Courant and Fritz, Construction of the real numbers

[GodfreyHW]

In chapter 1, page 10, real numbers are found by confining them to an interval that shrinks to "zero" length (we consider subintervals $I_0,\,I_1,...,\,I_n$). Basically, if $x$ is between $c$ and $c+1$, then we can divide that interval into ten subintervals, and we can, then, have $c+\frac{1}{10}c_1\leq x\leq c+\frac{1}{10}c_1+\frac{1}{10}$, where $c_1$ is a digit from zero to nine.

Repeating this process, and making $n\to\infty$ subdivisions, we'll eventually get $x=c+0.c_1c_2c_3...$

I am confused by this, though:

1) How is he constructing that $I_{n+1}$ interval?
If I suppose $c_0-1\leq x$, then I can do $c_0-1+\frac{1}{10}c_1+...+\frac{1}{10^n}c_n\leq x$, and if I say that $n\to\infty$, then I can also write $\underbrace{(c_0-1+\frac{1}{10}c_1+...+\frac{1}{10^n}c_n)}_x-\frac{1}{10^{n+1}}\leq x$, so that $\frac{1}{10^{n+1}})$ is like an infinitesimal.
Would this be correct for $\left[x-\frac{1}{10^{n+1}},x\right]$? If so, then I guess that it should be the same with $x\leq c_0+1$ by subtracting $\frac{c_i}{10^i}$s and adding $\frac{1}{10^{n+1}}$.

2) How is he getting the second representation with $c_n-1$? (typo?)
I see that, maybe, we'd do $c_0\leq x$, and get it to $c_0+\frac{1}{10}c_1+...+\frac{1}{10^n}c_n\leq x$, but we would have $c_{i>0}=9$.

 

[anuttarasammyak]

As for 2)

For an example
0.123450000...
c_5=5
c_5-1=4
so it is equal to
0.123449999...

 

[GodfreyHW]

As for 2)

For an example
0.123450000...
c_5=5
c_5-1=4
so it is equal to
0.123449999...

Yes, thank you!