When we were in primary school, teachers showed us graphs of 'continuous' functions and said something like

"Continuous functions are those you can draw without lifting your pen"

With this in mind I remember thinking (something along the lines of)

"Oh, that must mean that if the function takes two values $f(y)<f(z)$ then for every $c$ between $f(y), f(z)$ there must be some $x\, (y<x<z)$ such that $f(x)=c$"

And that's what I thought a continuous function was. But then the $\epsilon$-$\delta$ definition appeared, which put a more restrictive condition on what a continuous function was.

So, my question is, given the fact that Darboux functions "seem continuous" (in some subjective sense, I guess), why wasn't this used as the definition of continuity? More generally, how did today's (analytical) definition of continuity appear?

  • $\begingroup$ Somebody did propose that as the definition of continuity. I don't remember exactly who, maybe Maxwell. It turned out not to work. $\endgroup$ Apr 6, 2016 at 19:10
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    $\begingroup$ May I ask why the downvote? $\endgroup$ Apr 6, 2016 at 19:15
  • $\begingroup$ @DavidMitra Yup, I know. My question has nothing to do with that (I even said that in the part of the "more restrictive condition $[\dots]$"). What I'm asking is why wasn't the term "Continuous function" used for "Darboux function". $\endgroup$ Apr 6, 2016 at 19:17
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    $\begingroup$ I suspect you meant to write: "Oh, that must mean that if the function takes $2$ values $f(y)<f(z)$ then for every $c$ between $f(y)$ and $f(z)$, there must be some $x$ between $y$ and $z$ such that $f(x)=c$". Or maybe you're just reflecting accurately the troubles you had with with quantifier order when you were in primary school. :0) $\endgroup$ Apr 6, 2016 at 20:59
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    $\begingroup$ "Continuous functions are those you can draw without lifting your pen" is FALSE unless you require first that the domain is an interval in $\mathbb R$. For example $x\mapsto 1/x$ of $x\mapsto {\tan x}$ or $x\mapsto 1/{\sin x}$ are continuous, but their domains are not intervals, so their graphs consist of separate components and can not be drawn without lifting a pen. (They can not be drawn at all because they are infinite, but that's another story.) $\endgroup$
    – CiaPan
    Apr 7, 2016 at 10:27

5 Answers 5


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.

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    $\begingroup$ Why can't $x^3\sin (1/x)$ be drawn? $\endgroup$
    – Navin
    Apr 7, 2016 at 4:31
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    $\begingroup$ @Navin There's a bit of ambiguity here in terms of what can be drawn, but because $x^3\sin(1/x)$ has infinitely many local maxima and minima near 0, I would argue you can't draw all of them (even though they do become arbitrarily small). $\endgroup$
    – ktoi
    Apr 7, 2016 at 6:00
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    $\begingroup$ I will add that functions with all real values in every open interval have been mentioned in several other posts on this site. See, for example, the links given here. $\endgroup$ Apr 7, 2016 at 7:44
  • $\begingroup$ Another possible example of a continuous function that, in some sense "can't be drawn" is $$x \in (0,1/\pi) \mapsto \sin\left(\frac{1}{x}\right).$$ The problem, really, is that it takes an infinite amount of ink and/or time to actually draw it. $\endgroup$ Apr 8, 2016 at 13:38

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:

  1. $f$ is $\varepsilon$-$\delta$ continuous
  2. $f$ 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).


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.


I want to focus on your question "why wasn't this used as the definition of continuity?" In other words, you're asking why we don't interpret continuity to mean "having the intermediate value property."

One answer that hasn't been mentioned yet is that continuity can be a purely local phenomenon. That is, it makes sense to ask whether a function is continuous or discontinuous at a point. By contrast, one cannot ask whether a function possesses the intermediate value property at a point. My understanding is that in the early history of analysis, the local and global aspects of continuity often weren't carefully distinguished -- or rather, they weren't even recognized as different. This is why, for example, many could confuse continuity and uniform continuity. The latter is a strictly global phenomenon, too: it does not make sense to ask whether a function is uniformly continuous at a point. But as it became clear that continuity could be interpreted as a pointwise phenomenon, these different shades of regularity -- local versus global -- came into view.


Of course, since I mentioned shades of regularity, I might as well point out that the intermediate value property tells you nothing about smoothness (differentiability). If you want continuity to be the first step in a hierarchy of degrees of differentiability, you need to look for a different notion of continuity than the IVP. I don't know exactly how the history played out, but you can see why, as it became clear that differentiability too is a pointwise phenomenon, one would have looked for a notion of pointwise continuity to fit into this framework. (Compare this to the search for the right notion of differentiability for multivariable functions: you want to pick out the right definition to guarantee continuity.)


Yet another (quite fatal) problem with Darboux functions is that the class of such functions isn't closed under many operations. Continuous functions form a vector space, but Darboux functions do very much not:

Every function from the reals to the reals is the difference of two Darboux functions.

This result can be found in the delightful book [Van Rooij, Arnoud CM, and Wilhelmus Hendricus Schikhof. A second course on real functions. Cambridge University Press, 1982.].


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