Proving a is Supremum of Set E

[jaqueh]

Homework Statement

Prove that for every a in ℝ, and the set E = {rεQ : r<a}
the following equality holds:
a = sup(E)

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?

 

[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.

 

[SteveL27]

Homework Statement

Prove that for every a in ℝ, and the set E = {rεQ : r<a}
the following equality holds:
a = sup(E)

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?

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.

 

[jaqueh]

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.

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

 

[jaqueh]

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.

i did not realize that; thanks, thinking now

 

[jaqueh]

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

 

[SteveL27]

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

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?

 

[micromass]

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

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

 

[Dick]

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

 

[jaqueh]

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

 

[micromass]

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..."

 

[jaqueh]

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..."

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

 

[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.

 

[SteveL27]

got it
rough pf-
Suppose all the stuff in the question
i) a is a maximum because of the nature of the set.

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?

 

[jaqueh]

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?

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

 

[micromass]

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

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.

 

[jaqueh]

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.

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

 

[micromass]

Therefore this implies that b is in the set

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.

 

[jaqueh]

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.

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