I am reading "Mathematical Analysis 2nd Edition" by Tom M. Apostol.
Theorem 1.20. Assume $x\geq 0.$ Then for every integer $n\geq 1$ there is a finite decimal $r_n=a_0.a_1a_2\cdots a_n$ such that $$r_n\leq x<r_n+\frac{1}{10^n}.$$
Proof. Let $S$ be the set of all nonnegative integers $\leq x.$ Then $S$ is nonempty, since $0\in S,$ and $S$ is bounded above by $x$. Therefore $S$ has a supremum, say $a_0=\sup S.$ It is easily verified that $a_0\in S,$ so $a_0$ is a nonnegative integer. We call $a_0$ the greatest integer in $x$, and we write $a_0=[x].$ Clearly, we have $$a_0\leq x<a_0+1.$$
Now let $a_1=[10x-10a_0],$ the greatest integer in $10x-10a_0.$ Since $0\leq 10x-10a_0=10(x-a_0)<10,$ we have $0\leq a_1\leq 9$ and $$a_1\leq 10x-10a_0<a_1+1.$$
In other words, $a_1$ is the largest integer satisfying the inequalities $$a_0+\frac{a_1}{10}\leq x<a_0+\frac{a_1+1}{10}.$$
$\cdots$
The author wrote
$a_1$ is the largest integer satisfying the inequalities $$a_0+\frac{a_1}{10}\leq x<a_0+\frac{a_1+1}{10}.$$
Since $\#\{z\in\mathbb{Z}:a_0+\frac{z}{10}\leq x<a_0+\frac{z+1}{10}\}=1,$ and $a_1\in\{z\in\mathbb{Z}:a_0+\frac{z}{10}\leq x<a_0+\frac{z+1}{10}\}$, $a_1$ is certainly the largest integer satisfying the inequalities $$a_0+\frac{a_1}{10}\leq x<a_0+\frac{a_1+1}{10}.$$
So, the author is correct.
But I feel very strange.
Did the author mean the following?
$a_1$ is the largest integer satisfying the inequality $$a_0+\frac{a_1}{10}\leq x.$$