# Cantor's Theorem

• November 27th 2010, 04:38 PM
thamathkid1729
Cantor's Theorem
Prove by induction on n: If a set A has n elements, its power set has 2^n elements.

*Given a set A, the power set of A is the set of all subsets of A
• November 27th 2010, 04:51 PM
Quote:

Originally Posted by thamathkid1729
Prove by induction on n: If a set A has n elements, its power set has 2^n elements.

*Given a set A, the power set of A is the set of all subsets of A

P(k)

The power of a set with k elements is $2^k$

P(k+1)

The power of a set with k+1 elements is $2^{k+1}$

Show that P(k+1) must be true if P(k) is true.

If a set of k elements has a power of $2^k$

then an added element will form $2^k$ new subsets as it can be placed with all $2^k$ existing subsets.

Therefore a set of k+1 elements has a power of $(2)2^k$ if a set of k elements has power $2^k.$

Finally prove for the base case.
• November 27th 2010, 05:47 PM
Plato
This proof turns on a simple fact: $2^n+2^n=2(2^n)=2^{n+1}.$

Any subset of $\{1,2,\cdots,n\}$ is a subset of $\{1,2,\cdots,n,n+1\}.$

So by uniting any subset of $\{1,2,\cdots,n\}$ with $\{n+1}\}$ we get a subset of $\{1,2,\cdots,n+1\}$.