Prove that an upper bound a is the least upper bound

[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).

 

[Dick]

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.

 

[exclamationmarkX10]

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.

 

[chipotleaway]

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.

 

[micromass]

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

 

[Dick]

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

 

[chipotleaway]

@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.

 

[micromass]

@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.

 

[chipotleaway]

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&gt;sup(A), then a-sup(A)&gt;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.

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.

 

[micromass]

Hmm...here's an attempt:

If a&gt;sup(A), then a-sup(A)&gt;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 $A$.

 

[chipotleaway]

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}&gt;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.

 

[micromass]

Seems ok to me.