Solving the Set A: Proving cl(A) = R

[j0k3R_]

Homework Statement

Let A = {m + nx | m, n in Z} and x irrational. Show cl(A) = R.

 

[cellotim]

(1) what have you tried so far? I suggest writing down the definition of closure and what you know about A.
(2) Are you referring to the extended reals? The real line is usually not a closed set, but $$(-\infty,\infty)$$.

 

[j0k3R_]

cellotim

i think i am coming close to the solution

basically we need to show there exists (y - (m + nx), y + (m + nx)) C (y - e/2, y + e/2) for any e > 0. So take y - e/2 which is in R. since x is in R as well, we can apply the standard result from R is an archimedean ordered field to see that there is a unique nx, with n in Z, such that nx < y - e/2 < (n + 1)x. let m be the least integer greater than x, which exists by the AP. then, y - e/2 < (n + 1)x < nx + m since x is in R - Q.

etc. we can do this on other side and then show y - (m + nx) < y + (m + nx) which i think maybe completes solution by showing y in closure of A, and closure of A is R.

is this okay

edit: no maybe i think i need to show first on e/2, then show since e/2 < nx + m, y - e/2 < y - (nx + m) so this part is good. then on other side we see there is nx < e/2 < (n + 1)x. how we can fit this right side into y + e/2?

 

[cellotim]

To prove the closure of a set we must show that, for every $$y\in R$$ and $$\varepsilon > 0$$, the open set $$(y-\varepsilon,y+\varepsilon)$$ contains a point in A. So let us find it. We are given $$y,\varepsilon$$. Can we construct a point that is in A and closer to $$y$$ than $$\varepsilon$$? We can subtract $$m$$ and divide by $$n$$, so we only need to show that $$x$$ is within $$(y' - \varepsilon',y' + \varepsilon')$$, where primes denote the open set after subtracting and dividing. We can assume $$y'$$ is rational, since if it were irrational we would be done. If we take a $$\delta<\varepsilon'$$, irrational, can we show that $$y'+\delta$$ is irrational?

 

[cellotim]

And if you want to prove $$\delta$$ exists, just let it be $$\sqrt{2}/N$$ for sufficiently large N.

 

[j0k3R_]

okay i am understanding your argument. in previous problem i proved closure R - Q is R. so, we can choose 0 < delta < e', and if y' + delta is rational then we obtain simple contradiction.

is it okay?

 

[cellotim]

Sure, let $$y'=\frac{a}{b}$$. Take $$\frac{a}{b} + \frac{\sqrt{2}}{N}$$, for example. You can show that if this is rational, then it proves that $$\sqrt{2}$$ is rational, which is a contradiction.

 

[j0k3R_]

yes

also one more thing, iam feeling little bit uneasy when you give "subtract then divide everything in interval"; it is valid to say we have x in (changed interval), then x(operation) in (interval * operation)? because maybe i am thinking "there exists some pair (m, n)"

 

[cellotim]

If x is in the interval $$(y' - \varepsilon',y' + \varepsilon')$$, then it must be the case that m + nx is in the interval $$(y - \varepsilon,y+\varepsilon)$$. You can do it explicitly with some algebra on the inequalities.

 

[Dick]

yes

also one more thing, iam feeling little bit uneasy when you give "subtract then divide everything in interval"; it is valid to say we have x in (changed interval), then x(operation) in (interval * operation)? because maybe i am thinking "there exists some pair (m, n)"

Do feel uneasy with this 'proof'. Feel VERY uneasy. Dividing by an integer changes numbers of the form m+nx (m,n integers) to p+qx (p,q rationals). The simplest proof of the denseness I know starts by saying for every integer n, there is an integer m(n) such that m(n)+nx is in [0,1]. (Basically you just add or subtract an integer to get rid of the whole number part of nx). Now if n and n' are distinct integers then m(n)+nx and m(n')+n'x are also not equal, right? (Why?). Now you have an infinite set of distinct numbers of the form m+nx in [0,1]. Can you see where to go from there? (Hint: think pigeonhole principle and try to show the numbers are dense on [0,1]).

BTW, cellotim, the real line IS a closed set. Ask around.

 

[cellotim]

My proof is correct. I was not changing the number m + nx, m and n integers into p + qx, p,q rationals, but subtracting then dividing to get x. The proof of denseness only requires that every open set around a member of the reals contains a member of A, one member, which I constructed in my proof.

Furthermore, the reals are only a closed set if they contain -infinity and +infinity, which, in my graduate courses was referred to as the extended reals. A set is closed if and only if it contains all its limit points. Therefore, whether the reals are a closed set depends on how you define them.

