Register to reply

Prove that the closure of Q is equal to R

by Tetrinity
Tags: closure, rational numbers, real numbers
Share this thread:
Tetrinity
#1
Feb23-11, 01:34 PM
P: 6
1. The problem statement, all variables and given/known data

As the title states, the problem asks to prove that the closure of the set of rational numbers is equal to the set of real numbers. The problem includes the standard definition of the rationals as {p/q | q ≠ 0, p,q ∈ Z} and also states that the closure of a set X ⊂ R is equal to the set of all its limit points.

Our lecturer suggested the following method:
  • Show that the closure of Q is a subset of R.
  • Show that R is a subset of the closure of Q.
  • Therefore the closure of Q is equal to R

2. Relevant equations

I'm not sure if there's anything particularly useful here. My definition of R states that Q is a subset of R, it is ordered, the standard arithmetic operations work and it is complete (i.e. every bounded set S ⊂ R has a suprenum sup(S) ∈ R and infinum inf(S) ∈ R). It also states that every real number can be defined as a convergent sequence of rationals.

3. The attempt at a solution

I'm going to use the notation cl(X) to represent the closure of the set X.

The first step is pretty easy:

Q ⊂ R
⇒ cl(Q) ⊂ cl(R)
⇒ cl(Q) ⊂ R

(the closure of N, Z and R are already known to us at this point)


My problem lies with the second step. I know that the closure of a set X is the set of all limit points of X, and that a is a limit point to X if a is an adherent point to X\{a}. I'm just not really sure how to begin to prove the second step, and I've racked my brains to the point where I'm probably missing something obvious.

Any help with this, be it using this method or a different one, would be appreciated. Thanks.
Phys.Org News Partner Science news on Phys.org
'Office life' of bacteria may be their weak spot
Lunar explorers will walk at higher speeds than thought
Philips introduces BlueTouch, PulseRelief control for pain relief
micromass
#2
Feb23-11, 01:43 PM
Mentor
micromass's Avatar
P: 18,346
Well, it isn't obvious. Some even think of this as an axiom of R. Thus something that cannot be proven, but that is assumed. So whatever proof you'll find, it will be quite informal. (unless you're working with Dedekind cuts, but I don't assume that this is the case here).

To prove that Q is dense in R, we must show that for every x in R, there exists a sequence [tex](q_n)_n[/tex] in Q, that converges to x.
To find the sequence, take x and take as qn the first n decimals of x. This is a rational numher. For example, pi is the limit of following sequence
[tex]3,~3.1,~3.14,~3.141,~3.1415,...[/tex]

I'll leave it to you to formalize the argument...
Robert1986
#3
Feb23-11, 02:02 PM
P: 828
Quote Quote by micromass View Post
Well, it isn't obvious. Some even think of this as an axiom of R. Thus something that cannot be proven, but that is assumed. So whatever proof you'll find, it will be quite informal. (unless you're working with Dedekind cuts, but I don't assume that this is the case here).

To prove that Q is dense in R, we must show that for every x in R, there exists a sequence [tex](q_n)_n[/tex] in Q, that converges to x.
To find the sequence, take x and take as qn the first n decimals of x. This is a rational numher. For example, pi is the limit of following sequence
[tex]3,~3.1,~3.14,~3.141,~3.1415,...[/tex]

I'll leave it to you to formalize the argument...

Couldn't he do something like this:

The closure of a set S is defined as the intersection of all closed sets containing S. Now show that the only closed set containing Q is R. Then the intersection of these sets is R.

micromass
#4
Feb23-11, 02:06 PM
Mentor
micromass's Avatar
P: 18,346
Prove that the closure of Q is equal to R

Quote Quote by Robert1986 View Post
Couldn't he do something like this:

The closure of a set S is defined as the intersection of all closed sets containing S. Now show that the only closed set containing Q is R. Then the intersection of these sets is R.
Possibly, but I don't know if it is any easier to prove...
Robert1986
#5
Feb23-11, 02:27 PM
P: 828
Quote Quote by micromass View Post
Possibly, but I don't know if it is any easier to prove...
A closed set contains all of its limit points. Let [tex]F[/tex] be a closed set containing [tex]Q[/tex]. Let [tex]x[/tex] be a limit point of [tex]Q[/tex]. Then [tex]x[/tex] is also a limit point of [tex]F[/tex] and hence is in [tex]F[/tex]. Sine everything in R is a limit point of Q, we must have that [tex]F \subset R[/tex]. Clearly [tex]F \subset R[/tex] we have that [tex]F=R[/tex].
micromass
#6
Feb23-11, 02:33 PM
Mentor
micromass's Avatar
P: 18,346
Quote Quote by Robert1986 View Post
Sine everything in R is a limit point of Q,
Yes, and how does one prove this?
Tetrinity
#7
Feb23-11, 02:40 PM
P: 6
Quote Quote by micromass View Post
Well, it isn't obvious. Some even think of this as an axiom of R. Thus something that cannot be proven, but that is assumed. So whatever proof you'll find, it will be quite informal. (unless you're working with Dedekind cuts, but I don't assume that this is the case here).

