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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?

user264745
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    It doesn't make sense to ask other people if your proof is correct. Whether it's correct or not follows from the logic and reasoning. Writing something down and then asking if it's correct is just blind guessing, and means that you don't even understand what you yourself have written. – g.s Jan 05 '22 at 19:32
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    In approach 2: that's not de Morgan's law but simple associativity of $\cap$. – Henno Brandsma Jan 05 '22 at 23:31

1 Answers1

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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.

Henno Brandsma
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  • Thank you so much for the answer and for giving constructive criticism(last part). Let me know if I understand your last few lines of reasoning of (c): $(O\cap U)\cap A\neq \phi$. Which implies $\exists z\in X$ such that $z\in O\wedge z\in U \wedge z\in A$. Since $U\cap B=\phi$, $z\in O\wedge z\in U \wedge z\in A \Rightarrow z\in O \wedge z\in A \wedge z\notin B \Rightarrow z\in O \wedge z\in (A-B) $. So $O\cap (A-B) \neq \phi$. – user264745 Jan 06 '22 at 05:59
  • @user264745 Still not nice, but a formula dump. We can also avoid mentioning $z$ by saying: $U \cap B = \emptyset$ means $U \subseteq B^\complement$ so for $O$ we have $\emptyset \neq (U \cap O) \cap A \subseteq (A \cap B^\complement) \cap O = O \cap (A-B)$, if you like set theory. But talking about such a $z$ makes the proof more readable, concrete as it were. A proof is a story not a sequence of logical formulae (only for machines and formal proof theory in FOM studies, not in practice). – Henno Brandsma Jan 06 '22 at 09:47
  • I feel like, it’s my habit to use math symbol. The way you above showed $O\cap (A-B)\neq \phi$ is elegant, but it require little bit of set theory, which is fine. – user264745 Jan 06 '22 at 10:02
  • @user264745 tone down on the formulae. The set theory I used is almost trivial. It’s hardly worth the name. I know from experience that it can go way deeper into set theory once you start on non-trivial general topology. – Henno Brandsma Jan 06 '22 at 10:22
  • Yeah I agree it is trivial. I’m curious, what is non-trivial general topology? – user264745 Jan 06 '22 at 10:39
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    @user264745 go to the library and browse journals like “general topology and it’s applications” or read parts of the “ handbook of set-theoretic topology”. – Henno Brandsma Jan 06 '22 at 10:55