Proving a property of real numbers

[neutrino]

Homework Statement

Given x<y for some real numbers x and y. Prove that there is at least
one real z satisfying x<z<y

Homework Equations

This is an exercise from Apostol's Calculus Vol. 1. The usual laws of
algebra, inequalities, a brief discussion on supremum, infimum and
the Archimedean property preceeded this exercise.

The Attempt at a Solution

I'm trying to solve as many problems from the first chapter, which I had
just glossed over during a first read. Proving theorems in a rigorous
fashion is not exactly my forte. It becomes all the more difficult when
I'm asked to prove something "intuitively obvious."

I think I have the solution in fragments.

Let S be the set of all numbers greater than x. x is a lower bound for
this set, and therefore S has a greatest lower bound - inf S.

inf S \geq x

[The following argument assumes that x = inf S]

Consider some y \in S. I can always find at least one z \in S such that z < y, because if there are no z's satisfying that inequality then y would be inf S.

The only problem here seems to be to prove that inf S = x. The first choice is a proof by contradiction.

Assume inf S > x...

Am I on the right track?

 

[Dick]

How about proving some stuff about (x+y)/2?

 

[neutrino]

How about proving some stuff about (x+y)/2?

Yes, thank you...I really don't know why I was thinking about lower bounds and stuff.

Was my earlier "proof" correct? It will be complete when I prove x = inf S, right?

 

[Dick]

The thing is that I think you might find this property useful for proving x=inf S.

 

[neutrino]

The thing is that I think you might find this property useful for proving x=inf S.

I thought you were suggesting this...

x<(x+y)/2
y>(x+y)/2

Therefore x<(x+y)/2<y - this being a much shorter solution than what I had written down earlier. I'm not sure how (x+y)/2 relates to inf S.

 

[Dick]

Well, x is a lower bound of S. Suppose there is another lower bound y and y>x. Knowing there is a z such that x<z<y gives you a problem since then z is in S. Trying to prove inf S=x leads you to prove things that are an awful lot like what you are trying to prove to begin with.

 

[neutrino]

Ah...
Assume inf S > x. This implies that inf S belongs to S.

But then x<(inf S + x)/2 < inf S and it also belongs to S, which would imply inf S is not the infimum. A contradiction!

But trying to prove that x = inf S this way seems redundant since it relies on the fact that for all real x and y, x<(x+y)/2<y, which was what I set to prove in the first place.

And that's what I think you tried to convey in your last post.

 

[Dick]

Yes. If there were a gap in the real numbers between x and y, then inf S would be y. So you want to show there are no gaps before you prove inf S=x.

 

[neutrino]

Got it now. Thank you.

 

[HallsofIvy]

Well, x is a lower bound of S. Suppose there is another lower bound y and y>x. Knowing there is a z such that x<z<y gives you a problem since then z is in S.

?? No, it isn't. If y is a lower bounds on S, x< y, then x is NOT in s and neither is the z above.

Trying to prove inf S=x leads you to prove things that are an awful lot like what you are trying to prove to begin with.

In fact, "infimum" cannot have anything to do with this property: If x and y are rational numbers then there exist a rational number between them (again (x+y)/2 is a rational number) but bounded sets of rational numbers do not necessarily have rational infimum.

 

[Dick]

?? No, it isn't. If y is a lower bounds on S, x< y, then x is NOT in s and neither is the z above.

I didn't say x in is S. S is defined to be the set of all y such that y>x. I find it difficult to accept that z is not in S if x<z<y.

In fact, "infimum" cannot have anything to do with this property: If x and y are rational numbers then there exist a rational number between them (again (x+y)/2 is a rational number) but bounded sets of rational numbers do not necessarily have rational infimum.

I really don't think you read the thread.