Results 1 to 10 of 10

Math Help - Dense subset

  1. #1
    Newbie
    Joined
    Jan 2010
    Posts
    13

    Dense subset

    Hi,

    If  a is irrational, show that the set {m a+n; m \inN, n \inZ} is dense in R.

    Thanks!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor Drexel28's Avatar
    Joined
    Nov 2009
    From
    Berkeley, California
    Posts
    4,563
    Thanks
    21
    Quote Originally Posted by rhemo View Post
    Hi,

    If  a is irrational, show that the set {m a+n; m \inN, n \inZ} is dense in R.

    Thanks!
    Let x\in\mathbb{R} consider B_{\delta}(x). Show that there exists some element of your set in there. Without doing it for you that's as far as I can go. What work have you done?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Jan 2010
    Posts
    13
    I canít explain in English very well, but I will try:
    A={m \alpha+n}

    I make ( or do?):

    (1) \forall \epsilon >0 A \cap(0, \epsilon) is true, if x \inA \cap(0, \epsilon) then, sequence {x, 2x,3x...}is arithmetic progression d= \epsilon. \forallr \inR, \existsk \inZ; kx \in(r,r+ \epsilon).


    (1) \Rightarrow \existsu and v in A, |u-v|< \epsilon. A is closed for addition. Then, \existsw in A, |w|< \epsilon and \Rightarrow -w in A.

    Ufff....
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member
    Joined
    Jan 2010
    Posts
    26
    Quote Originally Posted by Drexel28 View Post
    Let x\in\mathbb{R} consider B_{\delta}(x). Show that there exists some element of your set in there. Without doing it for you that's as far as I can go. What work have you done?
    I tried to show it this way but i couldn't even start. Then i tried to find successive approximations for the real number x using the fact that there is a natural number k such that k\alpha\leq x\leq(k+1)\alpha. But i got stuck when i found p such that k\alpha+p\leq x\leq k\alpha+(p+1).

    What i thought was that i could make the first coefficient (the one that multiplies the irrational number) arbitrarily big if i made the second coefficient small enough, and, that way, i could build a sequence converging to x.

    But, uh, it didn't work. A few things came to my mind: what if \alpha\in\mathbb{Q}\backslash\mathbb{Z}? How would it be different? What if in the set definition you could use m\in\mathbb{Z}?

    Oh, yes, and i assumed that \alpha>0.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Banned
    Joined
    Mar 2009
    Posts
    256
    Thanks
    1
    Quote Originally Posted by Drexel28 View Post
    Let x\in\mathbb{R} consider B_{\delta}(x). Show that there exists some element of your set in there. Without doing it for you that's as far as I can go. What work have you done?
    Sorry for asking ,but do you really know how to do this problem??

    Because if you do ,please if you wish ,show us .

    I mean the whole problem to save us the anxiety .
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Junior Member
    Joined
    Jan 2010
    Posts
    26
    Ok, I'm not him, but i'll show you the progress i've made.

    First, we need to prove two facts.

    Lemma 1. (modified euclidean algorithm) Let x, a be real numbers. Then there is an integer q and a non-negative real number r such that x=a*q+r and r<a.

    Proof. Not hard and hasn't got much to do with the problem.

    Lemma 2. Let G be a non-trivial (i.e., not {0}) subgroup of \mathbb{R} with respect to addition. Then one and only one of the following is valid:
    • \exists g\in G such that G = g\mathbb{Z}
    • G is dense in \mathbb{R}


    Proof. Let G^+ denote G\cap\mathbb{R}^+. Clearly G^+ is non-empty and bounded below. Let a=\inf{G^+}. Then either a>0 or a=0.

    Suppose, first, that a>0. Then a\in G, because otherwise there would be g, h in G^+ such that a<g<h<a+(a/2), and that would imply \tfrac{a}{2}>g-h\in G^+, a contradiction.

    Furthermore, let g\in G. Then, by Lemma 1, there is an integer q and a non-negative real number r such that g=aq+r with r<a. Since both a and g are in G and q is an integer, then 0\leq r=g-aq\in G. But if r>0 then r\in G^+. Therefore 0<r<a=\inf{G^+}, a contradiction. Therefore the only possibility is r=0, i.e., g=aq.

