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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.

James
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    Almost, but you have too many pairs of curly braces: $\varnothing$ is the only subset of $\varnothing$, so $\wp(\varnothing)={\varnothing}$. $\wp({\varnothing})={\varnothing,{\varnothing}}$. – Brian M. Scott Sep 10 '12 at 07:13
  • I see, so the empty set is just $\emptyset$ – James Sep 10 '12 at 07:15
  • ${\emptyset}$ is not the same as ${{\emptyset}}$. ${\emptyset}$ is the one element set containing only the $\emptyset$. ${{\emptyset}}$ is the one element set containing ${\emptyset}$. – William Sep 10 '12 at 07:18
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    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. – Brian M. Scott Sep 10 '12 at 07:18

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As noted in the comments, $P(\emptyset)=\{{\emptyset\}}$.

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

Gerry Myerson
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