This is a definition:
$\Bbb D $ is the set of finite decimals
A mapping $f: \Bbb D^{n} \to \Bbb D\quad$ will be called finite decimal contiuous ($\Bbb D$-continuous) if for all intergers $N$ and $k$, there exists an integer $l$ such that if $(x_{1}, ...,x_{n})$ and $(y_{1}, ...,y_{n})$ are two elements of $\Bbb D^{n}$ with all $|x_{i}|$, $|y_{i}|$ < $N$, and if $|x_{i} - y_{i}| < 10^{-l}$ for all $i = 1, ..., n\quad$ then:
$$|f(x_{1}, ..., x_{n}) - f(y_{1}, ..., y_{n})| < 10^{-k}$$
At first glance it seems this is a n-dimensional generalization of the definition of continuity. Thinking it over a bit more I'm not sure what use $N$ has in this definition ? Could someone please provide a simpler or more intuitive explanation of what this definition is telling me?
I'm five years late to this discussion, but I believe my post may benefit other readers of the book where this problem appears.
The statement is from Hubbard's Vector Calculus
(https://www.amazon.com/Vector-Calculus-Linear-Algebra-Differential/dp/0136574467)
The main idea: We pick any integer $k$ and make $10^{-k}$ as small as we like. After we have fixed $k$, if $f$ is $\mathbb{D}$-continous, we can always find $l$, such that if we have two inputs of $f$, whose components differ by no more than $10^{-l}$, the output of the function will always be a number less than $10^{-k}$.
It's important to recognize that we are talking about decimal notation here. $10^{-k}$ is $0.00000...m$ (where $m$ is a digit) and will have k positions after the decimal point. $10^{-l}$ will have $l-1$ positions that are zeroes after the decimal point.
To further clarify the above definition, we will solve some of the exercises, proposed in the book.
First, we will to prove that addition is $\mathbb{D}$-continous. We can present addition as functions of two variables - $A(x,y)=x+y$. Then if we fix arbitrary integer $k$ we want to find $l$ such that
if $|x_1-y_1|<10^{-l}$ and $|x_2-y_2|<10^{-l}$ then $|A(x_1,x_2)-A(y_1,y_2)| = |(x_1 +x_2)-(y_1 + y_2)|=|(x_1-y_1)+(x_2-y_2)|<10^{-k}$
But this is easy. It's enough to take $l=k+1$. Then no matter what are the signs of $x_1-y_1$ and $x_2-y_2$, their sum cannot exceed $10^{-k}$.
For example if we take $k=4$ then $10^{-4}=0,0001$. If we have $|x_1- y_1|<0.00001$ and $|x_2-y_2|<0.00001$, then $|(x_1-y_1)+(x_2-y_2)|$ is less than $0.00002$ no matter what the signs of $(x_1-y_1)$ and $(x_2-y_2)$ are. And $0.00002$ is in turn less than $0.0001$.
Next we try to prove that $M(x,y)=xy$ is also $\mathbb{D}$-continous.
To set up things, we start with two pairs of numbers $(x_1,x_2)$ and $(y_1,y_2)$ and let $s_1=x_1-y_1$, $s_2=x_2-y_2$. We can then express $|x_1x_2-y_1y_2|$ as $|x_1x_2-(x_1-s_1)(x_2-s_2)| = |s_1x_2 + s_2x_1 - s_1s_2|$. This will be always smaller than $|s_1x_2|+|s_2x_1|+|s_1s_2|$. We might want to make each of these summands be less than $10^{-k-1}$ Then we are certain that the whole sum will be less than $10^{-k}$.
But what value of $l$ we should pick? Whatever value we pick, $s_1$ and $s_2$ must be less than it for the above to hold. But there is a problem, in two of the summands we have the numbers $x_1$ and $x_2$. Here we will use the N from the definition. It's upper bound for all x-es and y-s, so we can count the number of it's digits BEFORE the decimal point. Let's call this number $n$.
Now we shall make our $l$ account for the fact that our s-es are multiplied by a number with at most n digits. We will require l to be $k+1+n$.
So, if $l=k+n+1$, we will have $s_1<10^{-(k+n+1)}, s_2<10^{-(k+n+1)}$ So
$|x_1x_2-y_1y2|= |s_1x_2 + s_2x_1 - s_1s_2| <|s_1x_2|+|s_2x_1|+|s_1s_2|$