 

[Dick]

My proof is correct. I was not changing the number m + nx, m and n integers into p + qx, p,q rationals, but subtracting then dividing to get x. The proof of denseness only requires that every open set around a member of the reals contains a member of A, one member, which I constructed in my proof.

Furthermore, the reals are only a closed set if they contain -infinity and +infinity, which, in my graduate courses was referred to as the extended reals. A set is closed if and only if it contains all its limit points. Therefore, whether the reals are a closed set depends on how you define them.

I define the 'reals' as the 'reals'. The usual one. Standard topology. The real line contains all of it's limit points without being extended. I reread the proof. As near as I can tell it looks like you are taking the set A to be {m+nx : m and n integers, x ANY irrational}. That would make A the set of ALL irrationals. We already know that is dense. The set A they are referring to is for x a fixed irrational. Like all numbers of the form m+n*sqrt(2).

 

[j0k3R_]

"Can you see where to go from there? (Hint: think pigeonhole principle and try to show the numbers are dense on [0,1])."

can you illustrate for me how ot use pigeonhole principle here

 

[cellotim]

I define the 'reals' as the 'reals'. The usual one. Standard topology. The real line contains all of it's limit points without being extended. I reread the proof. As near as I can tell it looks like you are taking the set A to be {m+nx : m and n integers, x ANY irrational}. That would make A the set of ALL irrationals. We already know that is dense. The set A they are referring to is for x a fixed irrational. Like all numbers of the form m+n*sqrt(2).

Yes, I see my mistake. In which case we are looking for an m such that mx - floor(mx) is in the open set $$(y-\varepsilon,y+\varepsilon), y\in[0,1]$$. On that point you are correct.

On the second, I think not. This is largely semantics, but perhaps important. Let me give you a quote from Dudley's Real Analysis. "[T]he two points $$-\infty$$ and $$+\infty$$ are adjoined to R . . . The result is the so-called set of extended real numbers, $$[-\infty,\infty]$$." There is a wikipedia entry on it: http://en.wikipedia.org/wiki/Extended_real_number_line" .

 

[cellotim]

"Can you see where to go from there? (Hint: think pigeonhole principle and try to show the numbers are dense on [0,1])."

can you illustrate for me how ot use pigeonhole principle here

Take an integer n and the numbers 0*x, 1*x, ..., n*x of which there are n+1. Now partition [0,1] into [0, 1/n), [1/n,2/n), ... ,[n-1/n,1] intervals of which there are n. By the pigeonhole principle, two of the fractional parts of the n+1 numbers must fall into one of the n partitions.

 

[Dick]

Yes, I see my mistake. In which case we are looking for an m such that mx - floor(mx) is in the open set $$(y-\varepsilon,y+\varepsilon), y\in[0,1]$$. On that point you are correct.

On the second, I think not. This is largely semantics, but perhaps important. Let me give you a quote from Dudley's Real Analysis. "[T]he two points $$-\infty$$ and $$+\infty$$ are adjoined to R . . . The result is the so-called set of extended real numbers, $$[-\infty,\infty]$$." There is a wikipedia entry on it: http://en.wikipedia.org/wiki/Extended_real_number_line" .

I know you can extend the real line. But I'm talking about the un-extended real line. It's closed.

 

[Dick]

Take an integer n and the numbers 0*x, 1*x, ..., n*x of which there are n+1. Now partition [0,1] into [0, 1/n), [1/n,2/n), ... ,[n-1/n,1] intervals of which there are n. By the pigeonhole principle, two of the fractional parts of the n+1 numbers must fall into one of the n partitions.

That works. I was just thinking of dividing [0,1] into N equal intervals. Now since there are an infinite number of numbers of the form m+nx in [0,1], one interval contains at least two such numbers. j0k3R_, what you are trying to show with this is that for every epsilon>0, there is an m and n such that 0<m+nx<epsilon.

 

[cellotim]

For any $$x\in R$$, one can construct an open ball that is in R. This shows that R is open.

 

[Dick]

For any $$x\in R$$, one can construct an open ball that is in R. This shows that R is open.

Sure it's open. It's open AND closed.

 

[j0k3R_]

hi again

"
Take an integer n and the numbers 0*x, 1*x, ..., n*x of which there are n+1. Now partition [0,1] into [0, 1/n), [1/n,2/n), ... ,[n-1/n,1] intervals of which there are n. By the pigeonhole principle, two of the fractional parts of the n+1 numbers must fall into on"

thank you -- eventually i did realize precisely what Dick meant with pigeonhole argument so i wrote precisely this :)

"
That works. I was just thinking of dividing [0,1] into N equal intervals. Now since there are an infinite number of numbers of the form m+nx in [0,1], one interval contains at least two such numbers. j0k3R_, what you are trying to show with this is that for every epsilon>0, there is an m and n such that 0<m+nx<epsilon."

yes :)