# Dense subset

• 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 $G^+$ 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, 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. 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. 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.
• 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.