A and B are non-intersecting sets within the universal set U. If $(A\cup B)'=\varnothing$, what does this imply for sets A and B?

This is a question from my O' levels textbook. I figured out that if $(A\cup B)'$ is empty, it must mean A and B are the only elements in U, therefore $U = \{A, B\}=(A\cup B)$. However, according to the book's back end answers, $U=(A\cap B)$. This is contradictory, how can A and B have no intersection but at the same time have an intersection equal to U? I think either I'm missing some relation between sets and universal sets, or the book confused union with intersection.

Is it right to say non-intersecting sets have an intersection that's equal to the universal set? If so, what is the intuition behind this when using Venn diagrams? Thank you.

P.S. please comment if I forgot to include any key info, it's a bad habit of mine.

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    $\begingroup$ It seems, ironically, to be a typo. $\endgroup$ – Randall Oct 28 '18 at 5:08
  • $\begingroup$ Yeah, I had a hard time believing it was the book's fault because these books get revised frequently and I checked through two versions of the answer sheet, both of which had the wrong answer. It also doesn't help that so many of the symbols in sets are so easy confusable. $\endgroup$ – Typo Oct 28 '18 at 5:26

I'd rather say this as a comment but I lack the reputation. Sorry. However, you might want to check your claim that $\{A,B\} = \left(A\cup B\right)$.

Also yeah I agree with Randall that it's a typo.

  • $\begingroup$ I see, this is a nuance with $\{A, B\}$. If $A=\{1,2\}$ and $B=\{2, 3\}$, the former expression would be equal to $\{\{1,2\},\{2,3\}\}$, which includes the same number/element twice (and also includes second, inner brackets). Side note: is there any operation that removes the inner brackets from a set containing other sets? $\endgroup$ – Typo Oct 28 '18 at 5:55
  • $\begingroup$ That's actually exactly what union does. $\endgroup$ – user122495 Oct 28 '18 at 5:59
  • $\begingroup$ I didn't mean it like that, I meant just removing the brackets, without caring about repeating elements. It wouldn't really be useful now that I think about it. $\endgroup$ – Typo Oct 28 '18 at 6:06
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    $\begingroup$ That one, aptly named, would be disjoint union. Formally sets aren't really able to distinguish between repeated elements. But you can get around that by essentially "tying" something to each element to create a difference that way. $\endgroup$ – user122495 Oct 28 '18 at 6:10

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