Least upper bound/ greatest lower bound proof

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Homework Help Overview

The discussion revolves around a proof involving partial orders, specifically addressing the relationships between upper bounds and lower bounds within a set. The original poster presents two statements to prove: that every element of a subset B is a lower bound for the set of upper bounds U, and that if a certain element x is the greatest lower bound of U, then x is also the least upper bound of B.

Discussion Character

  • Conceptual clarification, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the logical structure of the proofs, particularly the implications of being a lower bound and an upper bound. There is discussion about the necessity of using part a) in the proof of part b) and the challenge of formalizing the logical relationships involved.

Discussion Status

Some participants have provided guidance on the proofs, suggesting that the original poster has made progress but still has uncertainties regarding the completeness of their arguments. There is an ongoing exploration of the implications of lower bounds not being in B and how that affects the proof structure.

Contextual Notes

Participants note potential confusion regarding the definitions and relationships between the sets involved, particularly concerning the existence of lower bounds that may not belong to B. This raises questions about the assumptions made in the problem setup.

Syrus
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Homework Statement


Suppose R is a partial order on A and B ⊆ A. Let U be the set of all upper bounds for B.
a) Prove that every element of B is a lower bound for U.
b) Prove that if x is the greatest lower bound of U, then x is the least upper bound of B.

Homework Equations



The Attempt at a Solution


I have proven a)

Let b ∈ B. Let u ∈ U. Then by the definition of upper bound, (b,u) ∈ R.


For the proof of b), you obviously assume the antecedent of the statement to be shown. I can't, however, seem to make out the logical form of this. Also, it appears you must use part a) in the proof of part b) as well. I have an intuitive understanding of the idea of the statement to be proven, I am just having a hard time (in-)formalizing it.

The logical form of the goal- i would think- is (∀b ∈ B)((b,x) ∈ R) and (∀u ∈ U)((x,u) ∈ R); I assume this means that x is an upper bound of B AND that x is the least upper bound of B (hence, x is the least upper bound of B).

So naturally, my proof thus far is:
Suppose x is the greatest lower bound of U. Let b ∈ B...
 
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Syrus said:

Homework Statement


Suppose R is a partial order on A and B ⊆ A. Let U be the set of all upper bounds for B.
a) Prove that every element of B is a lower bound for U.
b) Prove that if x is the greatest lower bound of U, then x is the least upper bound of B.

Homework Equations



The Attempt at a Solution


I have proven a)

Let b ∈ B. Let u ∈ U. Then by the definition of upper bound, (b,u) ∈ R.


For the proof of b), you obviously assume the antecedent of the statement to be shown. I can't, however, seem to make out the logical form of this. Also, it appears you must use part a) in the proof of part b) as well. I have an intuitive understanding of the idea of the statement to be proven, I am just having a hard time (in-)formalizing it.

The logical form of the goal- i would think- is (∀b ∈ B)((b,x) ∈ R) and (∀u ∈ U)((x,u) ∈ R); I assume this means that x is an upper bound of B AND that x is the least upper bound of B (hence, x is the least upper bound of B).

So naturally, my proof thus far is:
Suppose x is the greatest lower bound of U. Let b ∈ B...

You've proven the part a rightly but It's not complete I guess. to prove a, remember that U is the set of all upper bounds of B which means for any b in B we have: (b,u)∈R (u in U). now let u be any element of U and fix b. that means u is bounded by an element of b. now let b vary, that means every element of B is a lower bound for you and that means B is the set of lower bounds of U.

to prove the part b, It's easy to show if x is the glb of U then x is an upper bound of B. what you need to prove is that x is also the least upper bound. that means there exists no other element like y s.t y is an upper bound of B and (y,x)∈R. It's easy to show that, I leave the proof to you. prove it by reductio ad absurdum (indirect argument).
 
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Coulnd't there also be lower bounds which are not in B? Part of my trouble is specifying a set containing a lower bound of U, since b need not be in B.
 
Syrus said:
Coulnd't there also be lower bounds which are not in B? Part of my trouble is specifying a set containing a lower bound of U, since b need not be in B.

Maybe there are lower bounds of U that are not in B. I guess that might happen when B is left-bounded and I'm not 100% sure about that. but in our case, that doesn't matter. because if m is a lower bound for U that is not in B and we have: \forallb\forallu: b<l<u then that leads us to a contradiction. that's all we need to know to prove that glb(U)=lub(B). I don't think we need anything else to prove the problem.
 
AdrianZ said:
You've proven the part a rightly but It's not complete I guess. to prove a, remember that U is the set of all upper bounds of B which means for any b in B we have: (b,u)∈R (u in U). now let u be any element of U and fix b. that means u is bounded by an element of b. now let b vary, that means every element of B is a lower bound for you and that means B is the set of lower bounds of U.

But then B is not "the set of lower bounds of U" rather, A set of lower bounds of U.

Another thing... i am struggling with how to show that x is an upper bound of B... Even though x is the glb of U doesn't necesserily mean it is an element of U. Still somewhat confused i guess.
 

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