# Thread: Induction proof: Cardinality of power sets

1. ## Induction proof: Cardinality of power sets

I am trying to prove the following by induction:

The cardinality of the power set of a set with n elements has 2^n elements

Is there a property that says that the cardinality of a power set with x+1 elements is twice the cardinality of a set with x elements? Or how would I prove that this is the case?

Thanks!

2. Originally Posted by economanc
The cardinality of the power set of a set with n elements has 2^n elements
Is there a property that says that the cardinality of a power set with x+1 elements is twice the cardinality of a set with x elements? Or how would I prove that this is the case?
If $\mathcal{P}\left( {\left\{ {1,2,3, \cdots n} \right\}} \right)$ has $2^n$ elements then $\mathcal{P}\left( {\left\{ {1,2,3, \cdots n,n+1} \right\}} \right)$
would have the same subsets as the former plus each one of those united with $\{n+1\}$.
That does double the count.

3. Fair enough, I just wasn't sure if doing it that way would be "rigorous" enough of a proof. It does make sense however.

Thanks again!

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# power set proof by induction

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