Is P(A) ⊆ P(B) a Sufficient Condition for A ⊆ B?

[YamiBustamante]

There's what I have so far.
We assume that P(A) ⊆ P(B). This means that every element x that exists in P(A), also exits in P(B). By definition of a power set, x∈P(A) if x ⊆ A. Therefore, A∈P(A). Since P(A) ⊆ P(B), A∈P(B), meaning all x ⊆ A, x ∈ P(B). Furthermore, B∈P(B), meaning all x ⊆ B, x ∈ P(B). Since x ⊆ A and x ⊆ B and P(A) ⊆ P(B), A ⊆ B.

Is my proof correct?

 

[andrewkirk]

x∈P(A) if x ⊆ A. Therefore, A∈P(A).

You need to put in an extra step to justify this 'therefore'.

I can't make anything of what is written after that. It doesn't seem to prove the required conclusion.

You've shown (after adding the extra step) that A is in the powerset of A.
Can you prove that it's therefore in the powerset of B? (very easy)
If you can prove that, just use the definition of 'powerset' to get to your conclusion.

 

[FactChecker]

Your statements are not organized in the way a proof should be. So it is hard to follow your logic. Start with an arbitrary x in A and show, step-by-step that x is in B:
x∈A
Then what does that say about x and P(A)?
Then what does that say about x and P(B)?
Then what does that say about x and B?

 

[Ssnow]

Sorry but if $x\in A$ then $\{x\} \in \mathcal{P}(A)$ because is a singleton and by original assumption $\{x\}\in \mathcal{P}(B)$ so $\{x\}$ is a subset of $B$ that is (by transitivity of the order $\subseteq$) $x\in B$. This hold for every $x\in A$ so the conclusion. I have lost something?

 

[andrewkirk]

Sorry but if $x\in A$ then $\{x\} \in \mathcal{P}(A)$ because is a singleton

No. The reason $\{x\} \in \mathcal{P}(A)$ is that $\{x\}$ is a subset of $A$. Being a singleton has nothing to do with it.

 

[Ssnow]

ok, I used a bad expression sorry, sure the reason is what you said @andrewkirk ... thks