Discrete Math Proving some power sets

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Homework Help Overview

The discussion revolves around proving a statement in discrete mathematics regarding power sets. The original poster seeks to establish that if the power set of set A is a subset of the power set of set B, then set A must be a subset of set B, using proof by contradiction.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • The original poster attempts to understand how to start the proof and is unsure about the definitions of power sets. Some participants suggest a contradiction approach, while others question the assumptions made in the proof.

Discussion Status

Participants are actively engaging with the proof structure, discussing the validity of assumptions and the logic behind the contradiction method. There is a recognition of the need to clarify the assumptions involved in the proof, particularly regarding the negation of the original statement.

Contextual Notes

There is a focus on the definitions of power sets and the implications of the subset relationship, with participants exploring the nuances of proof by contradiction. The original poster expresses confusion and seeks guidance on how to proceed with the proof.

Servo888
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Ok so I need to prove (by contradiction) that... if the power set(A) is a subset of power set(B), then A is a subset of B.

I was given a hint to use proof by contradiction, but in general I'm lost as to what to do... I know the power set of (A) is {B|B subset A} and the powerset of (B) is {A|A subset B} but that's about it. As far as where to start; I'm drawing a blank. I need a point in the right direction of where to start, because as it stands now I've spent 2 hours looking at this problem.
 
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Ok I think I got it...

My proof is this; Assume that if P(A) is a subset of P(B), then A is not a subset of B (contradiction method). I then show that by definition of a power set; power set B must contain all sets that are in power set A. That would mean that powerset B must contain the set A, there for A is a subset of B (since P(B) contains the set A). Which contradicts the assumption.

Not sure if it's complete, but I think the logic is right (makes sense).
 
Unnecessary contradiction. A is an element of P(A), which is contained in P(B), ie A is an element of P(B), hence A is...
 
matt grime said:
Unnecessary contradiction. A is an element of P(A), which is contained in P(B), ie A is an element of P(B), hence A is...
a subset of B.
 
Assume that if P(A) is a subset of P(B), then A is not a subset of B
Also if you're doing it by contradiction, this is not the assumption you want to make. The negation of the statement "if P(A) is a subset of P(B), then A is not a subset of B" is "P(A) is a subset of P(B) and A is a subset of B," which you can't prove anyway. What you mean is "Assume P(A) is a subset of P(B) _and_ A is not a subset of B."
 
0rthodontist said:
Also if you're doing it by contradiction, this is not the assumption you want to make. The negation of the statement "if P(A) is a subset of P(B), then A is not a subset of B" is "P(A) is a subset of P(B) and A is a subset of B," which you can't prove anyway. What you mean is "Assume P(A) is a subset of P(B) _and_ A is not a subset of B."

Ah I understand what you are saying, so it's just assuming without implication (sort of). We are assuming P(A) is a subset of P(B) and we assume A is not a subset of B. That makes sense.
 

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