Example of a continuous function from $(0,1) \times (0,1)$ to $\mathbb{R}^2$

Question

The question is: Give an example of a function $f$, continuous on $S=(0,1) \times (0,1)$, such that $f(S)=\mathbb{R}^2$.

I'm getting stuck on $f(x)=\tan\left(\pi \left(x-\frac{1}{2}\right)\right)$, which covers all of $\mathbb{R}$ but not all of $\mathbb{R}^2$.

Why wouldn't $\tan\left(\pi\left(x-\frac12\right)\right)\tan\left(\pi\left(y-\frac12\right)\right)$ work? — J. M. ain't a mathematician, Oct 22 '10 at 02:30
Given the second line, I think the OP actually wants a function from $ (0, 1) $, not $ (0, 1)^2 $. Is this what you intended? (There's a theorem stating that such a mapping can't be continuous, but I can't remember the reference.) — Vandermonde, Oct 22 '10 at 02:33
@J.M. Haha! Thank you for pointing out the obvious for me... this is why we have math-friends — JimJones, Oct 22 '10 at 02:46
FWIW, $\ln\left(\frac{x}{1-x}\right)\ln\left(\frac{y}{1-y}\right)$ works also. — J. M. ain't a mathematician, Oct 22 '10 at 02:59
The statisticians among you might already be recognizing a pattern in the "suggestions" I've given so far... — J. M. ain't a mathematician, Oct 22 '10 at 03:53
@J.M. not sure I have the lingo right, but aren't they continuous cumulative probability distribution functions or something like that? — Ross Millikan, Oct 22 '10 at 04:43
Given bijections $f : A \longleftrightarrow U$ and $g : B \longleftrightarrow V$ you have $f\times g : A\times B \longleftrightarrow U \times V$. — , Oct 22 '10 at 15:57
@Anon: I don't know if I understood well, but it seems like f(x):$\mathbb R \rightarrow \mathbb R$ and g(x)=2x , also $\mathbb R \rightarrow \mathbb R$ are counterexamples: your image points will all be of the form (x,2x), so that, e.g., the point (1,3) will not be in the range. — gary, Aug 07 '11 at 02:51

score 5 · Answer 1 · answered Nov 08 '10 at 12:49

In what follows I assume that a map $f:\ ]0,1[\to \mathbb R^2$ is wanted. $$ $$ There are continuous surjective maps $p:[0,1]\to[0,1]^2$; they are called Peano curves. One may choose $p$ in such a way that the "curve" begins at $(0,0)$ and ends at $(1,1)$. Therefore it is possible to construct a continuous map $f:\ ]0,1]\to\mathbb R^2$ such that the restrictions of $f$ to the intervals $[{1\over n+1},{1\over n}]$ are Peano curves covering an increasing sequence of squares $Q_n$ whose union is $\mathbb R^2$. At the end, the point $1$ may safely be omitted from the domain of $f$.

Example of a continuous function from $(0,1) \times (0,1)$ to $\mathbb{R}^2$

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