# Function defined everywhere but continuous nowhere

I have been recently reading about the Weierstrass function, a function that is continuous everywhere but differentiable nowhere. It made me think of a similar puzzle involving functions: find $$f: \mathbb R \to \mathbb R$$ such that $$f$$ can be computed anywhere, is well defined, but is continuous nowhere.

I first thought of maybe mapping the reals on to a fractal and doing something with that point but that’s just a fuzzy idea and I doubt one could compute it everywhere. In my research I could find no such function that is defined for all real numbers, both rational and irrational. If anyone has a proof this is impossible (or even just an idea of how you might prove that), or an example of a function that has those properties, that would be great.

• We were challenged in high school by our math teacher to ask a question he couldn't answer outright and in third grade when we were learning derivatives I asked for a function that is continuous everywhere but differentiable nowhere and he couldn't answer it :D He was a marvelous teacher but didn't care that much for maths higher than our level.
– chx
Oct 24, 2019 at 23:55
• @chx Is this a copypasta? Oct 25, 2019 at 2:06
• No, this is my life. I am still proud and thought it's funny. Sorry if not.
– chx
Oct 25, 2019 at 6:19
• I'm assuming by third grade, you mean the third year of high school. Otherwise, you don't live on planet earth Oct 30, 2019 at 22:27

First off, the "majority" of functions (where majority is defined properly) have this property, but are insanely hard to describe. An easy example, though, of a function $$f:\mathbb R\to\mathbb R$$ with the aforementioned property is $$f(x)=\begin{cases}x&x\in\mathbb Q\\x+1&x\notin\mathbb Q\end{cases}$$This example has the added benefit of being a bijection!

• Get an upvote for "This example has the added benefit of being a bijection!", you made my day! :) Oct 23, 2019 at 13:42
• Wait, but is this function computable? How do you compute if $x\notin \mathbb Q$? Oct 24, 2019 at 3:52
• @eyeballfrog I don't think OP means computable in the technical sense. Oct 24, 2019 at 4:10
• And in fact all computable functions (in the technical sense) are continuous everywhere. Oct 24, 2019 at 12:38
• I apologize, I did not know “computable function” was a technical term. This answer fits what I meant by my question, even though it may not fit what I technically asked. Oct 25, 2019 at 15:37

Consider the function $$f:\mathbb{R} \rightarrow \mathbb{R}$$ defined by

$$f(x) = \begin{cases} 1, ~~ x \in \mathbb{Q} \\ 0, ~~ x \not\in \mathbb{Q} \end{cases}$$

Now let $$x \in \mathbb{R}$$. Then there exists a sequence $$(x_n)_{n \in \mathbb{N}}$$ with $$x_n \rightarrow x$$ which is entirely contained in $$\mathbb{Q}$$ and a sequence $$(y_n)_{n \in \mathbb{N}}$$ with $$y_n \rightarrow x$$ which is entirely contained in $$\mathbb{R} \setminus \mathbb{Q}$$. Then both sequences converge to $$x$$, however the images of the elements in the sequence converge to $$1$$ and $$0$$, respectively.

• Wow! This is brilliantly simple. I’m kind of disappointed I couldn’t find this myself. Oct 23, 2019 at 13:38

G. Chiusole's & Olivier's example is the standard one.

In fact, there are functions $$\Bbb R \to \Bbb R$$ that are not only discontinuous at every point but spectacularly so: More precisely, there are functions $$f : \Bbb R \to \Bbb R$$ for which $$f(I) = \Bbb R$$ for every (nonempty) open interval $$I$$ no matter how small; thus in a sense they are as far from being continuous as possible. (Functions with this property are called strongly Darboux functions.) The classic example is the Conway base $$13$$ function:

The Conway base $$13$$ function is a function $$f : \Bbb R \to \Bbb R$$ defined as follows. Write the argument $$x$$ value as a tridecimal (a "decimal" in base $$13$$) using $$13$$ symbols as 'digits': $$0, 1, \ldots, 9, \textrm{A}, \textrm{B}, \textrm{C}$$; there should be no trailing $$\textrm{C}$$ recurring. There may be a leading sign, and somewhere there will be a tridecimal point to separate the integer part from the fractional part; these should both be ignored in the sequel. These 'digits' can be thought of as having the values $$0$$ to $$12$$, respectively; Conway originally used the digits "$$+$$", "$$-$$" and "$$.$$" instead of $$\textrm{A}, \textrm{B}, \textrm{C}$$, and underlined all of the base $$13$$ 'digits' to clearly distinguish them from the usual base $$10$$ digits and symbols.

• If from some point onward, the tridecimal expansion of $$x$$ is of the form $$\textrm{A} x_1 x_2 \cdots x_n \textrm{C} y_1 y_2 \cdots$$, where all the digits $$x_i$$ and $$y_j$$ are in $$\{0, \ldots, 9\}$$, then $$f(x) = x_1 \cdots x_n . y_1 y_2 \cdots$$ in usual base $$10$$ notation.
• Similarly, if the tridecimal expansion of $$x$$ ends with $$\textrm{B} x_1 x_2 \cdots x_n \textrm{C} y_1 y_2 \cdots$$, then $$f(x) = -x_1 \cdots x_n . y_1 y_2 \cdots$$
• Otherwise, $$f(x) = 0$$.
• (this question/answer thread recently rose to to top of the MSE queue) Regarding "as far from being continuous as possible", I believe it's slightly worse (but I haven't thought about it carefully), namely at each point the image of each unilateral neighborhood of that point is dense in ${\mathbb R},$ a property I pointed out for another function here. On the other hand, the Conway base $13$ function is a Baire $2$ function (continued) Sep 12, 2022 at 19:30
• (see here and here), and so it is a very simple Borel measurable function $(2$ is a very small countable ordinal). By the way, your profile says you're at Guilford College. I grew up somewhat near there (rural Cabarrus country; at various times in the late 1970s to early 1990s I was at UNCC, UNC, NCSU) and I recall that, despite its small size, Guilford was well known (at least in the 1970s) for its success in engaging undergraduate math students (and general topologists in nearby universities). Sep 12, 2022 at 19:40

