Show that $\inf \{ m+n\omega: m+n\omega>0,and,m,n\in{Z}\}= 0$, where $\omega>0 $ is irrational.

Question

Let $\omega\in\mathbb {R}$ be an irrational positive number. Set $$A=\{m+n\omega: m+n\omega>0,and,m,n\in{Z}\}.$$

Show that $\inf{A}=0.$

How should I start this problem? I don't get this problem.

Answer 1

What $\inf A =0$ means is that, for any $\epsilon >0$ there is some multiple of $\omega$ that is within distance $\epsilon$ of some integer.

Hint 1: Pigeonhole principle.

For simplicity assume $\omega <1$. Then each interval $[0,1], [1,2], [2,3]$ contains a multiple of $\omega$, call them $n \omega, m \omega$ and $r\omega$. Each of those three intervals is divided into a "first half" and a "second half". For example $[0,1] = [0,1/2] \cup [1/2, 1]$. So two of $n \omega, m \omega$ and $r\omega$ are in the same half. For example $n\omega \in [1/2,1]$ and $r\omega \in [2+1/2,3]$. That implies $r\omega -2 \in [1/2,1]$ and so $r\omega -2$ and $n\omega$ are distance less than $1/2$ apart. Meaning $|(r\omega -2)-n\omega| < 1/2 \implies |(r-n)\omega -2| < 1/2$. So that $\inf A < 1/2$. How does this generalise?

Hint 2: (independent) Compactness of the circle

Answer 2

Hint

In this context ($\omega$ irrational) it can be shown that the set $\{n\omega-\lfloor n\omega\rfloor\mid n\in\mathbb Z\}$ is a dense subset of $(0,1)$. Also it can be shown to be a subset of $A$.

Answer 3

It boils down to showing that given any $\epsilon > 0$ there are integers $m, n$ such that $$0 < m + n\omega < \epsilon\tag{1}$$ Clearly this means we need to find good rational approximations to irrational $\omega$. This follows easily from a theorem that there are infinitely many approximations of the form $p/q$ with $$\left|\omega - \frac{p}{q}\right| < \frac{1}{q^{2}}$$ and then $|q\omega - p| < 1/q$. Since there are infinitely many such values of $p, q$ it follows that $q$ can take arbitrarily large values and we just need $q$ such that $1/q < \epsilon$ and then $n = q, m = -p$ satisfies $(1)$.

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It is better to supply the proof of the theorem mentioned in italics in the last paragraph. Thus let $\omega$ be any irrational and $n$ be a positive integer. Consider the following $(n + 1)$ real numbers $$0, \omega - [\omega], 2\omega - [2\omega], \ldots, n\omega - [n\omega]\tag{2}$$ and consider the $n$ intervals $$[0, 1/n), [1/n, 2/n), \ldots, [(n - 1)/n, 1)\tag{3}$$ whose union is $[0, 1)$.

The numbers in $(2)$ clearly lie in (0, 1) and hence each number in $(2)$ must lie in some interval listed in $(3)$. Moreover there are $(n + 1)$ numbers in the list $(2)$ and only $n$ intervals in list $(3)$ so by Pigeonhole Principle at least one of the intervals in $(3)$ must contain two members of list $(3)$. Let these two numbers be $n_{1}\omega - [n_{1}\omega]$ and $n_{2}\omega - [n_{2}\omega]$ and $0 \leq n_{1} < n_{2} \leq n$. Since they lie in an interval of type $[j/n, (j + 1)/n)$ it follows that their difference is less than $1/n$. Hence we have $$|(n_{1}\omega - [n_{1}\omega]) - (n_{2}\omega - [n_{2}\omega])| < \frac{1}{n}$$ or $$|(n_{2} - n_{1})\omega - ([n_{2}\omega] - [n_{1}\omega])| < \frac{1}{n}$$ and this means that $|q\omega - p| < 1/n$ where $q = (n_{2} - n_{1})$ and $p = ([n_{2}\omega] - [n_{1}\omega])$.

Also note that $q \leq n$ and hence $$\left|\omega - \frac{p}{q}\right| < \frac{1}{nq} \leq \frac{1}{q^{2}}\tag{4}$$ Now we need to show that there are infinitely many $p, q$ which satisfy $(4)$. This we do by contradiction. Let's assume that there are only finitely many such numbers $$\frac{p_{1}}{q_{1}},\frac{p_{2}}{q_{2}},\ldots,\frac{p_{r}}{q_{r}}$$ such that $$\left|\omega - \frac{p_{i}}{q_{i}}\right| < \frac{1}{q_{i}^{2}}$$ for $i = 1, 2, \ldots, r$.

Let $\epsilon$ be the minimum of the numbers $$\left|\omega - \frac{p_{1}}{q_{1}}\right|, \left|\omega - \frac{p_{2}}{q_{2}}\right|, \ldots, \left|\omega - \frac{p_{r}}{q_{r}}\right|$$ and chose an integer $n$ such that $0 < 1/n < \epsilon$. Then proceeding as before we can find another rational $p/q$ such that $$\left|\omega - \frac{p}{q}\right| < \frac{1}{nq} \leq \frac{1}{n} < \epsilon$$ and hence $p/q$ is different from the numbers $p_{i}/q_{i}$ and yet satisfies $|\omega - p/q| < 1/q^{2}$. This contradiction proves that there are infinitely many rationals of the form $p/q$.

Answer 4

Fix $\epsilon>0$. Fix an integer $N>1/\epsilon$. Consider the fractional parts of the numbers $k\omega, 0<k\le N$. These are commonly denoted $$ \{k\omega\}:=k\omega-\lfloor k\omega\rfloor. $$ Because $\omega$ is irrational the numbers $\{k\omega\}\in(0,1)$, $k=1,2,\ldots,N$, are all distinct. Because there are $N$ of them some two of them, say $\{k\omega\}$ and $\{\ell\omega\}$ are within $\epsilon$ of each other (you may use the pigeonhole principle to see this).

If $k>\ell$ (resp. if $k<\ell$) then $0<\{t\omega\}<\epsilon$ for $t=k-\ell$ (resp. $t=\ell-k$). This means that $$0<n\omega+m<\epsilon$$ for $n=t$ and $m=-\lfloor t\omega\rfloor$.

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Making this CW for this is surely a duplicate, but I don't have time to look for one.