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It seems obvious to me that the statement is true. If you are looking at the elements that are not in the universal set, there are no elements left, thus you are left with the empty set.

However, when I try to prove it, I unpack the definition of U' down to: x∈U ^ x∉U

That's a contradiction. What can I do to prove this statement? Or did that contradiction just disprove the statement?

William
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    The definition of $U'$ is the set of x such that $x\in U$ and $x\notin U$. Since, obviously, as you say, no $x$ can satisfy such an absurdity, this set is empty. – Andrés E. Caicedo Sep 24 '12 at 05:31

2 Answers2

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Ø'$=U$

Take complements both sides and use $(A')'=A$ which gives

(Ø')'$=U'\implies $ Ø$=U'$

Alternatively,

Let $U'\neq \phi\implies \exists x\in U'\implies x\notin U$ which contradicts the definition of U(universal set)

Aang
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  • Thank you. That makes sense. Related question, what is the mathematical definition of the universal set? I know the definition in words. – William Sep 24 '12 at 05:34
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    In set theory, a universal set is a set which contains all objects, including itself. see http://en.wikipedia.org/wiki/Universal_set – Aang Sep 24 '12 at 05:36
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    Under the most widely used axioms, there is no universal set. –  Sep 24 '12 at 05:39
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    @Shahab: That's right. Really, there is no point in discussing about $U'$ when there is no $U$ – Aang Sep 24 '12 at 05:43
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    @William: In restricted contexts you can talk about complements (and the unoiversal set), e.g. when discussing real numbers, the irrationals are the complement of the rationals and the transcendentals are the complement of the algebraics. Then the universal set is just the set $\mathbb R$ of real numbers. However, then all your sets are sets of real numbers - you have intersections, unions (and complements) but for example no notion of power set. When you want to rather talk about prime numbres and composites, you'd choose a different "universal set", but in its most general sense, ... – Hagen von Eitzen Sep 24 '12 at 06:28
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The set $U$ cannot exist. If it did, you could select a subset $r$ of $U$ such that:

$\forall x (x\in r \leftrightarrow x\in U \wedge x \notin x)$

Applying the definition of $r$ to $r$ itself and we can obtain the contradiction:

$r\in r \wedge r\notin r$

Since $U$ cannot exist, $U'$ is undefined.

Dan Christensen
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