There is a very simple example, the characteristic function of $$\mathbb{Q}$$, defined as follows :
$$f : x \mapsto \left\{ \begin{matrix} 1 & \textrm{if } x \in \mathbb{Q} \\0 & \textrm{otherwise} \end{matrix} \right.$$

You can get a whole bunch of functions like this (and some with even worse properties!) by inspecting the decimal representation of a number. To make sure these functions are well defined, we'll consider the decimal expansion of a terminating decimal to always end with $$00...$$ rather than the other possibility of ending in $$99...$$. The condition of continuity at non-terminating decimals $$x$$ means precisely that, for any bound $$\varepsilon$$, there is some $$N$$ such that every number $$x'$$ with the same first $$N$$ digits as $$x$$ has $$f(x)-f(x') < \varepsilon$$ (and, indeed, if $$f(x)$$ was also non-terminating, we can replace the $$\varepsilon$$ by a similar condition of agreement of digits). The case where $$x$$ is a terminating decimal is slightly different and annoying, so I won't talk about it.

As a starter, we can define a function $$f(x)$$ that writes $$x$$ in decimal, then counts how many $$9$$'s it has. If the count is finite, $$f(x)$$ is the count. If the count is infinite, $$f(x)=-1$$. This is discontinuous everywhere because knowing that $$x$$ and $$x'$$ share $$N$$ digits for any $$N$$ can, at best, tell you that they share some finite number of $$9$$'s - but the function takes into account every $$9$$ and we have no control after some point in the decimal expansion.

We can make the previous example somewhat worse by choosing a bijection $$k:\{-1,0,1,2,\ldots\}\rightarrow \mathbb Q$$ and then considering $$k\circ f$$ which now, one can check, has the property that the image of any open set is dense in $$\mathbb R$$. That's not very continuous at all!

Another fun example, along a similar line, would be to define $$f(x)$$ to be the number of places after the decimal point that the last $$9$$ in the representation of $$x$$ appears - or $$-1$$ if there are infinitely many $$9$$'s. You could even do worse and let $$f(x)$$ be the $$-1$$ if there are infinitely many $$9$$'s. If there is a last $$9$$, erase all the digits prior to it, leaving an infinite sequence of digits in $$\{0,1,\ldots,8\}$$. Write $$0.$$ before this sequence and interpret it in base $$9$$. Now, the image of every open set is $$[0,1]$$. That's pretty nasty. If you choose a bijection between $$[0,1]$$ and $$\mathbb R$$, now the image of every open set is $$\mathbb R$$.

There's also some examples that people actually do care about. For instance, there's a thing called irrationality measure which basically asks "How hard is this number to approximate by rationals?" The irrationality measure of $$x$$ is defined as the infimum of the $$\mu$$ such that $$0 < \left|x - \frac{p}q \right| < \frac{1}{q^{\mu}}$$ for infinitely many coprime pairs of integers $$(p,q)$$. This might be infinite, but you can always fix it up by mapping $$\infty$$ to some real number. This is $$1$$ at every rational, $$2$$ at algebraic irrationals, and can be anything at least $$2$$ elsewhere. This is actually useful as a tool for showing that things like that are like Liouville numbers (but not quite as extreme) are irrational - but the image of every open set is $$\{1\}\cup [2,\infty]$$, so a pretty nasty function.

Also: bonus, if you take any continuous function and add it to any discontinuous everywhere function, you get a discontinuous everywhere function - and if you take a discontinuous everywhere function and multiply it by a non-zero constant, it is still discontinuous everywhere. It turns out that, in the grand scheme of things, if you choose a function at random, the probability that it is continuous is $$0$$ - it's like randomly choosing a point on a plane and hoping that it lies on a line, except that instead of a "plane" you have an infinite dimensional space which is way bigger than the line.

• Bonus exercise: You can "improve" one of the examples to get a function $f$ with the property that, for open set $U$ and for any $x\in\mathbb R$, the set of $y\in U$ for which $f(y)=x$ is in bijection with $\mathbb R$. Oct 23, 2019 at 14:15

Consider the function $$f:\mathbb{R} \rightarrow \mathbb{R}$$ defined by $$f(x)=\begin{cases}x&x\in\mathbb Q\setminus\{0\}\\ -x&x\notin\mathbb Q \\ \sqrt{3}&x=0 \end{cases}$$

This function is not continuous for any $$x\in\mathbb R$$. Suppose $$x_0 \neq 0,$$ then by taking a sequence of rational numbers converging to $$x_0$$ and then a sequence of irrational numbers converging to $$x_0$$, you can see that $$\lim_{x\to{x_0}}f(x)$$ doesn't exist. As zero is a rational number, it is also a bijection.

• The title of the cited paper is "The Blancmange Function Continuous Everywhere but Differentiable Nowhere" (emphasis mine). Nov 17, 2019 at 19:06
• @XanderHenderson: Yes, I misread the question and have updated my answer. Nov 17, 2019 at 21:37