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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.

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  • $\begingroup$ 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
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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.

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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.

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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....]

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  • $\begingroup$ 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
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    $\begingroup$ Gerry, \varnothing or \emptyset. (\not0? really?) $\endgroup$ – Asaf Karagila Sep 27 '13 at 5:47
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    $\begingroup$ @Asaf, you grew up speaking TeX, I had to learn it as an adult. $\endgroup$ – Gerry Myerson Sep 27 '13 at 7:24
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$B$ must be $\emptyset$ since $\emptyset \subset A$. But then $\{ 1 \} \subset \aleph_0 \subset A$ and $\{ 1\} \neq \mathcal{P}(A)$, a contradiction.

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