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Does there exist a function $f:[0,1]\to [0,1]$ such that the graph of $f$ is dense in $[0,1]\times[0,1]$?

Not necessarily continuous.

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The image is a subset of [0,1]. Then you want it to be dense in [0,1]. – alpha.Debi May 2 '12 at 7:46
@alpha.Debi: Perhaps OP means the "graph" of $f$, so $$\{\big(x,f(x)\big):x\in[0,1]\}\subset [0,1]\times[0,1].$$ – anon May 2 '12 at 7:51
Yes, is in the graph, sorry. I have a mistake when I was typing. – Gastón Burrull May 2 '12 at 8:06
Would Hilbert curves be something of interest? – Jean-Luc Bouchot May 2 '12 at 8:49
@Jean-Luc: No, those are (images of) functions $[0,1]\to[0,1]\times[0,1]$. – anon May 2 '12 at 9:12

Yes. Let $Q=[0,1]\cap\Bbb Q$, the set of rational numbers in $[0,1]$; $Q$ is countable, so we can enumerate it as $Q=\{q_n:n\in\Bbb Z^+\}$. Let $P$ be the set of prime numbers, and let $A=\{\sqrt{p}:p\in P\}$; $A$ is a countably infinite set of real numbers with the property that $a-b$ is irrational whenever $a,b\in A$ with $a\ne b$.

For each $a\in A$ let $Q_a=\{x\in[0,1]:x-a\in\Bbb Q\}$; For $n\in\Bbb Z^+$ let $a_n=\sqrt{p_n}$, where $p_n$ is the $n$-th prime, and let $Q_n=\{x\in[0,1]:x-a_n\in\Bbb Q\}$; the sets $\{Q_n:n\in\Bbb Z^+\}$ are pairwise disjoint, and each is dense in $[0,1]$. Now let $$f:[0,1]\to[0,1]:x\mapsto\begin{cases} q_n,&\text{if }x\in Q_n\\\\ 0,&\text{if }x\in[0,1]\setminus\bigcup_{n\in\Bbb Z^+}Q_n\;; \end{cases}$$

the construction ensures that the graph of $f$ is dense in $[0,1]\times[0,1]$, since it is dense on a dense set of horizontal slices through the square.

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Graph of any discontinous linear function is dense in $\mathbb{R}^2$, let $f :\mathbb{R}\rightarrow\mathbb{ R}$ is a function such that$ f(x + y) = f(x) + f(y)\forall x,y\in\mathbb{R}$. If $f$ is cont, then of course it has to be linear. But here $f$ is NOT cont. Then show that the set $\{(x, f(x)) : x\in\mathbb{R}\}$ is dense in $\mathbb{R}^2.$ take $x_1\neq 0$, If $f$ is not of the form $f(x)=cx$, then there exist $x_2\neq 0$ such that $f(x_1)/ x_1\neq f(x_2) /x_2$ In other words if you write in the determinant form in which first and second row are $(x_1, f(x_1))$ , $(x_2, f(x_2))$ respectively, then the determinant will be non zero. So the vectors $v_1 =(x_1, f(x_1))$ and $v_2 =(x_2, f(x_2))$ are linearly indipendent and thus span the whole plane $\mathbb{R}^2$. From Now on I guess denseness of rationality will help!

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See also here. – Martin Sleziak May 2 '12 at 14:26

Conway's base-13 function takes on every real number in any closed interval $[a,b]$. So you just have to restrict its range to $[0,1]$ to get your function $f$.

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Some of the examples given as answers to this question or in a related MO thread have the same property; so they might be interesting for the OP too. – Martin Sleziak May 2 '12 at 14:28

If $x$ is a terminating decimal, $x=.d_1d_2\dots d_n$, let $f(x)=.d_n\dots d_2d_1$. Define $f$ any way you want to elsewhere. Can you show that the image is dense?

EDIT: As Robert points out in his comment, this doesn't quite work. Perhaps this works instead: if $x=.d_1d_2\dots d_{2n}$ is a terminating decimal with an even number of digits (not counting any terminating zeros), let $f(x)=.d_{n+1}d_{n+2}\dots d_{2n}d_1d_2\dots d_n$. Again, for other $x$, define $f(X)$ arbitrarily.

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Yes, I can. Is a nice function. But is dense in the image I'm not sure if graph is dense in [0,1]x[0,1], I have a mistake when I wrote question. I need a function such graph is dense in [0,1]x[0,1]. Sorry. – Gastón Burrull May 2 '12 at 8:06
@GerryMyerson: This almost works. The problem is that if trailing zeros are part of the terminating decimal, the function is not well-defined; if they are not, the first digit of $f(x)$ can't be $0$. – Robert Israel May 2 '12 at 8:56
It works if you take (for $n > 1$, and not counting trailing zeros) $f(x) = .d_{n-1}d_{n-2} \ldots d_2 d_1$. – Robert Israel May 2 '12 at 18:03

For a probabilistic example, let $\{U_n\}$ be an independent and identically distributed sequence of random variables with the uniform distribution on $[0,1]$. Enumerate the rationals as $\{q_n\}$, and set $f(q_n) = U_n$ and $f(x) = 0$ for irrational $x$. I claim $f$ has dense graph with probability 1.

Consider an open square $(a,b) \times (c,d) \subset [0,1]^2$, with $a,b,c,d$ rational (and half-open intervals at the boundary, like $(a,1]$, etc, also allowed). Choose infinitely many rationals $\{q_{n_k}\}$ lying in $(a,b)$. For each $k$, the probability that $U_{n_k} \in (c,d)$ is $d-c > 0$. By independence, the probability that at least one of $U_{n_1}, \dots, U_{n_m}$ is in $(c,d)$ is $1-(1-(d-c))^m$. As $m \to \infty$, this tends to 1, and so with probability 1, there exists some $k$ with $f(q_{n_k}) = U_{n_k} \in (c,d)$, i.e. the graph of $f$ contains a point of $(a,b) \times (c,d)$.

Since there are countably many rational squares, the event that the graph of $f$ contains a point in every rational square is a countable intersection of events of probability 1, hence itself has probability 1. But every open set in $[0,1]^2$ contains a rational square, so on this event, the graph of $f$ is dense.

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