Prove that $\inf A = 0$ for $A = \{ m + nx: m,n, \in \mathbb{Z}~\text{and}~m+ nx >0 \}$ with positive irrational $x$.

Question

Let $x$ be a positive irrational number, and let $A = \{ m + nx: m,n, \in \mathbb{Z}~\text{and}~m+ nx >0 \}$. Prove that $\inf A = 0$

So far I have the following:

So $0 < a$ for all $a \in A$, so $0$ is a lower bound.

Let $z$ be a positive real number. I will show $z$ is not a lower bound.

Assume $z$ is a lower bound

$$z \leq a \quad \forall a \in A \quad\implies a-z \geq 0$$

This is where I'm stuck. Any hints are appreciated.

Answer 1

Let $n$ be a positive integer such that $z>\frac{1}{n}$.

By Dirichlet's Approximation Theorem there are integers $p$ and $q$ such that $$|xq-p|<\frac{1}{n}<z.$$

By altering the signs of both $p$ and $q$, if necessary, we can suppose $$0<xq-p<z.$$

Therefore $z$ cannot be a lower bound.

Answer 2

Let $x\in\mathbb{R} $ and consider the set $$A=\{m+nx\mid m\in\mathbb {Z}, n\in\mathbb {Z}, m+nx>0\}$$ Clearly $A$ is non-empty (check by putting $m=1,n=0$) and bounded below by $0$. Thus $a=\inf A$ exists and $a\geq 0$.

Let's assume that $a>0$ and prove that $x$ is rational and then our job is done here. If $x=0$ then the analysis is trivial as $A=\mathbb{N}, a=1$. Let us then assume that $x\neq 0$.

We begin by first showing that $a\in A$. Suppose that $a\notin A$. Then we have a sequence $x_n\in A$ such that $x_n$ is strictly decreasing and $x_n\to a$. Now comes the surprise! The sequence $y_n=x_{n}-x_{n+1}>0$ and hence $y_n\in A$ and we have $y_n\to 0$ and this contradicts that $a=\inf A>0$.

It follows that $a\in A$ and we next show every member $b\in A$ can be written as $b=na$ for some $n\in\mathbb{N} $. Suppose that it is not the case then we have some member $b\in A$ and a unique positive integer $m$ such that $$a<2a<\dots<ma<b<(m+1)a<(m+2)a<\dots$$ Then the number $c=b-ma$ is such that $0<c<a$ and $c\in A$. This is a contradiction as $a$ is the least member of $A$.

And now we see that $1=1+0\cdot x\in A, |x|=0\pm 1\cdot x\in A$ and hence we have positive integers $m, n$ such that $|x|=ma, 1=na$ or $|x|=m/n$ and hence $x$ is rational.

Answer 3

Proposition: For all $k \in \mathbb N$, for all $b \notin \mathbb Q$, there exists $n,m \in \mathbb Z$ such that $|mb−n|< \frac 1 k$.

Proof: Let $k \in \mathbb N$. We divide $[0,1)$ into $k$ subintervals of the form $[\frac{i−1} k, \frac i k)$, for $i=1, \dots,k$, such that the length of each one equals $\frac 1 k$. Define the sequence $x_l = lb−[lb]$, for every $l \in \mathbb N$. Then there exist indices $s,t$ such that $|x_s − x_t|< \frac 1 k$ or $|(s−t)b−([sb]−[tb])|< \frac 1 k$ (the assertion is true, because there are $k$ subintervals and $k+1$ terms; so at least two terms have distance less than $\frac 1 k$). Define $m =s−t, \ n =[sb]−[tb]$, and the proposition's proof is complete.

It is obvious that this proposition proves that $\inf \{ma-[ma] \mid m \in \mathbb Z \}=0$.