# Is it true that $\mathcal{P}(A^c) = \mathcal{P}(U) − \mathcal{P}(A)$?

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?

• 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. Commented Mar 16, 2014 at 16:12
• Surely you meant "$A^c$ refers to the complement of A", relative to the "universal set" $U$? Commented Mar 16, 2014 at 20:39
• oh sorry, yes I meant complement not transpose Commented Mar 17, 2014 at 13:25

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.

• Thanks a bunch, I was having difficulty getting my head around the power set stuff and this helped a lot. Commented Mar 16, 2014 at 6:04

$$\emptyset \in \mathcal{P}(A^c)$$ $$\emptyset \in \mathcal{P}(A) \implies \emptyset \notin \mathcal{P}(U) - \mathcal{P}(A)$$ So: $$\mathcal{P}(A^c) \neq \mathcal{P}(U) - \mathcal{P}(A)$$

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!

• What I think is, even for $(n,k)=(2,1)$ these two sets are different, but there is a bijection between them. Commented Jun 25, 2017 at 16:19