. Hence $(A\cup B)’ = A’ \cup B’$. So $\overline{A\cup B}= (A\cup B) \cup (A\cup B)’= (A\cup A’) \cup (B\cup B’)= \overline{A} \cup \overline{B}$. Is this proof correct? In proof of $(A\cup B)’ \subseteq A’ \cup B’$ what if, for one neighbourhood of $x$,$(U-\{x\}) \cap A$ is non empty and for some other neighbourhood it’s empty?
Approach(2): let $x\in \overline{A\cup B}$. Then $\forall U\in \mathcal{N}_x, U\cap (A\cup B)\neq \phi$. So $(U\cap A)\cup (U\cap B)\neq \phi$. Rest of the proof is similar to approach(1). I’m facing the same problem as approach(1). Is this proof correct?
Approach(3): To show $\overline{A\cup B} \subseteq \overline{A}\cup \overline{B}$. $\overline{A}$ and $\overline{B}$ is closed in $X$. So $\overline{A}\cup \overline{B}$ is closed in $X$. $A\subseteq \overline{A}$ and $B\subseteq \overline{B}$. So $A\cup B\subseteq \overline{A}\cup \overline{B}$. Hence $\overline{A\cup B} \subseteq \overline{A}\cup \overline{B}$.
Approach(4): Since $A\cup B\subseteq \overline{A} \cup \overline{B}$, by part(a) $\overline{A\cup B} \subseteq \overline{ \overline{A} \cup \overline{B} }$. Since $\overline{A} \cup \overline{B}$ is closed, it’s equal to it’s closure. Thus $\overline{A\cup B} \subseteq \overline{A} \cup \overline{B}$.
(C) Let $x\in \bigcup \overline{A_\alpha}$. $x\in \overline{A}_j$. So $\forall U\in \mathcal{N}_x, U\cap A_j \neq \phi$. Since $A_j \subseteq \bigcup A_\alpha$, we have $U\cap A_j \subseteq U\cap (\bigcup A_\alpha)$. Since $\exists p\in U\cap A_j$, $p\in U\cap (\bigcup A_\alpha)$. So $U\cap (\bigcup A_\alpha)\neq \phi$. Thus $x\in \overline{\bigcup A_\alpha}$. Hence $\overline{ \bigcup A_{\alpha}} \supset \bigcup \overline{A_\alpha }$.
Best Answer
So Munkres defines (or shows the equivalence of) three definitions of $\overline{A}$ that section.
The original definition is $\overline{A}$ is the intersection of all closed sets containing $A$ (where closed=complement is open, and the topology is given).
He then goes on to show that $x \in \overline{A}$ iff every neighbourhood of $x$ intersects $A$ which is a handy view to actually find the closure of specific sets in a space.
Finally theorem 17.6 tells us $$\overline{A} =A \cup A'$$ which is added (I think) mainly for historical reasons, it's one of the oldest way of seeing the closure (as $A'$ was already defined way back by Cantor).
Your proof of the exercise alternates between these definitions, which is fine. The limit point one is often the least convenient for proofs though because you're often forced to make separate arguments for points in $A$ and in $A'$, and that's often more work.
(a): your way is fine: $\overline{B}$ is closed and contains $A$ so it's one of the sets in the defining intersection in def. 1 so $\overline{A} \subseteq \overline{B}$ then trivially holds. No "jump" at all: any intersection is trivially a subset of each of its intersectees (to coin a word), so of $\overline{B}$ in particular.
(b):approach 1 is messy, so avoid it; see my comments above though. An approach using def.1 could go:
$\overline{A} \subseteq \overline{A \cup B}$ by part (a). Likewise $\overline{B} \subseteq \overline{A \cup B}$. So for the union too: $$\overline{A} \cup \overline{B} \subseteq \overline{A \cup B}$$ and we have one inclusion. OTOH: $\overline{A} \cup \overline{B}$ is closed (finite union) and contains $A \cup B$ so by the same argument as before: $$\overline{A \cup B} \subseteq \overline{A} \cup \overline{B}$$ and we have the second inclusion and equality.
The point based approach using neighbourhoods and def 2: Suppose $x \in \overline{A \cup B}$. Then $x \in \overline{A}\cup \overline{B}$: because if $x \notin \overline{A}$ and $x \notin \overline{B}$ we'd have $O_1 \in \mathcal N_x$ so that $O_1 \cap A = \emptyset$ and we'd also have $O_2 \in \mathcal N_x$ so that $O_2 \cap B = \emptyset$ (def 2 negated!). But then $O_1 \cap O_2 \in \mathcal N_x$ and moreover $$(O_1 \cap O_2) \cap (A \cup B) \subseteq (O_1 \cap A) \cup (O_2 \cap B)= \emptyset$$ contradicting that $x \in \overline{A \cup B}$, so the claimed inclusion holds. The other inclusion $\overline{A} \cup \overline{B} \subseteq \overline{A \cup B}$ again follows from (a) as before (a union is a superset of each of its unionees etc.).
For (c) use def 1 and note that for each fixed $a'$, $A_{a'} \subseteq \bigcup_{a \in A} A_a$ and so part (a) gives $\overline{A_{a'}} \subseteq \overline{\bigcup_{a \in A} A_a}$ and as this holds for all $a' \in A$ we get again $\bigcup_{a \in A} \overline{A_a} \subseteq \overline{\bigcup_{a \in A} A_a}$ as required.
The approach you do via def 2 is also possible: let $x \in \bigcup_{a \in A} \overline{A_a}$. Then for some $a' \in A$, $x \in \overline{A_{a'}}$. Now let $O \in \mathcal N_x$ be arbitrary. Then $O \cap A_{a'} \neq \emptyset$ and so trivially $O \cap (\bigcup_{a \in A} \overline{A_a}) \neq \emptyset$ too (as $A_{a'}$ is a subset of the union). As $O$ was arbitrary, $x \in \overline{\bigcup_{a \in A} A_a}$ and we're done.