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So I'm no Mathematician, but I do need some help. Is there any way to make a power set that contains a set containing a set containing some element? The form that I'm looking for is: $\{\{b\}\}$

Doing something like: $\mathcal{P}(\{b\})$ gives me: $\{\{\},\{b\}\}$ which is close.

Doing something like: $\mathcal{P}(\{\{b\}\})$ gives me: $\{\{\}, (b)\}$ which I do not understand.

The full context of the problem is: I'm looking for a the smallest set whose power set contains the collection of sets given. The collection of sets I'm working on looks like: $\{\{a\},\{\{b\}\},\{a,b\}\}$. I've tried $\mathcal{P}(\{a,b\})$ and $\mathcal{P}(\{a,\{b\}\})$ and $\mathcal{P}(\{a,b,\{\}\})$ but I can't see how to get $\{\{b\}\}$. Any help would be appreciated!


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The elements of the powerset of the powerset of $\{b\}$ are the subsets of the powerset of $\{b\}$, hence sets of subsets of$\{b\}$. Since $\{b\}$ is among the subsets of $\{b\}$, $\{\{b\}\}$ is an element of th epowerset f the powerset o f$\{b\}$. – Hagen von Eitzen Oct 5 '12 at 15:42
When you say "contains", do you mean that {{a},{{b}},{a,b}} must be an element of the power set, or that it must be a subset of the power set? – Henning Makholm Oct 5 '12 at 15:49
it must be a subset of the powerset – atb Oct 5 '12 at 15:53
I don't like answering a question, and then realizing that the question was completely different. Try and provide a complete context the next time, it could save a lot of time for someone. – Asaf Karagila Oct 5 '12 at 16:01
up vote 2 down vote accepted

There is no set $S$ whose power set is $\{\{b\}\}$ for any $b$, because the power set of $S$ will always have the empty set as element (because the empty set is subset of every set), but $\{\{b\}\}$ doesn't contain the empty set.

For any set of sets, the smallest set $S$ whose power set has as elements at least those sets is the union of those sets. This is easily seen as follows:

Since $P(S)$ must contain each of the sets, each of the sets must be subsets of $S$ (because $M\in P(S)$ iff $M\subset S$). Therefore the union of all those sets must be a subset of $S$. And of course the smallest superset of the union of those sets is the union of those sets itself.

In your special case, the union of $\{\{a\},\{\{b\}\},\{a,b\}\}$ is $S=\{a,b,\{b\}\}$. The power set of that set has $8$ elements, three of which are the sets you listed.

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Thank you. I was wondering if there was any form or method to solving this. So if I have another problem with a different collection of sets, the union of all the sets in the collection should give me the smallest set S such that the collection is a subset of P(S)? – atb Oct 5 '12 at 15:57
@alexthebake: Exactly. That's what I wrote in the second paragraph and proved in the third. – celtschk Oct 5 '12 at 16:09

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