    Now suppose a=0. Let x be any real number. Suppose, first, that x is positive. Given \epsilon>0, there is g\in G^+ such that 0<g<\epsilon. Then there should be n\in\mathbb{Z} such that ng\in (x-\epsilon,x+\epsilon)\cap G. In fact, let A=\{t\in\mathbb{N}: tg>x-\epsilon\}. The well ordering principle assures that there's n=\min{A}. It's obvious that ng\in G. Let's show that ng\in (x-\epsilon,x+\epsilon).

    It is sufficient to show that x-\epsilon<ng<x+\epsilon. The first inequality is valid due to the definition of A. Now suppose ng\geq x+\epsilon. Then ng\geq x+\epsilon>x+g\implies (n-1)g>x, but that contradicts the definition of n.

    Now let x be any negative real number. We have shown that, since -x>0, \exists h\in (-x-\epsilon,-x+\epsilon)\cap G. Equivalently, there's g in G such that -x-\epsilon < h < -x+\epsilon. But since G is a group, -h is also in G and:

    -x-\epsilon < h < -x+\epsilon \Leftrightarrow x-\epsilon < (-h) < x+\epsilon\Leftrightarrow (-h)\in(x-\epsilon,x+\epsilon)

    Therefore (since 0 is in G), G is dense in R if 0=\inf{G^+}.
    QED.

    -------------

    Ok. Now the problem gets easier: first, we show that S_1 = \{m\alpha+n|m,n\in\mathbb{Z}\} is an additive subgroup of \mathbb{R}. Then it is easy to get a contradiction by supposing that S_1 = k\mathbb{Z}. That is equivalent, by Lemma 2, to asserting that S_1 is dense in \mathbb{R}. We then use this fact to prove that S_1\supset S_2=\{m\alpha+n|m\in\mathbb{N},n\in\mathbb{Z}\} is dense in \mathbb{R}. These proofs are all easy and I won't write them, since i've already written the dirty part.
    Last edited by JoachimAgrell; February 3rd 2010 at 06:22 AM.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Banned
    Joined
    Mar 2009
    Posts
    256
    Thanks
    1
    Quote Originally Posted by JoachimAgrell View Post
    Ok, I'm not him, but i'll show you the progress i've made.

    First, we need to prove two facts.

    Lemma 1. (modified euclidean algorithm) Let x, a be real numbers. Then there is an integer q and a non-negative real number r such that x=a*q+r and r<a.

    Proof. Not hard and hasn't got much to do with the problem.
    .
    Suppose  x=\pi and a= e ,what would be the q and r such that:

    \pi = eq +r??

    \pi =3,14...... and e= 2,718...........
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Super Member
    Joined
    Aug 2009
    From
    Israel
    Posts
    976
    Quote Originally Posted by xalk View Post
    Suppose  x=\pi and a= e ,what would be the q and r such that:

    \pi = eq +r??

    \pi =3,14...... and e= 2,718...........
    q=1, ~ r = \pi - e ?
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Banned
    Joined
    Mar 2009
    Posts
    256
    Thanks
    1
    Quote Originally Posted by Defunkt View Post
    q=1, ~ r = \pi - e ?
    yes yes i am sorry i thought r is an integer .

    But can we in general prove that theorem??
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Junior Member
    Joined
    Jan 2010
    Posts
    26
    Quote Originally Posted by xalk View Post
    yes yes i am sorry i thought r is an integer .

    But can we in general prove that theorem??
    If you agree that my previous post is correct, then there are, as i said, only a few basic things to do. Let S=\{m\alpha+n:m\in\mathbb{Z^{+}}, n\in\mathbb{Z}\}.

    Then you can show that \inf{S\cap\mathbb{R^+}}=\sup{S\cap\mathbb{R^{-}}}=0 and use an argument similar to the one used in Lemma 2 to show that an open interval with arbitrary radius centered at an arbitrary element of R contains at least one element of S.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 2
    Last Post: August 28th 2011, 03:33 AM
  2. Dense subset
    Posted in the Differential Geometry Forum
    Replies: 19
    Last Post: January 12th 2011, 04:41 AM
  3. Dense subset
    Posted in the Differential Geometry Forum
    Replies: 2
    Last Post: April 1st 2010, 09:47 AM
  4. Dense subset of R^n
    Posted in the Math Challenge Problems Forum
    Replies: 9
    Last Post: January 28th 2010, 10:12 AM
  5. Dense subset
    Posted in the Calculus Forum
    Replies: 2
    Last Post: December 3rd 2008, 11:38 AM

Search Tags


/mathhelpforum @mathhelpforum