Least upper bound proof (again)

In summary, the conversation discusses a question about proving that every element of a set is a lower bound for the set of all upper bounds, and how to prove that the greatest lower bound is also the least upper bound. The conversation also addresses confusion about the logical form of the antecedent and how to formalize it.
  • #1
Syrus
214
0

Homework Statement



Okay, this is essentially the same question I had in an earlier thread, but i am trying to make my questions and uncertainties more clear for more accurate assistance:


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



My trouble is in proving part b). Also, I feel part a)- which i have proven below- is used in the part b) proof, I just can't seem to piece the two together appropriately.

The Attempt at a Solution



Here is my proof for part a)

Let b ∈ B. Let u ∈ U. Then by the definition of upper bound, (b,u) ∈ R. Since u was arbitrary, b is a lower bound of U. Since b was arbitrary, every element of B is a lower bound of U.

To prove part b) obviously you assume the antecedent of the statement to be proven. That is, x is the greatest lower bound of U. You can also assume the result obtained in part a). Now our goal is x is the least upper bound of B. Logically this translates to for all b in B, (b,x) is an element of R. Also, for all u in U, (x,u) is an element of R. So I am trying to prove these separately.

My confusion begins here:

Does the antecedent mean x is a lower bound of U AND x is the greatest element of the set of lower bounds of U (which, as far as i can come up with, is A\U)?

How do you show x is an upper bound of B? I feel this is a start to the rest of the proof.
 
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  • #2
Syrus said:

Homework Statement



Okay, this is essentially the same question I had in an earlier thread, but i am trying to make my questions and uncertainties more clear for more accurate assistance: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



My trouble is in proving part b). Also, I feel part a)- which i have proven below- is used in the part b) proof, I just can't seem to piece the two together appropriately.

The Attempt at a Solution



Here is my proof for part a)

Let b ∈ B. Let u ∈ U. Then by the definition of upper bound, (b,u) ∈ R. Since u was arbitrary, b is a lower bound of U. Since b was arbitrary, every element of B is a lower bound of U.

To prove part b) obviously you assume the antecedent of the statement to be proven. That is, x is the greatest lower bound of U. You can also assume the result obtained in part a). Now our goal is x is the least upper bound of B. Logically this translates to for all b in B, (b,x) is an element of R. Also, for all u in U, (x,u) is an element of R. So I am trying to prove these separately.

So far so good.

My confusion begins here:

Does the antecedent mean x is a lower bound of U AND x is the greatest element of the set of lower bounds of U

Correct.

(which, as far as i can come up with, is A\U)?

This is not true if you're only working with partial orders. Let A=B=[0,1] with the usual ordering and adjoin an element y that has no relationship with elements of A. Then U={1} is the only upper bound of A. But y is not a lower bound of U.

How do you show x is an upper bound of B? I feel this is a start to the rest of the proof.

Take [itex]b\in B[/itex]. We must prove that [itex]b\leq x[/itex]. But b is a lower bound of U and x is the greatest lower bound.
 
  • #3
I guess this was my "intuitive" understanding of the theorem. The only reason I can't seem to formalize it is that translating the antecedent into its logical form poses a problem:

Since x is the greatest lower bound of U, x is a lower bound of U AND x is the greatest element of the set of lower bounds of U. The first conjunct can be expressed by: for all u in U, (x,u) is an element of R. But for the second conjunct... for all z in (what set then?) (x,z) is an element of (what set)?

Is that making sense?
 
  • #4
Syrus said:
I guess this was my "intuitive" understanding of the theorem. The only reason I can't seem to formalize it is that translating the antecedent into its logical form poses a problem:

Since x is the greatest lower bound of U, x is a lower bound of U AND x is the greatest element of the set of lower bounds of U. The first conjunct can be expressed by: for all u in U, (x,u) is an element of R. But for the second conjunct... for all z in (what set then?) (x,z) is an element of (what set)?

Is that making sense?

The second could be formalized as follows
if for all u in U it holds that (y,u) in R, then (y,x) is in R.
 

FAQ: Least upper bound proof (again)

What is a least upper bound?

A least upper bound is the smallest number that is greater than or equal to all the numbers in a given set. It is also known as the supremum or the least upper bound.

Why is a least upper bound important in mathematics?

A least upper bound is important because it helps to prove that a set has a maximum value and it is useful in many mathematical proofs, particularly in real analysis and topology.

How do you prove the existence of a least upper bound?

The existence of a least upper bound can be proven using the least upper bound property, which states that any non-empty set of real numbers that is bounded above must have a least upper bound. This can be proven using the completeness axiom of the real numbers.

Can a set have more than one least upper bound?

No, a set can only have one least upper bound. If a set has more than one number that satisfies the definition of a least upper bound, then those numbers must be equal to each other.

How is the least upper bound property different from the supremum?

The least upper bound property and the supremum are essentially the same concept, but they are often used in different contexts. The least upper bound property is used to prove the existence of a least upper bound, while the supremum is the actual least upper bound of a given set.

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