Proving the density of a function in an interval.

Question

I am reading Steven G. Krantz's Real Analysis and Foundations and came across this problem.

Problem: Let $\lambda$ be a positive irrational real number. If $n$ is a positive integer, choose by the Archimedean property an integer $k$ such that $k\lambda \leq n < (k+1)\lambda$. Let $\phi(n) = n-k\lambda$. Prove that the set of all $\phi(n)$ is dense in the interval $[0,\lambda]$. (Hint: Examine the proof of the density of the rationals in the reals)

My Approach: We want to show that for all $x,y\in[0,\lambda]: x<y$ there exist $m: x<\phi(m)<y$.

After examining the proof of the density of the rationals in the reals, I came up with the following.

Since $y>x$ we can choose A such that $A(y-x)>1$. Which we can simplify to $Ax+1<Ay$, and by adding $k\lambda$ to both sides we get $Ax+k\lambda+1<Ay+k\lambda$. Where $k$ is an integer and $\lambda$ is an irrational number.

Now by using the Archimedean property, we can get $M\in \mathbb N$ such that $M>Ax+k\lambda$. Let $m$ be the smallest natural number which satisfies the above property. Therefore, we can get $m>Ax+k\lambda\geq m-1$. By rearranging we get, $m+1>Ax+k\lambda+1\geq m$.

Now, let's analyse what we have. First, we have $m>Ax+k\lambda$. Then we have $Ax+k\lambda+1\geq m$. COmbining we get, $Ax+k\lambda+1\geq m>Ax+k\lambda$. We also have $Ay+k\lambda>Ax+k\lambda+1$. We can combine all to get, $Ay+k\lambda>m>Ax+k\lambda$. Which can be simplified to, $Ay>m-k\lambda>Ax$.

How can I eliminate the $A$ in my last statement?
Also, I didn't use the fact, $k\lambda \leq n < (k+1)\lambda$ in my solution. Where can I use it?

Answer 1

Dividing by $\lambda$, you wish to show $\{\frac{n}{\lambda} : n \ge 1\}$ is dense in $[0,1]$, where $\{x\}$ denotes the fractional part of $x$. Since $\lambda$ is irrational iff $\frac{1}{\lambda}$ is irrational, we wish to show $\{\lambda n : n \ge 1\}$ is dense in $[0,1]$ for any irrational $\lambda$. Fix some $\epsilon > 0$, and take $N \ge 1/\epsilon$. By pigeonhole, there are $n_1 \not = n_2$ so that $\{\lambda n_1\},\{\lambda n_2\}$ are in the same interval $[\frac{j}{N},\frac{j+1}{N}]$ ($0 \le j \le N-1$) (since there are infinitely many choices for $n$ and of course only finitely many intervals of the form $[\frac{j}{N},\frac{j+1}{N}]$ with $0 \le j \le N-1$). Say $\{\lambda n_2\} > \{\lambda n_1\}$. Then $\{\lambda (n_2-n_1)\} \in [0,\frac{1}{N}]$, and therefore, for any $0 \le i \le N-1$, we have that $\{\lambda (kn_2-kn_1)\} = \{k\lambda (n_2-n_1)\} \in [\frac{i}{N},\frac{i+1}{N}]$ for some $k \ge 1$.

Answer 2

Prove that the set of all $\phi(n)$ is dense in the interval $[0,\lambda]$.

For such claims I have the following simple proof. Let $0\le x<y\le\lambda$ be any numbers and $\varepsilon=y-x>0$. Pick any natural $N\ge \tfrac \lambda\varepsilon$. Then among $N+1$ points $\phi(1),\dots,\phi(N+1)$ of the segment $[0,\lambda]$ there exist two, say $\phi(n’)$ and $\phi(n’’)$ with $n’<n’’$ such that $0\le \phi(n’’)-\phi(n’)<\varepsilon$. By the definition of the function $\phi$, we have $\phi(n’’-n’)-(\phi(n’’)-\phi(n’))=s\lambda$ for some integer $s$. On the other hand, $|\phi(n’’-n’)-(\phi(n’’)-\phi(n’))|<\lambda$, so $s=0$ and $\phi(n’’-n’)=(\phi(n’’)-\phi(n’))<\varepsilon$. Put $n=n’’-n’$. If $\phi(n)=0$ then $n-k\lambda=0$ for some integer $k$ and so $\lambda$ is rational, a contradiction. Similarly to the above and by the induction with respect to $\ell$ we can show that $\phi(\ell n)=\ell\phi(n)$ for each natural $\ell$ such that $\ell\phi(n)<\lambda$. Since the distance between two consecutive points $\phi(\ell n)$ and $\phi((\ell+1) n)$ is $\phi(n)<\varepsilon$, we have $\phi(\ell n)\in (x,y)$ for some $\ell$.