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To clarify, $A^c$ refers to the transpose of $A$ while $\mathcal{P}(A)$ refers to the power set of $A$.

I was practicing, trying to answer this question and I was confused as to whether I could apply the logic that since $A^c = U - A$,

let $X$ be an arbitrary element of $\mathcal{P}(A^c)$ and so,

$X$ is a subset of $A^c$ and since $A^c = U-A$, $X$ is a subset of $U-A$. After this, the main part I had a problem with is whether it is correct to say that since $X$ is an element of $\mathcal{P}(U-A)$ , $X$ is an element of $\mathcal{P}(U) - \mathcal{P}(A)$. Additionally, if the initial statement isn't true, how would I provide a counterexample?

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    $\begingroup$ The main observation: if a set is not a subset of $A$, it does not mean it sits completely in the complement of $A$, it can members of both. $\endgroup$ – Henno Brandsma Mar 16 '14 at 16:12
  • $\begingroup$ Surely you meant "$A^c$ refers to the complement of A", relative to the "universal set" $U$? $\endgroup$ – hardmath Mar 16 '14 at 20:39
  • $\begingroup$ oh sorry, yes I meant complement not transpose $\endgroup$ – Bob Marley Mar 17 '14 at 13:25
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Let $U=\{a,b,c\}$ and let $A=\{a\}$. Then $\mathcal{P}(U)-\mathcal{P}(A)$ contains the set $\{a,b\}$, which is not part of $\mathcal{P}(A^c)$.

Therefore, the answer to your question is no.

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  • $\begingroup$ Thanks a bunch, I was having difficulty getting my head around the power set stuff and this helped a lot. $\endgroup$ – Bob Marley Mar 16 '14 at 6:04
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$$\emptyset \in \mathcal{P}(A^c)$$ $$ \emptyset \in \mathcal{P}(A) \implies \emptyset \notin \mathcal{P}(U) - \mathcal{P}(A)$$ So: $$ \mathcal{P}(A^c) \neq \mathcal{P}(U) - \mathcal{P}(A)$$

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You can also see this by a counting argument if $U$ is finite. If $|U|=n$, then $|\mathcal P(U)|=2^n$, and if $|A|=k$, then $|\mathcal P(A)|=2^k$. Then

$$|\mathcal P(A^c)| = 2^{n-k}, |\mathcal P(U)-\mathcal P(A)| = |\mathcal P(U)|-|\mathcal P(A)| = 2^n-2^k=2^k(2^{n-k}-1)$$ These are different unless $(n,k)=(2,1)$ - in other words, your sets are almost always different!

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  • $\begingroup$ What I think is, even for $(n,k)=(2,1)$ these two sets are different, but there is a bijection between them. $\endgroup$ – Bumblebee Jun 25 '17 at 16:19

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