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I have some home work with problems such as...

Determine whether each of these sets is the power set of a set:

  1. {∅, {a}}

So yes 1 is the power set of {a}.

But what about 2? Since power sets have to have $2^k$ members then no it can't be a power set. It would have to be P(∅) = {∅, {∅}} correct? Since it must contain 2 members. However, when I try to check myself with Wolfram Alpha, it says P({}) = {{}}... why? I thought it must have $2^k$ members!

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3  
WolframAlpha is correct. The power set of the empty set has one element, which is the empty set. –  Qiaochu Yuan Oct 8 '12 at 4:25
9  
$1 =2^0{}{}{}$. –  MJD Oct 8 '12 at 4:25
    
Is WolframAlpha ever wrong!? –  sidht Oct 8 '12 at 4:25
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If you have to ask if W|A is ever wrong, you haven't been using it enough! –  anon Oct 8 '12 at 4:26
    
The first option is the power set of $\{ a \}$, not $a$. If $a$ is a set it may have a much larger power set. –  Qiaochu Yuan Oct 8 '12 at 4:37

3 Answers 3

up vote 15 down vote accepted

We want to make a power set, so start with the set bars: $$ P(\emptyset) = \{\text{stuff goes here}\}. $$ What do we fill this set with? As with any other power set, we fill it with all subsets of the big set in question. Here, the big set in question is $\emptyset$, which has exactly one subset (namely, $\emptyset$). So, we write that one subset into the power set to get $$ P(\emptyset) = \{\emptyset\}. $$ Notice the empty set contains $0$ elements, while its power set contains $2^0 = 1$ elements.

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I thought that might be the reasoning! I was correct that 2 is not a power set but wrong about the reason. Thank you! –  Rice Newman Oct 8 '12 at 4:33
    
Isn't $\{\emptyset\}$ different from $\emptyset$? –  Shashwat Oct 8 '12 at 6:56
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@Shashwat that's the point –  OrangeDog Oct 8 '12 at 9:24

One way you can see that $\emptyset$ cannot be a power set is that by Cantor's theorem the power set of any set $A$ has strictly greater cardinality than $A$. Now which set $A$ could have strictly smaller cardinality than the empty set?

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1  
+1 for an actual proof, rather than answering a different question –  OrangeDog Oct 8 '12 at 9:26

Every powerset must contain at least one member, namely $\emptyset$. The set $\emptyset$ itself contains no members, so it cannot be the powerset of any set.

The powerset of $\emptyset$ is $\{\emptyset\}$, which has exactly $1 = 2^0$ members. It is the smallest powerset, in the strict sense that it is a proper subset of every other powerset, all of which contain at least two members. (Specifically, the powerset of any set $S \ne \emptyset$ has at least $\emptyset$ and $S$ as members.)

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