To prove that Q is dense in R, we must show that for every x in R, there exists a sequence [tex](q_n)_n[/tex] in Q, that converges to x.
To find the sequence, take x and take as qn the first n decimals of x. This is a rational numher. For example, pi is the limit of following sequence
[tex]3,~3.1,~3.14,~3.141,~3.1415,...[/tex]

I'll leave it to you to formalize the argument...
I am not working with Dedekind cuts. I must admit my understanding of them is somewhat limited, but the topic has not arisen in my course thus far, so I would assume they are not required.

I can find such a sequence by setting qn = a1a2a3...ak-1.akak+1...an where ai is the ith digit of x.

Then if we assume that qn is known, we can find an+1 by setting it such that qn+1 < x and setting it to be 1 higher causes qn+1 ≥ x. This can be done for any x ∈ R.

Does this complete the proof that the closure of Q is R? I know it's not amazingly clear, but I can't think of any other way to represent the decimal.

Quote Quote by Robert1986 View Post
Couldn't he do something like this:

The closure of a set S is defined as the intersection of all closed sets containing S. Now show that the only closed set containing Q is R. Then the intersection of these sets is R.
Interesting... I wasn't actually aware of this definition. I guess solving the problem this way would involve a proof by contradiction with a closed set X ≠ R that contains Q?

And as a side-note: wrong gender. :)
Robert1986
#8
Feb23-11, 02:46 PM
P: 828
Quote Quote by Tetrinity View Post
I am not working with Dedekind cuts. I must admit my understanding of them is somewhat limited, but the topic has not arisen in my course thus far, so I would assume they are not required.

I can find such a sequence by setting qn = a1a2a3...ak-1.akak+1...an where ai is the ith digit of x.

Then if we assume that qn is known, we can find an+1 by setting it such that qn+1 < x and setting it to be 1 higher causes qn+1 ≥ x. This can be done for any x ∈ R.

Does this complete the proof that the closure of Q is R? I know it's not amazingly clear, but I can't think of any other way to represent the decimal.



Interesting... I wasn't actually aware of this definition. I guess solving the problem this way would involve a proof by contradiction with a closed set X ≠ R that contains Q?
Oh, what is the def of closure that you're working with? But if you can show that the only closed set containing Q is R, then the intersection of these sets is just R. So R \in cls(Q). Clearly cls(Q) \in R so R = cls(Q).

And as a side-note: wrong gender. :)
Ohh, my mistake; though "he" is the neuter pronoun in English, so technically, I was correct :)
micromass
#9
Feb23-11, 02:47 PM
Mentor
micromass's Avatar
P: 18,346
Quote Quote by Tetrinity View Post
I am not working with Dedekind cuts. I must admit my understanding of them is somewhat limited, but the topic has not arisen in my course thus far, so I would assume they are not required.

I can find such a sequence by setting qn = a1a2a3...ak-1.akak+1...an where ai is the ith digit of x.

Then if we assume that qn is known, we can find an+1 by setting it such that qn+1 < x and setting it to be 1 higher causes qn+1 ≥ x. This can be done for any x ∈ R.

Does this complete the proof that the closure of Q is R? I know it's not amazingly clear, but I can't think of any other way to represent the decimal.
Yes, I think that this is about the best you can do. However, you might want to prove that the qn converge to x. It isn't that hard, I think...
Robert1986
#10
Feb23-11, 02:51 PM
P: 828
Quote Quote by micromass View Post
Yes, and how does one prove this?
Every neighborhood of any point r \in R has at least one point of Q distinct from r. Since we are dealing in R, this just ammounts to saying that there is always a rational number between any two real numbers.
Char. Limit
#11
Feb23-11, 02:53 PM
PF Gold
Char. Limit's Avatar
P: 1,957
Quote Quote by Robert1986 View Post
Ohh, my mistake; though "he" is the neuter pronoun in English, so technically, I was correct :)
Actually, "it" is the neuter pronoun. However, I use "he" when referring to someone whose gender I do not know, as I am a "he" and so I tend to think I'm talking to other "he"s.

