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

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In summary: And vice versa. You seem to be saying that the real line contains some of the extended reals and not others????In summary, the conversation is discussing how to prove that cl(A) = R, where A = {m + nx | m, n in Z} and x is irrational. The conversation discusses using the definition of closure and applying it to the set A, as well as using the standard result from R being an archimedean ordered field. The conversation also mentions constructing a point in A that is closer to y than any given epsilon, and using the proof of denseness to show that the real line is a closed set. However, there is some disagreement on whether the real line should be considered
  • #1
j0k3R_
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Homework Statement



Let A = {m + nx | m, n in Z} and x irrational. Show cl(A) = R.
 
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  • #2
(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 [tex](-\infty,\infty)[/tex].
 
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  • #3
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?
 
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  • #4
To prove the closure of a set we must show that, for every [tex] y\in R[/tex] and [tex]\varepsilon > 0[/tex], the open set [tex](y-\varepsilon,y+\varepsilon)[/tex] contains a point in A. So let us find it. We are given [tex]y,\varepsilon[/tex]. Can we construct a point that is in A and closer to [tex]y[/tex] than [tex]\varepsilon[/tex]? We can subtract [tex]m[/tex] and divide by [tex]n[/tex], so we only need to show that [tex]x[/tex] is within [tex](y' - \varepsilon',y' + \varepsilon')[/tex], where primes denote the open set after subtracting and dividing. We can assume [tex]y'[/tex] is rational, since if it were irrational we would be done. If we take a [tex]\delta<\varepsilon'[/tex], irrational, can we show that [tex]y'+\delta[/tex] is irrational?
 
  • #5
And if you want to prove [tex]\delta[/tex] exists, just let it be [tex]\sqrt{2}/N[/tex] for sufficiently large N.
 
  • #6
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?
 
  • #7
Sure, let [tex]y'=\frac{a}{b}[/tex]. Take [tex]\frac{a}{b} + \frac{\sqrt{2}}{N}[/tex], for example. You can show that if this is rational, then it proves that [tex]\sqrt{2}[/tex] is rational, which is a contradiction.
 
  • #8
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)"
 
  • #9
If x is in the interval [tex](y' - \varepsilon',y' + \varepsilon')[/tex], then it must be the case that m + nx is in the interval [tex](y - \varepsilon,y+\varepsilon)[/tex]. You can do it explicitly with some algebra on the inequalities.
 
  • #10
j0k3R_ said:
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.
 
  • #11
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.
 
  • #12
cellotim said:
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).
 
  • #13
"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
 
  • #14
Dick said:
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 [tex](y-\varepsilon,y+\varepsilon), y\in[0,1][/tex]. 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 [tex]-\infty[/tex] and [tex]+\infty[/tex] are adjoined to R . . . The result is the so-called set of extended real numbers, [tex][-\infty,\infty][/tex]." There is a wikipedia entry on it: http://en.wikipedia.org/wiki/Extended_real_number_line" [Broken].
 
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  • #15
j0k3R_ said:
"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.
 
  • #16
cellotim said:
Yes, I see my mistake. In which case we are looking for an m such that mx - floor(mx) is in the open set [tex](y-\varepsilon,y+\varepsilon), y\in[0,1][/tex]. 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 [tex]-\infty[/tex] and [tex]+\infty[/tex] are adjoined to R . . . The result is the so-called set of extended real numbers, [tex][-\infty,\infty][/tex]." There is a wikipedia entry on it: http://en.wikipedia.org/wiki/Extended_real_number_line" [Broken].

I know you can extend the real line. But I'm talking about the un-extended real line. It's closed.
 
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  • #17
cellotim said:
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.
 
  • #18
For any [tex]x\in R[/tex], one can construct an open ball that is in R. This shows that R is open.
 
  • #19
cellotim said:
For any [tex]x\in R[/tex], 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.
 
  • #20
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 :)
 

What is cl(A) and R in the context of solving Set A?

In mathematics, cl(A) refers to the closure of a set A, which is the smallest closed set that contains all elements of A. R represents the set of real numbers. In this context, we are trying to prove that the closure of set A is equal to the set of real numbers.

Why is proving cl(A) = R important?

Proving that the closure of set A is equal to the set of real numbers is important because it helps us understand the properties and characteristics of both sets. It also allows us to make connections between different mathematical concepts and to solve more complex problems.

What are some strategies for proving cl(A) = R?

One strategy is to show that every element in R is also in the closure of set A, and vice versa. Another strategy is to use the definition of closure, which states that the closure of A is the intersection of all closed sets that contain A.

Can cl(A) ever be a proper subset of R?

No, the closure of set A can never be a proper subset of R because by definition, the closure must contain all elements of A and R is the set of all real numbers. This means that cl(A) must be equal to or a superset of R.

Are there any real-life applications of solving cl(A) = R?

Yes, this concept is commonly used in topology, which is a branch of mathematics that studies the properties of spaces and the relationships between them. It is also used in engineering and computer science, particularly in the analysis of algorithms and data structures.

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