Let $m$ be arbitrary positive rational number. Show that the sequence $(mn)_{n=1}^{\infty}$ is not $\epsilon$-steady for any $\epsilon>0$

Question

The proposition I would like to prove:

Proposition 1. For arbitrary positive rational $m$, the sequence $(nm)_{n=1}^{\infty}$ is not $\epsilon$-steady for any positive rational $\epsilon$.

Couple of notes:

1. I will use notation $d(x,y)$ to refer to the distance between rational numbers $x,y$, i.e., $d(x,y) = |x-y|$.

2. The definition of $\epsilon$-steadiness presented in the book: For some $\epsilon > 0$, the sequence $(a_{n})_{n=m}^{\infty}$ is $\epsilon$-steady if and only if for all natural numbers $j,k$ larger or equal to $m$, $d(a_{j},a_k) ≤ \epsilon$

My attempt:

We will first prove the following lemma:

Lemma 1. For arbitrary positive rational $m$, the sequence $(nm)_{n=1}^{\infty}$ is not $\epsilon$-steady for any positive natural number $\epsilon$.

We will use induction. For the base case, consider $\epsilon := 1$ and suppose that the sequence is $\epsilon$-steady. If $m ≥ 1$, then $d(3m,m) = 2m > 1$, a contradiction. Suppose $m < 1$. Since $m$ is non-zero, there must be some natural number $k$ such that $\frac{1}{m} < k$, and so we have $1 < mk$. But then $d(m(k+1),m)= mk > 1$, a contradiction. So for any positive rational $m$, the sequence cannot be $1$-steady. Now inductively suppose that for some positive natural number $\epsilon$, the sequence is not $\epsilon$-steady. We want to show that the sequence cannot be $(\epsilon+1)$-steady. Since the sequence is not $\epsilon$-steady, it follows that the sequence must contain some elements $a_j,a_k$ (without loss of generality, let's suppose that $a_j > a_k$), such that $|a_j - a_k| = a_j - a_k > \epsilon$. Since $a_j,a_k$ are multiples of $m$, the rationals $3a_{j},3a_{k}$ will also be multiples of $m$, and thus $3a_{j},3a_{k}$ are also elements of the sequence $(nm)_{n=1}^{\infty}$. Furthermore, we have $d(3a_j,3a_k) = |3a_j - 3a_k| = 3a_j - 3a_k > 3\epsilon$, and since $\epsilon ≥ 1$, $3\epsilon > \epsilon + 1$, which implies that $d(3a_j,3a_k) > \epsilon + 1$, but that means that the sequence is not $(\epsilon+1)$-steady. $\Box$

Now back to the proposition 1:

Take arbitrary positive rationals $m,\epsilon$ and suppose that the sequence $(nm)_{n=1}^{\infty}$ is $\epsilon$-steady. Since $\epsilon$ is rational, we know that there must exist some natural number $k$ such that $\epsilon < k$. By lemma 1, we know that the sequence is not $k$-steady, implying that there are some elements $a_i, a_j$ such that $d(a_i,a_j) > k$. But then clearly, $d(a_i,a_j) > \epsilon$, a contradiction. $\Box$

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Question 1.

Is the proof correct?

Question 2.

Are there other alternatives to proving the proposition?

Answer 1

Possibly an easier approach: If $(a_n)_{n=m}^\infty$ is $\epsilon$-steady, then $(a_n)_{n=m}^\infty$ is bounded. That implies $(a_n)_{n=1}^\infty$ is bounded. But in your situation, $(nm)_{n=1}^\infty$ is unbounded.