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?


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.

  • $\begingroup$ In more formal situations, using the usual Kuratowski definition of ordered pairs, $( \emptyset , \emptyset ) = \{\,\{ \emptyset \} , \{ \emptyset , \emptyset \}\,\} = \{\,\{ \emptyset \}\,\} \neq \emptyset$. $\endgroup$ – user642796 Sep 3 '14 at 19:36
  • $\begingroup$ However, with the Quine-Rosser definition of ordered pairs, it so happens that $(\varnothing,\varnothing)=\varnothing$. (Mathematical arguments that make a modicum of effort not to be confusing will be constructed such that this doesn't actually make a difference). $\endgroup$ – Henning Makholm Jul 17 '17 at 9:29

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