# Supremum proof

1. Apr 23, 2012

### jaqueh

1. The problem statement, all variables and given/known data
Prove that for every a in ℝ, and the set E = {rεQ : r<a}
the following equality holds:
a = sup(E)

3. The attempt at a solution

I'm not sure where to go? should i do this by contradiction that a≠sup(e) or should i do a traditional supremum proof or should i even do an epsilon proof?

I have let x be an element of R then E is the set {r element Q : r<x}. MY QUESTION is how do I say that x is the greatest number of the set?
do i do:
case 1: y is an element of E such that y=x, then x is a rational
case 2: y is an element of E such that y<x, then x is irrational?

Last edited: Apr 23, 2012
2. Apr 23, 2012

### Dick

You'll want to use that the rationals Q are dense in ℝ. Given any two elements a and b in ℝ there is a rational between them. Try a proof by contradiction.

3. Apr 23, 2012

### SteveL27

Why do you think x is the greatest element of E? It might not be. That's the entire point of the exercise. Review the definition of sup.

Secondly, the best way to approach a problem like this is to start by considering a specific example so you can visualize what they're talking about.

Take a = pi, say. Then what would E be? E is the set of rationals that are less than pi. What are some of them? Well the ones of interest are, for example, 3, 3.1, 3.14, 3.141, etc. You agree that each of those is a rational less than pi, right? So clearly the set E is going to contain a sequence of rationals that approach pi as a limit from below; but pi is NOT an element of E, because E consists only of rationals.

Now see if you can use this example to try to build a proof.

And do review the definition of sup, because that's the point of the exercise. The sup need not be an element of the set it's the sup of. Pi is the sup of the set of rationals less than pi; but pi is not an element of that set.

What you are being asked to prove is that pi is the sup of the set of rationals less than pi. Can you prove that? Then generalize to an arbitrary real.

4. Apr 23, 2012

### jaqueh

i tried contradiction, but the one i had was not a good contradiction. here is a straight forward one:

rough pf - Let the interval (b,c) be an interval where b is in the set E but not c. Then since the interval is a subset of ℝ, there exists either a rational number or irrational number such that b<n<c that is the last element of the set E. Therefore, since a is an element of ℝ, n=a such that a is the last element of E. Thus, sup(E)=a

5. Apr 23, 2012

### jaqueh

i did not realize that; thanks, thinking now

6. Apr 23, 2012

### jaqueh

MOMENT OF BRILLIANCE, i make a an element of the rationals and proceed my proof from there

7. Apr 23, 2012

### SteveL27

Well it's a moment of convenience! But unfortunately it's not quite good enough, because a is allowed to be an arbitrary real.

So you can start by letting a be rational, and see if you can get the proof to work. But after that you're still going to have to let a be irrational and see if you can get that to work too.

What did you think of the pi example? Can you prove that pi is the sup of the set of rationals less than pi?

8. Apr 23, 2012

### micromass

Staff Emeritus
But a is not (necessarily) an element of the rationals...

9. Apr 23, 2012

### Dick

Suppose a=sqrt(2) or pi? What are you thinking?

10. Apr 23, 2012

### jaqueh

well i would do it by contradiction, but i realized im supposing ~P->Q instead of P->~Q

11. Apr 23, 2012

### micromass

Staff Emeritus
To prove that a=sup(E), I would always prove it in two steps:

1) a is an upper bound of E. That is, for all x in E, we have that x≤a
2) a is the least upper bound of E. That is, if b is an other upper bound (that is: if for all x in E, we have x≤b), then we have that a≤b

Can you start by proving (1)?? This is quite easy.

Prove (2) by contradiction: "Assume b<a, then..."

12. Apr 23, 2012

### jaqueh

ok i will try that, i know the traditional way of proving a supremum, just this one throws me for a loop because its rationals and irrationals

13. Apr 23, 2012

### jaqueh

got it
rough pf-
Suppose all the stuff in the question
i) a is a maximum because of the nature of the set.
ii) suppose not, suppose there's a 'b' such that sup(e)=b so b<a. Then the interval b<a is dense so that b<q<a for some rational q. Therefore this implies that b is in the set and that b isn't greater than or equal to all the elements of the set E. Thus if sup(e)=b then sup(e)=b=a.

14. Apr 24, 2012

### SteveL27

Is pi a member of the set of all rationals less than pi?

Did you miss the part where I explained above that the sup need not be a member of the set it's the sup of?

15. Apr 24, 2012

### jaqueh

no i understood the pi example, but a is defined to be greater than all elements of the set in the definition. i guess the only thing that's important is that it's the lowest

16. Apr 24, 2012

### micromass

Staff Emeritus
What Steve is getting at is that a is not the MAXIMUM of the set. By definition, the maximum is contained in the set.
Rather, you want to say that a is the UPPER BOUND of the set.

17. Apr 24, 2012

### jaqueh

oh ok, thanks for pointing that out i would have gotten that wrong on the assignment :tongue2:

18. Apr 24, 2012

### micromass

Staff Emeritus
This is also false (and unnecessary to the proof). Indeed, b is not necessarily rational, so it doesn't need to lie in the set.

19. Apr 24, 2012

### jaqueh

yeah i realized that, what i have on my paper is just that q is in the set which implies b is not a sup