
#1
Aug1611, 11:25 PM

P: 202

1. The problem statement, all variables and given/known data
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. 2. Relevant equations 3. 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.... 



#2
Aug1711, 02:30 AM

P: 320

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). 



#3
Aug1711, 09:50 AM

P: 202

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.




#4
Aug1711, 10:16 AM

P: 320

Least upper bound/ greatest lower bound proof 



#5
Aug1711, 11:32 AM

P: 202

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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