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

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The discussion centers on whether P(A) ⊆ P(B) is a sufficient condition for A ⊆ B. It is established that if P(A) ⊆ P(B), then every subset of A is also a subset of B, leading to the conclusion that A must be a subset of B. Participants emphasize the need for clarity in the proof structure, suggesting a step-by-step approach to demonstrate the relationship between elements of A and B. The importance of correctly applying the definition of power sets is highlighted, with clarification that being a singleton does not affect membership in P(A). Overall, the conversation aims to refine the proof to ensure logical coherence and correctness.
YamiBustamante
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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?
 
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YamiBustamante said:
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.
 
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?
 
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?
 
Ssnow said:
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.
 
ok, I used a bad expression sorry, sure the reason is what you said @andrewkirk ... thks
 
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