Proving Set Inclusion: P(A) ⊆ P(B) Implies A ⊆ B

AI Thread Summary
To prove that P(A) ⊆ P(B) implies A ⊆ B, start by assuming P(A) is a subset of P(B). For any element x in A, the singleton set {x} is an element of P(A). Since P(A) ⊆ P(B), it follows that {x} must also be in P(B), meaning {x} is a subset of B. Consequently, this implies that x must be an element of B, establishing that A is indeed a subset of B. The proof hinges on the relationship between elements and their corresponding singleton sets in the power sets.
LaMantequilla
Messages
8
Reaction score
0

Homework Statement



Prove that if P(A) \subseteq P(B) then A \subseteq B,
where A and B are two sets and P symbolizes the power set (set of all subsets) of a particular set.

Homework Equations





The Attempt at a Solution


Okay, so here goes.

Because it's a conditional, we suppose P(A)\subseteq P(B), and make it a "given."

From there, we look at the goal ( A\in B ), and let x be arbitrary such that x \in A \rightarrow x \in B. Because x is arbitrary, we suppose x \in A.

So far, we have:

Givens:
P(A) is a subset of P(B), or \forally( y \in P(A) \rightarrow y \in P(B)
x \in A

Goals:
x \in B

So this is where it falls apart. Looking at the given above, I see the opportunity for universal instantiation. However, in order to do that I need to know some variable that y \in P(A), or that y \subseteq A. I see neither. Can you help me?
 
Physics news on Phys.org
If x is an element of A then the set {x} is in P(A). Does that help?
 
Thanks!
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
Back
Top