Is the following proof correct? The Definition of $ε$-steady is given below.
Proposition. The sequence $\left(\frac{1}{10^n}\right)_{n=1}^{\infty}$ is $0.1$-steady.
Proof. Let $j,k\in\mathbf{Z}^+$, furthermore let $x=\max(j,k)$ and $y = \min(j,k)$ consequently $y=x+r$ for some $r\in\mathbf{Z}^+$. Now Consider the following $$d(j,k) = d(x,y) = \left|\frac{1}{10^x}-\frac{1}{10^y}\right| = \left|\frac{1}{10^x}-\frac{1}{10^{x+r}}\right| = \left|\frac{1}{10^x}\right|\cdot\left|1-\frac{1}{10^r}\right|.$$ Now since $x,r\in\mathbf{Z}^+$ consequently $a=\left|\dfrac{1}{10^x}\right|\leq \dfrac{1}{10}$ and $b=\left|1-\dfrac{1}{10^r}\right|<1$ thus $ab\leq\dfrac{1}{10}$. Since our choice of $j,k$ was arbitrary it follows that $d(x,y)\leq 0.1,\forall j,k\in\mathbf{Z}^+$.
$\blacksquare$
Definition. A sequence $(a_n)_{n=m}^{\infty}$ is $ε$-steady if and only if Given a $\epsilon>0$ we have$$d(j,k) = |a_j-a_k|\leq ε,\forall j,k\in\{n\in\mathbf{N}|n\ge m\}.$$