Please help: Is my proof correct?

[cedar bison]

For plomble #8:

(—>) Suppose A ⊆ B. Suppose C ∈ P(A). Then C ⊆ A. Suppose x ∈ C. Then since C ⊆ A and A ⊆ B,x ∈A and x ∈ B. Thus,since x was arbitrary,C ⊆ B,Therefore C ∈ P(B). But C was an arbitrary element of P(A),So P(A) ⊆ P(B).
(<—) Suppose P(A) ⊆ P(B). Let C ∈ P(A). Then C ∈ P(B). Let x ∈ C. Then x ∈ A and x ∈ B. Thus,since x was arbitrary, A ⊆ B.

I am feeling OK with the (—>) section (But if you find something wrong/handwavy about it,Please do tell me!) Most of my issue lies in the (<—) Section. That part feels really handwavy,if not plain false. Any help is appreciated,thank you

[cedar bison]

I got this part:

Then x ∈ A and x ∈ B.

From the fact that C ∈ P(A) and C ∈ P(B)

Meaning that C ⊆ A and C ⊆ B,by definition of the power set

I also noticed that since the goal was A ⊆ B,I could have right off the bat assumed x in A and proved x in B,but the pieces didn’t fit together very well that way,so I tried to start with x in C first,to be able to use the subset givens

But then I thought it was a bit sketchy going that way,and that it feels like it’s not fully justified

[cedar bison]

<@&286206848099549185>

[prisma tangle]

For the <= direction it is not enough to pick a C in ℘A and x in C

You also need to show that every x in A can be obtained this way, ie. every x in A belongs to C ⊂ A for some C

Here's a quicker alternative proof. Notice that ℘A contains A itself. Then ℘A ⊂ ℘B means A ∈ ℘B, ie. A ⊂ B.

[cedar bison]

That’s alot better than what I thought of,Thank you for this 👍 I’ll go and edit it

Suppose P(A) ⊆ P(B). Notice that P(A) contains A itself. Then P(A) ⊆ P(B) means A ∈ P(B),ie. A ⊆ B.

How’s that?

[prisma tangle]

sure

[cedar bison]

.solved