4
$\begingroup$

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?

$\endgroup$
0

1 Answer 1

Reset to default
17
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ In more formal situations, using the usual Kuratowski definition of ordered pairs, $( \emptyset , \emptyset ) = \{\,\{ \emptyset \} , \{ \emptyset , \emptyset \}\,\} = \{\,\{ \emptyset \}\,\} \neq \emptyset$. $\endgroup$
    – user642796
    Commented Sep 3, 2014 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). $\endgroup$
    – hmakholm left over Monica
    Commented Jul 17, 2017 at 9:29

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .