# Thread: Question with power sets

1. ## Question with power sets

How can I prove or disprove that no set is equal to its power set?
(I'm relatively new to Discrete Mathematics so this may be a very simple question.)

Things I think might help with proving this are:

The definition of a power set: Given a set S, the power set of S is the set of all subsets of the set S.
(I have trouble wrapping my head around what this means)

and

Every set is a subset of itself.

So if a power set is a set, then it contains a subset of itself...does that mean anything for this problem?

Any help is appreciated!

2. ## Re: Question with power sets

Let's start with a small discrete example: $\displaystyle S = \{1,2\}$. Then, the power set of $\displaystyle S$ would be the set $\displaystyle \{\emptyset, \{1\}, \{2\}, \{1,2\}\}$. So, it contains all subsets with zero elements (the empty set), all subsets with 1 element, and all subsets with two elements. How do you show that two sets are equal? You show that the first set is a subset or equal to the second set. Then, you show that the second set is a subset or equal to the first. So, since $\displaystyle S \subseteq S$, we know that $\displaystyle S \in P(S)$ (the power set of S) since $\displaystyle P(S)$ contains all subsets of $\displaystyle S$. But, $\displaystyle S \notin S$ since no set can contain itself (that is one of the axioms of set theory). So, $\displaystyle P(S)$ contains at least one element that $\displaystyle S$ does not contain. Hence, the two sets cannot be equal.

3. ## Re: Question with power sets

Ah, that makes perfect sense! Thank you so much

4. ## Re: Question with power sets

Originally Posted by neojac93
How can I prove or disprove that no set is equal to its power set?
(I'm relatively new to Discrete Mathematics so this may be a very simple question.
In think that that you mean equinumerous and not equal. This is Cantor's theorem.

Suppose $\displaystyle S$ is a set and there is a bijection $\displaystyle f:S\leftrightarrow\mathcal{P}(S)$
Because the function is onto if $\displaystyle G\in\mathcal{P}(S)$ then if $\displaystyle \exists x\in S$ such that $\displaystyle f(x)=G$.

What if $\displaystyle A=\{x\in S:x\notin f(x)\}$. Now because $\displaystyle A\in \mathcal{P}(S)$ then $\displaystyle \exists t\in S$ such that $\displaystyle f(t)=A$.

Question does $\displaystyle t\in f(t)=A~?~?~?$