Dense subset

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January 27th 2010, 05:11 PM
rhemo

Dense subset

Hi,

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

Thanks!
January 27th 2010, 05:14 PM
Drexel28

Quote:

Originally Posted by rhemo

Hi,

If $a$ is irrational, show that the set {m $a$ +n; m $\in$ N, n $\in$ Z} 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?
January 28th 2010, 03:27 AM
rhemo

I can’t 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 $\in$ A $\cap$ (0, $\epsilon$ ) then, sequence {x, 2x,3x...}is arithmetic progression d= $\epsilon$ . $\forall$ r $\in$ R, $\exists$ k $\in$ Z; kx $\in$ (r,r+ $\epsilon$ ).

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

Ufff....
January 31st 2010, 03:09 PM
JoachimAgrell

Quote:

Originally Posted by Drexel28

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 $\alpha>0$ .
February 2nd 2010, 05:27 PM
xalk

Quote:

Originally Posted by Drexel28

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 .
February 3rd 2010, 05:07 AM
JoachimAgrell
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 denote . Clearly is non-empty and bounded below. Let . Then either a>0 or a=0.

Suppose, first, that a>0. Then , because otherwise there would be g, h in such that a<g<h<a+(a/2), and that would imply , a contradiction.

Furthermore, let . 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 and q is an integer, then . But if r>0 then . Therefore , 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 , there is such that . Then there should be such that . In fact, let . The well ordering principle assures that there's . It's obvious that . Let's show that .

It is sufficient to show that . The first inequality is valid due to the definition of A. Now suppose . Then , but that contradicts the definition of n.

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

Therefore (since 0 is in G), G is dense in R if .
QED.

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

Ok. Now the problem gets easier: first, we show that is an additive subgroup of . Then it is easy to get a contradiction by supposing that . That is equivalent, by Lemma 2, to asserting that is dense in . We then use this fact to prove that is dense in . These proofs are all easy and I won't write them, since i've already written the dirty part.
February 3rd 2010, 02:57 PM
xalk

Quote:

Originally Posted by JoachimAgrell

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...........
February 3rd 2010, 03:10 PM
Defunkt

Quote:

Originally Posted by xalk

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...........

$q=1, ~ r = \pi - e$ ?
February 3rd 2010, 03:41 PM
xalk

Quote:

Originally Posted by Defunkt

$q=1, ~ r = \pi - e$ ?

yes yes i am sorry i thought r is an integer .

But can we in general prove that theorem??
February 8th 2010, 06:51 PM
JoachimAgrell

Quote:

Originally Posted by xalk

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.