Proving Power Set Inclusion: A Simple Proof for A⊆B and P(A)⊆P(B)

[im2fastfouru]

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

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

 

[Kurdt]

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.

 

[im2fastfouru]

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

 

[CRGreathouse]

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

That's probably a good way.

 

[HallsofIvy]

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

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

 

[Office_Shredder]

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

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

 

[im2fastfouru]

what is x? In particular, what set are all of x's elements in?

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!

 

[Office_Shredder]

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!

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