Hmm...
micromass
#12
Feb23-11, 02:58 PM
Mentor
micromass's Avatar
P: 18,346
Quote Quote by Robert1986 View Post
Every neighborhood of any point r \in R has at least one point of Q distinct from r. Since we are dealing in R, this just ammounts to saying that there is always a rational number between any two real numbers.
Yes, but the statement that between every two reals is a rational, is equivalent to the density of Q. So if you want to prove the density of Q in R, from scratch, then you will have to prove this somehow. And I see no easy method in proving this thing...
Tetrinity
#13
Feb23-11, 03:27 PM
P: 6
Quote Quote by Robert1986 View Post
Oh, what is the def of closure that you're working with? But if you can show that the only closed set containing Q is R, then the intersection of these sets is just R. So R \in cls(Q). Clearly cls(Q) \in R so R = cls(Q).
The only definition of closure I have is the one stated in the original post - that the closure of a set X is the set of all of its limit points. Looking further into this, there seem to be a LOT of definitions of "closure" in mathematics...

I think I may need to stick to the definition we've been given. I lost marks on a previous assignment for using a limit before they were formally introduced in the course, with no indication as to whether my proof was actually correct or not. It still seems a bit unfair, but it might be best to avoid that situation again!

Quote Quote by micromass View Post
Yes, I think that this is about the best you can do. However, you might want to prove that the qn converge to x. It isn't that hard, I think...
Alright, thank you so much for your help. :) I think I've spent more time on this single question than all of the others on the assignment put together...


Interestingly, I came across another proof of the density of Q in R while reading through my notes just now.

Let x, y ∈ R and x < y.

We can always find n ∈ N such that n > 1/(y-x)

Now let m be the smallest integer such that m/n > x (in other words, [m-1]/n ≤ x)

Then x < m/n = [1 + (m-1)]/n = 1/n + (m-1)/n

We know 1/n < y - x, and (m-1)/n ≤ x

Therefore 1/n + (m-1)/n < (y-x) + x = y

Therefore x < m/n < y (i.e. there is a rational number between any two reals)

Would that proof be better to use, since it avoids having to further prove qn converges to x? I'd been reading through my notes to remind myself of how to show a sequence converges to a limit...
Robert1986
#14
Feb23-11, 03:32 PM
P: 828
Quote Quote by micromass View Post
Yes, but the statement that between every two reals is a rational, is equivalent to the density of Q. So if you want to prove the density of Q in R, from scratch, then you will have to prove this somehow. And I see no easy method in proving this thing...

Let y < x be reals. Then x - y > 0. We can pick an n such that n > x-y => 1/n < 1/(x-y). Pick the largest m st m/n < y. Then (m+1)/n > y. Since 1\n < x-y, and m/n < x we have that (m+1)/n < x. Hence y < (m+1)/n < x.


I think the problem here is that our books have presented information in different orders and differntly. For example, the thing I just proved was an early result in the book I use.
micromass
#15
Feb23-11, 03:34 PM
Mentor
micromass's Avatar
P: 18,346
Quote Quote by Tetrinity View Post
Interestingly, I came across another proof of the density of Q in R while reading through my notes just now.

Let x, y ∈ R and x < y.

We can always find n ∈ N such that n > 1/(y-x)

Now let m be the smallest integer such that m/n > x (in other words, [m-1]/n ≤ x)

Then x < m/n = [1 + (m-1)]/n = 1/n + (m-1)/n

We know 1/n < y - x, and (m-1)/n ≤ x

Therefore 1/n + (m-1)/n < (y-x) + x = y

Therefore x < m/n < y (i.e. there is a rational number between any two reals)
Yes, this is also a good proof (in fact a better one). But if your professor is rigourous, he will want a proof of this:

We can always find n ∈ N such that n > 1/(y-x)
This is called the axiom of Archimedes. If it is in your course somewhere, then the above proof will be exactly what your prof wants. If it's not in your course, then maybe your prof would want a "proof" of this...
Tetrinity
#16
Feb23-11, 03:48 PM
P: 6
Quote Quote by micromass View Post
This is called the axiom of Archimedes. If it is in your course somewhere, then the above proof will be exactly what your prof wants. If it's not in your course, then maybe your prof would want a "proof" of this...
The axiom was stated, albeit rather informally ("there is no largest natural number"), in my Analysis I class from the previous semester (this homework is for Analysis II, which has Analysis I as a pre-requisite). I think I'll go with it.

Thanks for all of your help!


Register to reply

Related Discussions
Prove that two definite integrals are equal Calculus & Beyond Homework 17
Prove that this vector is equal to this? Introductory Physics Homework 1
Prove n^3 less than or equal to 3^n for n=1,2,... Calculus & Beyond Homework 4
Prove that the third invariant is equal to the determinant Calculus & Beyond Homework 4
Prove that the closure is the following set. Calculus & Beyond Homework 2