Why weren't continuous functions defined as Darboux functions?
Solution 1:
You can't actually draw many Darboux functions without lifting your pen from paper. The extreme example is the Conway base 13 function, which takes every value in every interval and hence is a Darboux function. You can't however begin to even imagine what this function looks like, let alone draw it with a pen.
So you may want to require the function to be bounded. However this still fails, since you can get functions like,
$$ f(x) = \begin{cases} \sin (1/x), & \text{if } x \neq 0 \\ 0 & \text{if } x = 0 \end{cases} $$
Which again, you can't draw. So one idea may be to require functions $f:[a,b] \rightarrow \Bbb R$ to be of bounded variation, meaning that,
$$ \sup\left\{\sum_{i=1}^n |f(x_i)-f(x_{i-1})| \;\middle|\; a=x_0<x_1<\dots<x_n \right\} < \infty $$
Intuitively, this 'measures' the graph of the line $\{(x,f(x)) \mid x \in [a,b] \}$ and requires it to be finite. Then we intuitively should get a finite length curve which we can actually draw without lifting your pen.
It turns out however, that Darboux functions which have bounded variations are actually continuous. So in attempt to define continuity in a more intuitive way, we have found a more restrictive definition.
Even worse, this still isn't enough. One can show that the function,
$$ f(x) = \begin{cases} x^3\sin (1/x), & \text{if } x \neq 0 \\ 0 & \text{if } x = 0 \end{cases} $$
is continuous and has bounded variation, but you can't really draw it. Note this function is also differentiable and has continuous first derivative. I'll spare you the details, but using similar ideas we can also construct infinitely differentiable functions of bounded variation, which you can't draw on paper.
From this, I think you can see why we don't try to model continuity off the intuitive definition. Instead, we adopt the usual definition because it's a much more useful and interesting class of functions to work with.
Solution 2:
I agree with ctoi's and vadim's answers. Let me just add a few things.
The $\varepsilon$-$\delta$ definition is rigged so that if $f : X \rightarrow Y$ is a function between metric spaces, then the following are equivalent:
- $f$ is $\varepsilon$-$\delta$ continuous
- $f$ preserves the "converges to" relation; meaning that from $x \rightarrow x_\infty$ we can deduce $f(x) \rightarrow f(x_\infty)$.
I think that Condition 2 is closer to what we really use in practice, e.g. we often want to commute $\frac{d}{dx}$ across an infinite summation, or change the order of integration, etc. So as a general rule, I think that when we're thinking about continuity, we should keep in mind its mathematical purpose, namely, the preservation of limits.
Taking this philosophy to its logical conclusion, convergence spaces are a very very natural structure to study. Modulo certain largely irrelevant size issues, the idea is basically that:
a convergence space is a set $X$ together with a distinguished collection of ordered pairs $(x,x_\infty)$ where $x$ is a net in $X$ and $x_\infty$ is an element of $X$. The idea is that $(x,x_\infty)$ is in this collection iff $x$ converges to $x_\infty$; we write $x \rightarrow x_\infty$. Certain axioms are imposed.
a morphism of convergence spaces is a continuous function, i.e. a function $f : X \rightarrow Y$ such that for all nets $x$ in $X$ and all elements $x_\infty$ in $X$, we have $$(x \rightarrow x_\infty) \rightarrow (f(x) \rightarrow f(x_\infty))$$
It was this viewpoint (namely, that a continuous function is, by definition, a morphism of convergence spaces) that allowed me to finally make peace with the $\varepsilon$-$\delta$ definition of continuity for mappings between metric spaces.
By the way, the category $\mathbf{Conv}$ of convergence spaces is better-behaved categorially than the category of topological spaces $\mathbf{Top}$; in particular, $\mathbf{Conv}$ is Cartesian closed, while $\mathbf{Top}$ famously isn't. Its almost as if category theory is trying to tell us that convergence is secretly the correct way of thinking about continuity (cue X-Files theme music).
Solution 3:
The Darboux definition does not correspond very well with our intuition about continuity. For example, the Conway function takes on every value in every interval, and is therefore Darboux. However it is not continuous, and I don't think we want it to be continuous, because it certainly doesn't agree with your teacher's definition of drawing without lifting your pencil.