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I need some help understanding this concept a little better. I understand the general power sets, but only worked nice and easy examples where only the set consisted of only single elements like $\{a,b,c\}$. I can do that power set, but I am confused on the power set of a set containing a set. The problem I am trying to work is $\{a,\{a,b\}\}$. Typically when you have a set $\{x,y\}$ the power set is $\{\{x\},\{y\},\{x,y\}\}$ so would this problem be similar to that where $\{y\}=\{a,b\}$? Or would I then need to perform a second power set on the inner set?

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  • $\begingroup$ What are you trying to prove? There's nothing stopping you from defining a set that itself contains single elements, as well as sets, as long as it doesn't violate the ZFC axioms. What's the problem? $\endgroup$
    – Mnifldz
    Commented Jan 29, 2015 at 21:40

2 Answers 2

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Let $X=\{a,\{a,b\}\}$. The set $X$ has two elements: nothing more, nothing less. Then, $$\mathcal P(X)=\big\{\emptyset,\{a\}, \{\{a,b\}\}, \{a,\{a,b\}\}\big\}$$

If things don't become clear, you can try "encapsulating" the element $\{a,b\}$ calling it $c$, so you can now write $X=\{a,c\}$.

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  • $\begingroup$ I tried encapsulating and that did help out! Thanks a lot! $\endgroup$ Commented Jan 29, 2015 at 23:47
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Actually you are not completely right when you compute the power set of $\{x,y\}$:

The power set of $\{x,y\}$ is $\{\emptyset, \{x\},\{y\},\{x,y\}\}$ i.e the empty set is always included as it is a subset of any set.

If you have the set $A = \{a, \{a,b\}\}$ and want to compute the power set, we need to find all subsets of $\{a,\{a,b\}\}$.

The power set must have $2^2$ elements as there are 2 elements of the set.

What are these elements?

$\emptyset$ is always an element, so we include that one

$\{a\}$ is another subset as all its elements are in $A$.

Same reasoning applies to $\{\{a,b\}\}$ as $\{a,b\}$ is an element in $A$.

Finally $\{a,\{a,b\}\}$ is a subset as the entire set is always a subset.

Final powerset $\mathcal{P}(X) = \{\emptyset, \{a\}, \{\{a,b\}\},\{a,\{a,b\}\}\}$

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  • $\begingroup$ Okay that makes sense. Yeah after looking back I realized I forgot the empty set, I just don't know the character code for it. Thank you for the help! $\endgroup$ Commented Jan 29, 2015 at 23:49

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