4
$\begingroup$

I am stuck on a problem in my discrete mathematics textbook at the moment. The problem, as written in the textbook, is:

For a certain set $A$, the power set of $A$ is $\mathcal{P}(A) = \{\aleph_0, \{0\}, B\}$, where $B$ is a set. What is $A$?

My confusion here is that I was under the impression that for any set, let's say $D$, that $|\mathcal{P}(D)| = 2^n$. If this is the case, I don't really understand how the power set of $A$ from the problem above can contain only three elements. If set $A$ has two elements, then its power set will have fourelements. If set $A$ has one element, then its power set will have two elements. I know I missing something here, and I appreciate any hints, help, or guidance.

Improve this question
$\endgroup$
  • $\begingroup$ Which textbook, please? Which page? $\endgroup$ – Gerry Myerson Sep 21 '13 at 1:09
  • $\begingroup$ Could this problem be using some weird notation for unions? That's the only explanation I could think of. $\endgroup$ – William Ballinger Sep 21 '13 at 1:10
  • $\begingroup$ DISCRETE MATHEMATICS - Gary Chartrand, Ping Zhang Ch.2 pg 76 $\endgroup$ – SherMM Sep 21 '13 at 1:14
7
$\begingroup$

You are perfectly correct: there is no such set, for precisely the reason that you give. If the power set of $A$ is finite, then so is $A$, and in that case $|\wp(A)|=2^{|A|}$; $3$ is finite and not a finite power of $2$, so $\{\aleph_0,\{0\},B\}$ is not the power set of any set.

Improve this answer
$\endgroup$
5
$\begingroup$

Even worse, if $\aleph_0$ is taken as a representative of the first infinite cardinality (e.g., the usual $\aleph_0=\omega$), then since $\aleph_0\in P(A)$, $\aleph_0\subset A$, so $A$ is infinite. An infinite set cannot have a finite power set.

Improve this answer
$\endgroup$
3
$\begingroup$

I typed, "for a certain set A, the power set of A is" into Google Books, and it seems that where you have written $\aleph_0$, what's actually in the book is $\emptyset$ (thanks, Asaf). Can you see how to do the problem now?

EDIT: OP having confirmed that what was posted as $\aleph_0$ was really meant to be the empty set, we still have to work out how a power set can have 3 elements. Well, it can't have 3 distinct elements, so two of the listed elements have to be the same. Since all we are told about $B$ is that it's a set, we are both able and required to take $B$ to be either $\{\,0\,\}$ or the empty set --- it doesn't matter which. Now the power set has only 2 distinct elements, $A$ has to be $\{\,0\,\}$, and all is right in the world.

[Of course, in the light of the clarification by OP, all the other answers are wrong....]

Improve this answer
$\endgroup$
  • $\begingroup$ Yes, for some reason I typed aleph null instead of the empty set. I still don't understand how to solve the problem, though. From the equation I wrote above, I don't see how a power set could have an odd number of elements. $\endgroup$ – SherMM Sep 27 '13 at 1:40
  • 1
    $\begingroup$ Gerry, \varnothing or \emptyset. (\not0? really?) $\endgroup$ – Asaf Karagila♦ Sep 27 '13 at 5:47
  • 1
    $\begingroup$ @Asaf, you grew up speaking TeX, I had to learn it as an adult. $\endgroup$ – Gerry Myerson Sep 27 '13 at 7:24
1
$\begingroup$

$B$ must be $\emptyset$ since $\emptyset \subset A$. But then $\{ 1 \} \subset \aleph_0 \subset A$ and $\{ 1\} \neq \mathcal{P}(A)$, a contradiction.

Improve this answer
$\endgroup$

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.