Can someone please help me prove this:

Show that |P(X)| = 2 ^ (|X|) for all finite sets X

Any help would be greatly appreciated!! Thank you!

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- Jan 6th 2010, 04:18 PMpseudonymcardinality of finite sets
Can someone please help me prove this:

Show that |P(X)| = 2 ^ (|X|) for all finite sets X

Any help would be greatly appreciated!! Thank you! - Jan 6th 2010, 04:39 PMtonio
- Jan 6th 2010, 04:46 PMDrexel28
Let $\displaystyle f(n)$ denote the number of subsets of $\displaystyle \left\{1\cdots,n\right\}$. Consider the power set of $\displaystyle \left\{1,\cdots,n,n+1\right\}$. Partition set into two blocks $\displaystyle B_1=\left\{A\in\wp\left(\left\{1,\cdots,n+1\right\ }\right):n+1\in A\right\}$ and $\displaystyle \wp\left(\left\{1,\cdots,n+1\right\}\right)-B_1=B_2$. Clearly, we have that $\displaystyle \left|B_2\right|=f(n)$. And, with the exception of $\displaystyle \{n+1\}$ we may (and can) only put $\displaystyle \{n+1\}$ into each of the subsets of in $\displaystyle B_1$. Thus, we have that $\displaystyle f\left(n+1\right)=f(n)+f(n)+1$...almost...we double counted one thing (what is it??). So $\displaystyle f(n+1)=f(n)+f(n)+1-1=2f(n)$. Thus, it follows by induction that the recurrence relation $\displaystyle f(n+1)=2f(n)$ has solution $\displaystyle f(n)=C\cdot 2^n$ for some $\displaystyle C$. Noting though that $\displaystyle f(0)$ is the number of subsets of the empty set we may conclude that $\displaystyle f(0)=1=C\cdot 2^0=C$. The conclusion follows.

EDIT: I am sorry**tonio**. I didn't see your post! - Jan 6th 2010, 06:36 PMBruno J.
Why so complicated? Just notice that if you want to construct a subset $\displaystyle S$ of $\displaystyle X$, you have two possibilities for every element of $\displaystyle X$: either you include it, or you exclude it. So in total you have $\displaystyle 2^{|X|}$ possibilities.

Another proof : clearly the number of subsets of $\displaystyle X$ is

$\displaystyle {n \choose 1} + {n \choose 2} + \dots + {n \choose n} = (1+1)^n = 2^n$ - Jan 6th 2010, 06:39 PMDrexel28
- Jan 6th 2010, 06:43 PMBruno J.
Which of the two proofs is exactly like yours?

- Jan 6th 2010, 06:45 PMDrexel28
- Jan 6th 2010, 06:50 PMBruno J.
Your proof relies on induction whereas my proof relies on a combinatorial argument... I don't think they're similar at all! (Giggle)

- Jan 6th 2010, 06:53 PMDrexel28
- Jan 6th 2010, 07:03 PMBruno J.
I'm just teasing you at this point. (Angel)

- Jan 6th 2010, 07:08 PMDrexel28
- Jan 6th 2010, 07:47 PMtonio

The "another proof" is not so unless you first prove that there are $\displaystyle \binom{n}{k}$ subsets with k elements out of a set with n elements, $\displaystyle k\le n$, and also you forgot the number $\displaystyle \binom{n}{0}$ in the sum and also, and perhaps most important, this another proof relies on "hidden induction": after all, we need to show that's true for all n, just like the formal proof of Newton's binomial theorem is usually, and as far as I am aware uniquely, done by induction.

Tonio