Suppose that R is a partial order on A, B1 ⊆ A, B2 ⊆ A, x1 is the least upper bound of B1, and x2 is the least upper bound of B2. Prove that if B1 ⊆ B2, then (x1,x2) ∈ R.
(looking for an informal proof here)
My attempt at solving:
I have been finding this problem somewhat difficult. After assuming B1 ⊆ B2, I have been trying to construct a proof by cases- however, I am not even certain that this is the best way to proceed. The cases I have attempted are x1 ∈ B2 or x1 ∉ B2, but i haven't found a way to complete the proof this way, and so i am beginning to doubt this proof strategy (by cases). As such, I have also attempted a proof by contradiction, but this doesn't seem to make it too far either.
My main questions are:
1) If this IS a proof that should proceed via cases, are there any suggestions as to which cases should be used?
2) If this is not a proof by cases, what method of proof should be used, and in what ways does the problem suggest this?
I have been practicing proofs in set theory for about a year and am now reviewing for a credit by examination test I will be taking this semester by performing some of the problems which occur later in the problem sets in Velleman's text on proofs/set thoery. If you have any insight, I would also like to ask your opinion as to the degree of difficulty you believe this problem is (I'm interested in assessing my mastery of the subject). The proofs which appear in the examples within the text and earlier within problem sets I can easily achieve, but the later ones usually leave me thinking for many hours, and sometimes even a day or two (oftentimes without much progress toward the solution).
Any help is appreciated
Complete Proof of Original post:
Lemma: Suppose that R is a partial order on A, B1 ⊆ A, B2 ⊆ A, x1 is the least upper bound of B1, and x2 is the least upper bound of B2. Prove that if B1 ⊆ B2, x2 is an upper bound of B1.
Proof: Assume B1 ⊆ B2. Let b1 ∈ B1. Then b1 ∈ B2. Thus, (b1, x2) ∈ R, so x2 is an upper bound of B1 (since b1 was arbitrary).
Problem: Suppose that R is a partial order on A, B1 ⊆ A, B2 ⊆ A, x1 is the least upper bound of B1, and x2 is the least upper bound of B2. Prove that if B1 ⊆ B2, (x1, x2) ∈ R.
Proof: By the lemma above, we know x2 is an upper bound of B1. But since x1 is the least upper bound of B1, it follows that (x1, x2) ∈ R.
Big thanks to Plato
Did you read reply #4? That is the meaning of least upper bound.
In this problem we are given three facts to work with.
The last condition means that is also an upper bound for .
By the definition of least upper bound that means that .
At that point the proof is done.
But, from the given it could be that
Also it could be that .
That does not change that proof in any way.
Yes, i understand what you're saying Plato. The text this exercise appears in, however, is rather elementary and prefers (I would guess, since it contains itself such) more introductory/ technical proofs like the one I have attempted (eg. working with the logical forms of the statements and writing out many of the particular details of the derivation).
Your expertise with the subject is most likely leagues ahead of this, though, which may explain why you're scratching your head as to why I'm making a meal out of this.