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If $P(S)$ denotes power set of a set $S$ then is $P(S) \cap P(P(S)) = \{\emptyset\}$?

My book says its true.

I considered a simple set $S=\{1,2\}$

Now $P(S) = \{\emptyset, \{1\}, \{2\}, \{1,2\}\}$ $$ P(P(S))= \{\emptyset,\{\emptyset\}, \{\{1\}\}, \{\{2\}\},\{\{1,2\}\}, \{\{1\}\,\{2\}\}, \{\{1\},\{1,2\}\}, \{\{2\},\{1,2\}\} , \ldots\} $$ Now $P(S) \cap P(P(S))=\emptyset$ and not $\{\emptyset\}$.

So I think that above claim is wrong.


If I go by theorem of power set intersection which says $P(A) \cap P(B)=P(A\cap B)$ and put $A=P(S)$ and $B=P(S)$, then I get $$ P(S) \cap P(P(S)) = P(S \cap P(S)) = P(\emptyset)={\emptyset}. $$ So if I go by theorem, then the claim seems to be correct.

Where am I making a mistake?

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    $\begingroup$ In your example, $P(S)$ has four elements so $P(P(S))$ should have $16$ and the last element has a repeated element within it, but that does not change the answer. $\endgroup$ – Ross Millikan Nov 17 '17 at 6:04
  • $\begingroup$ Yeah, I forgot to consider pairing the elements with $\phi$ $\endgroup$ – Zephyr Nov 17 '17 at 6:05
  • $\begingroup$ @Arthur, Yes, that is possible. Maybe the book didn't consider such cases. You can check the answer by user503979 below. He describes the same thing. $\endgroup$ – Zephyr Nov 17 '17 at 17:25
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In your first argument, you're claiming that $P(S)$ and $P(P(S))$ have empty intersection, but they have one element in common, namely $\emptyset$, so their intersection is $\{\emptyset\}$, as claimed.

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  • $\begingroup$ How is $\emptyset$ same as $\{\emptyset\}$ ? $\endgroup$ – Zephyr Nov 17 '17 at 6:12
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    $\begingroup$ It's not. We have $\emptyset$ in the intersection, but $\{\emptyset\}$ is the intersection. This works, because $\emptyset\in\{\emptyset\}$. $\endgroup$ – G Tony Jacobs Nov 17 '17 at 6:15
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    $\begingroup$ If $S=\{a,b,c\}$ and $T=\{a,e,i,o,u\}$, then $a\in S\cap T$, and $\{a\}=S\cap T$. Your example is basically the same, but we have $\emptyset$ for $a$. $\endgroup$ – G Tony Jacobs Nov 17 '17 at 6:17
  • $\begingroup$ Wew. Now I realize how stupid my question was :P. That $\phi$ just confused me. $\endgroup$ – Zephyr Nov 17 '17 at 6:19
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    $\begingroup$ That’s fair. It’s legitimately confusing! $\endgroup$ – G Tony Jacobs Nov 17 '17 at 6:20
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If $S=\{\emptyset, \{\emptyset\}\}$, then $P(S) = \{\emptyset, \{\emptyset\}, \{\emptyset, \{\emptyset\}\}, \{\{\emptyset\}\}\}$ and $P(P(S))$ will contain the whole of $P(S)$ as a subset.

So the statement as such does not generally follow. Basically whenever a set contains an element as well as a set containing only that element, its powerset will contain the set containing that element as well (obviously).

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Note that $\emptyset \in P(S)$ and $\emptyset \in P(P(S))$ so $\{\emptyset\}\subseteq P(S) \cap P(P(S))$

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  • $\begingroup$ Should that last "$\in$" be a "$\subset$"? $\endgroup$ – G Tony Jacobs Nov 17 '17 at 6:07
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    $\begingroup$ @GTonyJacobs: yes, the horizontal line crept upwards a bit. $\endgroup$ – Ross Millikan Nov 17 '17 at 6:10
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Here is a necessary and sufficient condition for this to hold:

$$S \cap P(S) = \emptyset \Leftrightarrow P(S) \cap P(P(S)) = \{\emptyset\}\;.$$

Clearly, any powerset contains $\emptyset$ as element. All other elements of $P(S)$ are non-empty subsets of $S$, all other elements of $P(P(S))$ are non-empty subsets of $P(S)$. Any common element among those must thus be made from shared elements between $S$ and $P(S)$, and if there is a shared element $x$ between $S$ and $P(S)$, $\{x\}$ is a shared element between $P(S)$ and $P(P(S))$.

An easy way to satisfy this is to have no sets as elements of $S$ since $P(S)$ has only sets as elements.

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