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I'm trying to show that class $C$ of all even-cardinality sets is not closed over powerset via counter-example.

Is it not closed because $|\wp(\{\emptyset\})|=1$ therefore it is not in $C$?

I was wondering mainly if the statement "$|\wp(\{\emptyset\})|=1$ because $\wp\{\emptyset\}=\{\{\emptyset\}\}$" was true.

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    $\begingroup$ Almost, but you have too many pairs of curly braces: $\varnothing$ is the only subset of $\varnothing$, so $\wp(\varnothing)=\{\varnothing\}$. $\wp(\{\varnothing\})=\{\varnothing,\{\varnothing\}\}$. $\endgroup$ – Brian M. Scott Sep 10 '12 at 7:13
  • $\begingroup$ I see, so the empty set is just $\emptyset$ $\endgroup$ – James Sep 10 '12 at 7:15
  • $\begingroup$ $\{\emptyset\}$ is not the same as $\{\{\emptyset\}\}$. $\{\emptyset\}$ is the one element set containing only the $\emptyset$. $\{\{\emptyset\}\}$ is the one element set containing $\{\emptyset\}$. $\endgroup$ – William Sep 10 '12 at 7:18
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    $\begingroup$ The empty set is $\varnothing$; it has no elements. $\{\varnothing\}$ is the set whose only member is the empty set, so it has one element. Where you write $\wp\{\varnothing\}$, you mean $\wp(\varnothing)$, the set of all subsets of the empty set; this is $\{\varnothing\}$, the one-element set whose only member is the empty set. $\endgroup$ – Brian M. Scott Sep 10 '12 at 7:18
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As noted in the comments, $P(\emptyset)=\{{\emptyset\}}$.

$P(\{{\emptyset\}})=\{{\emptyset,\{{\emptyset\}}\}}\ne\{{\{{\emptyset\}}\}}$

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