Let $A$, $B$, and $A_\alpha$ denote subsets of a space $X$. Prove the following:
(a)$\overline{A\cap B} \subset \overline{A}\cap \overline{B}$.
(b) $\overline{ \bigcap A_{\alpha}} \subset \bigcap \overline{A_\alpha}$. (c) $\overline{A-B} \supset \overline{A} – \overline{B}$.
My attempt: (a)
Approach(1): $\overline{A}$ and $\overline{B}$ is closed. So $\overline{A} \cap \overline{B}$ is closed in $X$. Since $A\cap B\subseteq \overline{A}\cap \overline{B}$, we have $\overline{A\cap B} \subset \overline{A} \cap \overline{B}$. Our desired result.
Approach(2): let $x\in \overline{A \cap B}$. $\forall U\in \mathcal{N}_x, U\cap (A\cap B)\neq \phi$. By associative Laws, $U\cap (A\cap B)=(U\cap A) \cap (U\cap B)\neq \phi$. So $U\cap A \neq \phi$ and $U\cap B\neq \phi$, holds for all neighbourhood of $x$. Which implies $x\in \overline{A} \cap \overline{B}$. Thus $\overline{A \cap B} \subset \overline{A} \cap \overline{B}$. Notice this time we don’t run into the same problem as we did in part(b) of Exercise 6, Section 17 of Munkres’ Topology. Is this proof correct?
(b) $\overline{A}_\alpha$ is closed in $X$, $\forall \alpha \in A$. So arbitrary intersection, $\cap_{\alpha \in A} \overline{A}_\alpha$, is closed in $X$. Since $A_\alpha \subseteq \overline{A_\alpha}, \forall \alpha \in A$, we have $\cap_{\alpha \in A} A_\alpha \subseteq \cap_{\alpha \in A} \overline{A_{\alpha}}$. Thus $\overline{ \cap A_{\alpha}} \subset \cap \overline{A_\alpha}$. Is this proof correct?
(c) let $x\in \overline{A}$ and $x\notin \overline{B}$. $\forall U\in \mathcal{N}_x, U\cap A\neq \phi$. $\exists V\in \mathcal{N}_x, V\cap B=\phi$. So $(V\cap A)\cap (V\cap B)=\phi = V\cap (A\cap B)$. Since $V\neq \phi$(it contain $x$), $A\cap B= \phi$. So $A-B=A$. $U\cap A= U\cap (A-B)\neq \phi$. Thus $x\in \overline{A-B}$. $\overline{A-B} \supset \overline{A} – \overline{B}$. Is this proof correct?
Best Answer
Better way to do approach 2 of (a) (approach 1 is easiest and shortest; leave it at that): Let $x \in \overline{A \cap B}$ let $O \in \mathcal{N}_x$, then $\emptyset \neq O \cap (A \cap B) \subseteq (O \cap A)$, so $x \in \overline{A}$ and $\emptyset \neq O \cap (A \cap B) \subseteq (O \cap B)$, so $x \in \overline{B}$. Hence $x \in \overline{A} \cap \overline{B}$ and we're done.
(b) is exactly the same as (a) really. Your approach (using the def 1 from before approach) is fine.
(c) needs a better write-up: let $x \in \overline{A}-\overline{B}$. As $x \notin \overline{B}$ we have some $U \in \mathcal{N}_x$ such that $U \cap B = \emptyset$. So if $O \in \mathcal{N}_x$ is arbitrary, $O \cap U \in \mathcal{N}_x$ so $x$ being in the closure of $A$ tells us that $(O \cap U) \cap A \neq \emptyset$. But $z$ in that intersection is thus in $O$ and in $A$ but not in $B$ as $z \in U$ also. So $z \in O \cap (A-B)$ and so (as $O$ was arbitrary) $x \in \overline{A-B}$. Don't make a proof about just writing formulae but do reasoning too. It's a common thing in your proofs, I noticed.