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I have to decide whether this statement is true, I think it is not. Since the power set of a set with cardinality $n$, will have $2^n$ subsets, however the power set of this set will include the subsets themselves and subsets of the subsets.

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    $\begingroup$ You are correct! The power set of the power set will actually have $2^{2^n}$ elements $\endgroup$ – Bib Apr 26 '15 at 20:23
  • $\begingroup$ I wanted to write that!! I need more confidence. Thank you. :) $\endgroup$ – user197848 Apr 26 '15 at 20:30
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You are right. You can start by $A=\{a\}$.

$P(A) = \{\emptyset, \{a\}\}$

$P(P(A)) = \{\emptyset, \{\emptyset\}, \{\{a\}\}, \{\emptyset, \{a\}\}\}$

To reduce confusion of $\emptyset$, you can do $B = \{a,b\}$. However $P(P(B))$ will be length to list and I am too lazy to do this. (Actually making clear on all those sets involves empty set will make you clearer on understanding power set of a power set.)

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  • $\begingroup$ Could I ask a follow up question? if $A = \{a\}, B = \{b\}$ would $A \X B = \{(a,b)\}$ be correct?, this may be painfully clear to you. $\endgroup$ – user197848 Apr 26 '15 at 20:49
  • $\begingroup$ What is $AxB$? and $A=a, B=b$ makes no sense on set notation. I guess your notation problem makes you confuse. $\endgroup$ – MonkeyKing Apr 26 '15 at 20:51
  • $\begingroup$ sorry, my latex capabilities suck. $\endgroup$ – user197848 Apr 26 '15 at 20:51
  • $\begingroup$ Cartesian product? Yes. The symbol is "\times". $\endgroup$ – MonkeyKing Apr 26 '15 at 20:58
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All the proofs given above are for finite sets, but what if it is an $\textit{arbitrary}$ set?

Well, $\mathcal{P}(\mathcal{P}(X))=\mathcal{P}(X)$ is still false if you were to assume a set theoretic axiom called Regularity Axiom, which forbids the existence of sets that belong to themselves. Since $\mathcal{P}(X)\in\mathcal{P}(\mathcal{P}(X))$, then we would have $\mathcal{P}(X)\in\mathcal{P}(X)$, a contradiction by this axiom.

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  • $\begingroup$ You don't need the Axiom of Regularity to prove that $\mathcal P(\mathcal P(X))\ne\mathcal P(X)$ or, more generally, $\mathcal P(Y)\not\subseteq Y,$ see the answer to this question. $\endgroup$ – bof Apr 13 '16 at 15:18
  • $\begingroup$ @bof you're right! there is no need to use this axiom to prove $\mathcal{P}(\mathcal{P}(X))\neq\mathcal{P}(X)$. $\endgroup$ – Daniel Muñoz Quintero Apr 13 '16 at 15:48
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You are correct, and just need to choose one strategy for a proof and stick to it. Either you take some concrete small set (say with one element, or maybe even no elements), compute the double power set and show it is different from the power set. Or you argue by cardinality, in which case state the formulas clearly and show that the power set of the power set (of any finite set) always contains more elements then the power set of the same set. Here again, it is better to fix some finite cardinality to consider, e.g., $1$ or $0$.

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