Least Upper bound proof

  • Thread starter Syrus
  • Start date
  • #1
214
0

Homework Statement



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.


Homework Equations





The Attempt at a Solution



This problem has been stumping me. After assuming B1 ⊆ B2, I have been trying to construct a proof by cases. I am not even certain that this is the best way to proceed, however. 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). 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 in the problem sets are easily achieved, 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).

Thanks in advance,

Syrus
 

Answers and Replies

  • #2
214
0
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).

Q.E.D.

_________________________________________________________________________________

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.

Q.E.D.
 
Last edited:

Related Threads on Least Upper bound proof

  • Last Post
Replies
3
Views
2K
  • Last Post
Replies
3
Views
2K
  • Last Post
Replies
4
Views
5K
  • Last Post
Replies
4
Views
1K
Replies
4
Views
6K
  • Last Post
Replies
11
Views
5K
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
4
Views
3K
  • Last Post
Replies
11
Views
1K
  • Last Post
Replies
2
Views
2K
Top