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Show that if $A$ and $B$ are subsets of a set $S$, then $\overline{A \cap B}=\overline{A}\cup \overline{B}$.

I tried to prove that $A \cap B=A \cup B$ because I didn't realize that the overline meant to prove it for the closure of the sets.

So, now I am confused about how to prove for closure. I cannot find it in my textbook, and by some "similar" proofs online led me to conclude that $\overline{A \cap B}=\overline{A \cup B}$ but I somehow don't know if this is true, or how to prove it exactly. So, now I am not sure if I understand this principle at all.

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    this is wrong: take $A= \emptyset$ and $B= S$. – 2017-01-23
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    I think here that it means the *complement* (everything not in the set). So if $A$ and $B$ are subsets of a set $X$, then $\overline{A} = \{x \in X : x \not \in A\}$. – 2017-01-23
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    Have you tried drawing a picture? Draw a Venn-diagram of A and B, and then see how both parts of the equality look in that venn-diagram. – 2017-01-23
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    What is $\overline{A}$? You're calling it the "closure"? The closure of a set doesn't mean anything unless you have a topology on the set. I'm fairly certain that $\overline{A}$ is supposed to be referring to the complement of $A$ in $S$. – 2017-01-23
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    If this is the complement then you should replace the tags first and have a look at [De Morgan's law](https://en.wikipedia.org/wiki/De_Morgan's_laws). – 2017-01-23

2 Answers 2

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Like I said in my comment, I'm pretty sure that $\overline A$ is referring to the complement of $A$ in $S$. The way to prove this problem is to just blindly "chase elements":

Let $x\in\overline{A\cap B}$. Then $x\in S$ but $x\notin A\cap B$. Therefore $x\notin A$ or $x\notin B$. This precisely means $x\in\overline A\cup\overline B$, so $\overline{A\cap B}\subseteq\overline A\cup\overline B$.

I would encourage you do the other direction on your own. Just follow the same procedure I did above, and follow the definitions to show $\overline A\cup\overline B\subseteq\overline{A\cap B}$.

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    Oh, I see. Sorry I didnt see your comment at first...I think you are right. I have worked with closures a little bit before, so what you said makes sense to me too, that the bar must mean compliment here. I never saw that notation other than for closures So....now I think I can do this problem for sure. Thanks for the input! – 2017-01-23
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    @PBJ no problem. – 2017-01-23
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Let M = (A ∩ B)' and N = A' U B'

Let x be an arbitrary element of M then x ∈ M ⇒ x ∈ (A ∩ B)'

⇒ x ∉ (A ∩ B)

⇒ x ∉ A or x ∉ B

⇒ x ∈ A' or x ∈ B'

⇒ x ∈ A' U B'

⇒ x ∈ N

Therefore, M ⊂ N …………….. (i)

Again, let y be an arbitrary element of N then y ∈ N ⇒ y ∈ A' U B'

⇒ y ∈ A' or y ∈ B'

⇒ y ∉ A or y ∉ B

⇒ y ∉ (A ∩ B)

⇒ y ∈ (A ∩ B)'

⇒ y ∈ M

Therefore, N ⊂ M …………….. (ii)

Now combine (i) and (ii) we get

M = N

(A ∩ B)' = A' U B'

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    Are you using the apostrophe to denote the closure of the set? – 2017-01-23
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    Yes $\overline{A \cap B} = (A \cap B)'$ – 2017-01-23
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    If you want me to change I can but its same thing – 2017-01-23
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    No, dont change it. That's fine. Thank you! – 2017-01-23
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    Mine pleasure :-) – 2017-01-23