4
$\begingroup$

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 $\mathbb{R}^2$.

  • 0
    @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.2011-08-07

1 Answers 1

5

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 $f$.