# Power set proof

1. Dec 3, 2007

### im2fastfouru

This seems like a simple proof but i'm not familiar with power set proofs

If A$$\subseteq$$B then P(A) $$\subseteq$$ P(B)

2. Dec 4, 2007

### Kurdt

Staff Emeritus
A good place to start might be with the definitions of a subset and a power set. So the general set A is a subset of another general set B if every element of A is contained in B.

The power set P(A) of a set A is defined as $P(A) = \{X:X\subseteq A\}$, that is the set of all the subsets of A.

3. Dec 4, 2007

### im2fastfouru

i'm more inclined to start with x $$\in$$ P(a), can i start the proof this way?

4. Dec 4, 2007

### CRGreathouse

That's probably a good way.

Last edited: Dec 4, 2007
5. Dec 5, 2007

### HallsofIvy

Staff Emeritus
Why do you say "more inclined"? That was exactly what was suggested.

6. Dec 5, 2007

### Office_Shredder

Staff Emeritus
If $$x \in P(A)$$ what is x? In particular, what set are all of x's elements in?

Last edited: Dec 5, 2007
7. Dec 5, 2007

### im2fastfouru

x is just an arbitrary element. And if A $$\subseteq$$ B then prove P(A) $$\subseteq$$ P(B). This need to be proved formally as well for my assignment!

8. Dec 5, 2007

### Office_Shredder

Staff Emeritus
Sorry, the latex got screwed up. Re-read it now