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If A is any set, then (A')' = A. Prove.

How does this work? Does a complement function somewhat like a negative integer? If all the elements not in A are multiplied by a complement how does that equal A?

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  • $\begingroup$ Complement.${}$ $\endgroup$ – Andrés E. Caicedo Nov 12 '16 at 15:19
  • $\begingroup$ A minor issue: you cannot simply take complement of any set, it must be a subset of some bigger set or otherwise the "complement" might not be a set itself. That said, what you want to prove is just definition chasing, what does it mean that something is complement of a set? $\endgroup$ – user160738 Nov 12 '16 at 15:25
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There isn't a 'complement-function' per se, don't think about negative integers right now.

Let $U$ be all the elements and let $A$ be some subset of these elements.

Then $A'$ ($A$'s complement) is $U/A = \{ e | e \in U and\ e \not\in A\}$

In plain words: $A$s complement is every element in U which is not in A.

The complement of $A$ can be seen as such:

enter image description here

The double complement would simply be all the elements in $U$ which are not in $A'$, which is $A$.

This question has already been asked here, and here a proof is provided: Double Complement of a set proof

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  • $\begingroup$ This helps. Thanks. $\endgroup$ – Jebussy Nov 12 '16 at 15:45
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For intuitive purpose, See the image below

enter image description here

$U$ is the universal set, The region which is colored white is your set $A$ and the yellow colored region is the complement.

Now, take the complement of the yellow colored part.

That is exactly the set $A$.

"complement laws give the fundamental properties of the somewhat inverse." You can read more about it on the following Wikipedia page Algebra of Sets

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