I am learning multi-var calculus using the textbook "Vector Calculus, Linear Algebra, and Differential Forms: A Unified Approach". Here is a definition that's puzzling me.
Let $\mathbb{D}$ be the set of finite decimals, which consists of finite decimal expansions.
A mapping $f: \mathbb{D}^{n} \to \mathbb{D}$ will be considered $\mathbb{D}$-continuous (or "finite decimal continuous") if, for any given integers $N$ and $k$, there exists an integer $l$ such that the following condition holds:
For any two elements $(x_1, \ldots, x_n)$ and $(y_1, \ldots, y_n)$ in $\mathbb{D}^{n}$, where $|x_i|, |y_i| < N$ for all $i = 1, \ldots, n$, and if $|x_i - y_i| < 10^{-l}$ for all $i = 1, \ldots, n$, then it must also satisfy:
$|f(x_1, \ldots, x_n) - f(y_1, \ldots, y_n)| < 10^{-k}$
Ex: $A(x,y) = x+y$ is $\mathbb{D}$ contineous.
I am confused about a few points: is $\mathbb{D}^n$ just the cartesian product of this set? So is the definition saying that for any $N,k$, we must find some $l$ such that if we take two distinct elements of the set $\mathbb{D}^n$, we have that the $i'th$ tuple of each element is less than $N$, and the diffrence between the $i'th$ element of each tuple with the other is less than $\frac{1}{10^l}$, we must get that the diffrence between the images of these two elements is less than $\frac{1}{10^k}$?
Here is an example to illustrate what I think this is saying:
Take $A: \mathbb{D}^2 \longrightarrow \mathbb{D}$ where $x,y \mapsto x+y$. Let $N,k \in \mathbb{Z}$, and let $l = \log_{10}(2)+k+1$. Then let $d_{1}, d_{2} \in \mathbb{D}^2$ be given by $d_{1} = (x_{1},x_{2})$ and $d_{2} = (y_{1},y_{2})$ such that $x_{1},x_{2} < N, y_{1},y_{2}<N$.Then $f(d_{1}) = x_{1} + x_{2}$ and $f(d_{2}) = y_{1} + y_{2}$. Then suppose further that $|x_{1} - y_{1}|<\frac{1}{10^l}$ and $|x_{2}-y_{2}|<\frac{1}{10^l}$. Then we get that $|f(x_{1},x_{2}) - f(y_{1},y_{2})| = |x_{1} + x_{2} - y_{1} - y_{2}|$.
Notice that $|x_{1} + x_{2} - y_{1} - y_{2}| \leq |x_{2}-y_{2}|+|x_{1} - y_{1}| < \frac{2}{10^l}=\frac{2}{10^{\log_{10}(2)+k+1}} = \frac{1}{10^{k+1}}<\frac{1}{10^k}$. Thus the function $A$ is $\mathbb{D}$ contineous.
I'm seeking an explanation as to what this means, and why are we developing this definition. Is this a generalization of continuity? Why focus on finite decimals?
Why the set $\mathbb D$, I have no idea. I assume the author plans to use it for something. Possibly to define a function only on such tuples and later extend to all of $\mathbb R$.
The definition you wrote is exactly uniform continuity on bounded sets. If you were working on $\mathbb R^n$ that would be equivalent to plain continuity, due to compactness. But $\mathbb D$ is not complete.
The functions that are uniformly continuous on bounded sets on $\mathbb D^n$ are precisely those that can be extended to a continuous function on $\mathbb R^n$. I'll do the one-dimensional case to simplify notation. If $f$ is $\mathbb D$-continuous and $x\in\mathbb R$, choose $\{x _j\}\subset\mathbb D$ with $x_j\to x$. Being convergent, the sequence is bounded. This implies that $\{f(x_j)\}$ is Cauchy. Indeed, given $\varepsilon>0$ choose $k$ such that $10^{-k}<\varepsilon$. As the sequence is Cauchy, there exists $j_0$ such that $|x_j-x_h|<10^{-l}$ for all $j,h>j_0$. Then $|f(x_j)-f (x_h)|<10^{-k}<\varepsilon$. Hence $\{f(x_j)\}$ is Cauchy. With almost the same argument as above one can see that the $\lim f(x_j)$ is the same for any sequence that converges to $x$. So we can define $\tilde f(x)=\lim f(x_j)$ and yet another similar argument shows that $\tilde f$ is continuous.