The proof for (A ∪ B)' = (A' ∩ B') is:

Let's say x ∈ (A ∪ B)'. This means x ∉ (A ∪ B), x ∉ A and x ∉ B. So, x ∈ A' and x ∈ B'. So x ∈ (A' ∩ B')

So (A ∪ B)' ⊆ (A' ∩ B'). Then we have to prove (A' ∩ B') ⊆ (A ∪ B)' and we are done.

Can someone please explain why we write (A ∪ B)' ⊆ (A' ∩ B') and not (A ∪ B)' = (A' ∩ B')?

  • $\begingroup$ In your (half) proof you wrote $(A\cup B)'\subseteq (A'\cup B')$ yourself, so detect you own motivation. Don't we write $(A\cup B)'=(A'\cup B')$? What makes you say that? $\endgroup$
    – drhab
    Oct 10 '15 at 8:20
  • 1
    $\begingroup$ It was shown that $x\in (A\cup B)'\implies x\in A'\cap B'.$ This means exactly $(A\cup B)'\subset A'\cap B'.$ $\endgroup$
    – mfl
    Oct 10 '15 at 8:21
  • $\begingroup$ Hard to judge for me. In short there are two things that must be proved, as you remarked yourself. If both have been done ($\subseteq$ and $\supseteq$) then the conclusion $=$ is justified. It is nicely worded in the answer of Alessandro. $\endgroup$
    – drhab
    Oct 10 '15 at 8:24

You have shown that $x\in (A\cup B)'\implies x\in (A'\cap B')$, in other words every element of the first set is contained in the second set, so the first set is contained into the second one, or $(A\cup B)'\subseteq(A'\cap B')$.

At this point you don't know that the $2$ sets are equal, there could be elements in the second set that are not contained in the first one, to rule out this possibility you have to prove $(A'\cap B')\subseteq(A\cup B)'$.

It is a very common strategy to prove $A\subseteq B$ and $B\subseteq A$ to show $B=A$ since it's usually easier to prove the $2$ statements separately then the equality directly

  • $\begingroup$ My discrete math class also say it's easier to prove 2 separately, but I never got to know why. Would you mind explaining that, please? $\endgroup$
    – RexYuan
    Oct 10 '15 at 10:47

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