# Open and closed intervals and real numbers

1. Oct 16, 2010

### reb659

1. The problem statement, all variables and given/known data

Show that:

Let S be a subset of the real numbers such that S is bounded above and below and
if some x and y are in S with x not equal to y, then all numbers between x and y are in S.

then there exist unique numbers a and b in R with a<b such that S is one of the intervals (a,b), [a,b), (a,b], or [a,b].

2. Relevant equations

3. The attempt at a solution

Assume if x and y are elements of S with x not equal to y, then all numbers between x and y are in S and S is bounded above and below.

Thus there exists a M, N such that M is greater than or equal to the maximal element of S and N is smaller than the minimal element of S. Also all elements between x and y are inside (M,N).

Last edited: Oct 16, 2010
2. Oct 16, 2010

### tiny-tim

hi reb659!

i think i'd start by proving that there must be a greatest lower bound and a least upper bound, and then call them a and b, and carry on from there.

3. Oct 16, 2010

### reb659

Good idea.

Isn't it an axiom that if a nonempty subset of R has an upper bound, then it has a least upper bound/sup(S)?

4. Oct 16, 2010

### tiny-tim

I don't think it's an axiom, I think it's the Dedekind-completeness theorem:

A bounded real-valued function has a least upper bound and a greatest lower bound.

(See Rolle's theorem in the PF Library )

5. Oct 17, 2010

### reb659

So far:

Since S is bounded above and below, by Dedekind completeness there exists a supremum of S. Call it b. Again by dedekind completeness we can say there exists an infimum of S. Call it b. By definition of sup and inf, a<b. We are left to show a,b are unique and that S is exactly one of the intervals in the OP.

To show this, can't we consider four simple different cases in which a,b are either in S or outside of it?

6. Oct 18, 2010

### tiny-tim

Yup, that's the proof!

7. Oct 19, 2010

### reb659

Yay!

How exactly does uniqueness follow though? It seems like its trivial to prove.

8. Oct 19, 2010

### tiny-tim

as you said, "by Dedekind completeness there exists a supremum of S" …

there can't be two supremums, can there?