1
$\begingroup$

To follow up on an earlier question of mine and in order to improve understanding I would like to ask the following: What is the power set of $ \{\{\emptyset\}\} $?

Is it $ \mathscr P(\{\{\emptyset\}\})= \{\emptyset,\{\{\emptyset\}\}\} $?

$\endgroup$
  • 1
    $\begingroup$ yes. this is one of the most important results in mathematics (kidding). $\endgroup$ – Elliot G Jul 19 '15 at 6:12
  • 4
    $\begingroup$ Yes, of course. Sets $\{\emptyset\}$ and $\{\{\emptyset\}\}$ both have only one element. And $P(\{a\})=\{\emptyset, a\}$. You should not be afraid of nested braces $\endgroup$ – Michael Galuza Jul 19 '15 at 6:12
  • 1
    $\begingroup$ Sometimes to check that you have all the elements in a power set, you can use the fact that if a set has cardinality $n$, then the cardinality of its power set is $2^n$. This tells you that for any set of one element, the power set has only two elements, which have to be the empty set and the set itself, which are always elements of the power set. $\endgroup$ – coldnumber Jul 19 '15 at 6:23
  • 1
    $\begingroup$ @MichaelGaluza Are you missing the braces around the $a$? What I mean is $P(\{a\})=\{\varnothing, \{a\}\}$. $\endgroup$ – coldnumber Jul 19 '15 at 6:28
  • $\begingroup$ @coldnumber, yes, it was a typo. $\endgroup$ – Michael Galuza Jul 19 '15 at 6:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.