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The empty set is a member of $P(\{a,b\}) \times P(\{p,q\})$. True or false?

My first instinct was false, since the empty set is a member of each power set individually, but when multiplied together, you get $(\emptyset,\emptyset)$, which I'm not sure represents the empty set. But my counter argument is that the empty set is a member of the power set of anything, right?

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up vote 16 down vote accepted

Your first instinct was right: $P(\{a,b\}) \times P(\{p,q\})$ contains $(\emptyset, \emptyset)$, but not $\emptyset$. $(\emptyset, \emptyset)$ is not the empty set, so $P(a,b) \times P(p,q)$ does not contain the empty set.

And yes, every powerset contains the empty set, but $P(a,b) \times P(p,q)$ is not a powerset, it's the cartesian product of two powersets.

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In more formal situations, using the usual Kuratowski definition of ordered pairs, $( \emptyset , \emptyset ) = \{\,\{ \emptyset \} , \{ \emptyset , \emptyset \}\,\} = \{\,\{ \emptyset \}\,\} \neq \emptyset$. –  Arthur Fischer Sep 3 at 19:36

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