# Prove that an upper bound a is the least upper bound

• chipotleaway
In summary: I guess you could argue that if a is not in A, then there must be an 'infinite' number of real numbers between a and sup(A)?
chipotleaway

## Homework Statement

Let A be a non-empty subset of R (real numbers) and a an upper bound in R for A. Suppose that every open interval I containing a intersects A (so the intersection is non-empty). Show that a is a least upper bound for A.

## The Attempt at a Solution

I've seen the prettier proof which involved a contradiction, but just wanted to know if my rather non-rigorous argument makes sense, and if it can be made more precise:

Let b=sup(A). Let an interval I=(x,y) satisfy the above (contains a, intersects A). If we take the limit as y approaches b, then a must approach b=sup(A) and so since the limit of this I is just another open interval, and a=sup(A) in this limit, then a=sup(A).

chipotleaway said:

## Homework Statement

Let A be a non-empty subset of R (real numbers) and a an upper bound in R for A. Suppose that every open interval I containing a intersects A (so the intersection is non-empty). Show that a is a least upper bound for A.

## The Attempt at a Solution

I've seen the prettier proof which involved a contradiction, but just wanted to know if my rather non-rigorous argument makes sense, and if it can be made more precise:

Let b=sup(A). Let an interval I=(x,y) satisfy the above (contains a, intersects A). If we take the limit as y approaches b, then a must approach b=sup(A) and so since the limit of this I is just another open interval, and a=sup(A) in this limit, then a=sup(A).

I would suggest you stick with the pretty proof. 'a' is supposed to be a fixed upper bound. Talking about it 'approaching' something really doesn't make any sense.

1 person
chipotleaway said:

## Homework Statement

Let A be a non-empty subset of R (real numbers) and a an upper bound in R for A. Suppose that every open interval I containing a intersects A (so the intersection is non-empty). Show that a is a least upper bound for A.

## The Attempt at a Solution

I've seen the prettier proof which involved a contradiction, but just wanted to know if my rather non-rigorous argument makes sense, and if it can be made more precise:

Let b=sup(A). Let an interval I=(x,y) satisfy the above (contains a, intersects A). If we take the limit as y approaches b, then a must approach b=sup(A) and so since the limit of this I is just another open interval, and a=sup(A) in this limit, then a=sup(A).

I don't think you can take the limit as y approaches b because how do you know that (x, y) would still contain a? what if b < a and then as you take y closer and closer to b, there's a point where the interval doesn't contain a anymore.

1 person
Yeah I guess I assumed a to be in the interval not matter how you took I, as long as it intersected with A (so not some fixed a).

Anyway, is this correct? I didn't write down it down but I remember the idea went like this:
Suppose a is not the least upper bound of A, then we can construct an open interval I around it such that the intersection of I and A is the empty set, but that contradicts the assumption therefore a must be the least upper bound.

That's the right idea, but I'd want to see how you construct the interval I (or why it exists).

1 person
And I'd want you to say exactly why the existence of the open interval contradicts a being a least upper bound.

1 person
@micromass: Hmm...I'm not sure of a rigorous justification, but I would say because if there are an 'infinite' number of real numbers between a and sup(A) if a is not sup(A), it's possible to construct an open interval around it. *

@Dick: Contradicts a being a least upper bound? Do you mean a not being a least upper bound? It contradicts the assumption because we've assumed that every open interval I containing also intersects with A. Assuming a is not sup(A), then the construction of an open interval I around it such that I does not intersect with A leads to a contradiction, therefore a must be sup(A).

*I've also been thinking about how to show that if a=sup(A), then it's not possible to have an open interval I containing a, NOT intersect with A (i.e. if it contains a=sup(A), it must intersect with A).
I think the case if a is in A is 'trivial' as they say, but I'm unsure as to how to make rigorous the argument for the case where a is not in A.

chipotleaway said:
@micromass: Hmm...I'm not sure of a rigorous justification, but I would say because if there are an 'infinite' number of real numbers between a and sup(A) if a is not sup(A), it's possible to construct an open interval around it. *

If ##\textrm{sup}(A)<a##, then think about an interval centered at ##a## with radius ##(a-\textrm{sup}(A))/2##.

*I've also been thinking about how to show that if a=sup(A), then it's not possible to have an open interval I containing a, NOT intersect with A (i.e. if it contains a=sup(A), it must intersect with A).
I think the case if a is in A is 'trivial' as they say, but I'm unsure as to how to make rigorous the argument for the case where a is not in A.

That's not really needed for the proof here. But you should proceed by contradiction. Assume there is an open interval ##(a-\varepsilon,a+\varepsilon)## that contains no elements of ##A##. Prove that ##a-\varepsilon## is an upper bound.

1 person
micromass said:
If ##\textrm{sup}(A)<a##, then think about an interval centered at ##a## with radius ##(a-\textrm{sup}(A))/2##.

Hmm...here's an attempt:

If $a>sup(A)$, then $a-sup(A)>0$. Let the the interval I be, for some ε>0 $I=(a-\frac{a-sup(A)}{2}, a+\epsilon)=(\frac{a+sup(A)}{2} a+\epsilon$.

micromass said:
That's not really needed for the proof here. But you should proceed by contradiction. Assume there is an open interval ##(a-\varepsilon,a+\varepsilon)## that contains no elements of ##A##. Prove that ##a-\varepsilon## is an upper bound.

Yeah, it was just out of curiosity :p
Still working on it though, I think I've got the idea but just refining it.

chipotleaway said:
Hmm...here's an attempt:

If $a>sup(A)$, then $a-sup(A)>0$. Let the the interval I be, for some ε>0 $I=(a-\frac{a-sup(A)}{2}, a+\epsilon)=(\frac{a+sup(A)}{2} a+\epsilon$.

It's probably obvious to you, but you also need to prove this doesn't intersect ##A##.

1 person
micromass said:
It's probably obvious to you, but you also need to prove this doesn't intersect ##A##.

Sorry for the delayed response, had - still have lots assessable work to do (procrastinating on that!).

Anyway, is it correct to say that since $inf(I)=\frac{a+sup(A)}{2}>sup(A)$, then I doesn't intersect with A?

And for the other part (show if a=sup(A) is in an open interval I, I must intersect with A).

Suppose a=sup(A) and that there exists an open interval I containing sup(A) that does not intersect with A. Let I=(sup(A)-ε, sup(A)+ε). Since the intersection of I and A is the null set, this implies sup(A)-ε is an upper bound for A which contradicts a=sup(A).

Or I think I could explain further and get a different contradiction: I not intersecting A implies sup(A)-ε is an upperbound because if it weren't, then there would be some element from A greater than sup(A)-ε and less than sup(A), i.e. in I, which would contradict the assumption that I does not intersect A.

Seems ok to me.

1 person

## 1. What is an upper bound?

An upper bound is a value that is greater than or equal to all elements in a set.

## 2. What does it mean for an upper bound to be the least upper bound?

The least upper bound is the smallest possible upper bound for a set. It is the smallest value that is greater than or equal to all elements in the set.

## 3. How do you prove that an upper bound a is the least upper bound?

To prove that an upper bound a is the least upper bound, you must demonstrate that a is an upper bound and that there is no smaller upper bound for the same set.

## 4. Can there be more than one least upper bound for a set?

No, there can only be one least upper bound for a set. If there were multiple least upper bounds, they would all have to be the same value.

## 5. Why is it important to prove that an upper bound is the least upper bound?

Proving that an upper bound is the least upper bound is important because it ensures that the value is the smallest possible upper bound for a set. This can be useful in a variety of mathematical and scientific contexts, such as optimization problems or analysis of data sets.

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