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Is there a continuous bijection from $[0,1]$ onto $[0,1] \times [0,1]$?
That is with $I=[0,1]$ and $S=[0,1] \times [0,1]$, is there a continuous bijection $$ f: I \to S? $$

I know there is a continuous bijection $g:C \to I$ from the Cantor set $C$ to $[0,1]$.
The square $S$ is compact so there is a continuous function $$ h: C \to S. $$ But this leads nowhere.
Is there a way to construct such an $f$?

I ask because I have a continuous functional $F:S \to \mathbb R$.
For numerical reason, I would like to convert it into the functional $$ G: I \to \mathbb R, \\ G = F \circ f , $$ so that $G$ is continuous.

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Any continuous $f:I\to S$ produces a continuous $G$. One can even arrange $f$ is surjective. Is there some further reason $f$ needs to be bijective? –  Bob Pego Oct 1 '13 at 0:47
    
@BobPego The bijection requirement was a first instinct. You made me realized I can forgo it. Thank you for your insightfull comment. –  Nicolas Essis-Breton Oct 1 '13 at 0:56
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By the way, there is no continuous bijection from $C$ to $I$ since $C$ is compact, $I$ is Hausdorff, and $I$ is connected while $C$ is not. –  Stefan Hamcke Oct 1 '13 at 1:30
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Side note: A functional maps functions to $\mathbb R$, e.g. $F: C^\infty(S)\to\mathbb R$ –  Tobias Kienzler Oct 1 '13 at 9:19
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Something can also be found in this older question. @TobiasKienzler I have removed my previous two comments. Since the older question is already closed, they seem to be obsolete now. –  Martin Sleziak Oct 1 '13 at 12:12

4 Answers 4

up vote 26 down vote accepted

No, such a bijection from the unit interval $I$ to the unit square $S$ cannot exist. Since $I$ is compact and $S$ is Hausdorff, a continuous bijection would be a homeomorphism. But in $I$ there are only two non-cut-points, whereas in $S$ each point is a non-cut-point.

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Hint: Consider what happens to the connected $[0,1]$ if the point $\frac12$ is removed. What happens to $[0,1]\times[0,1]$ when $f(\frac12)$ is removed?

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No. The easiest was to see this is to first notice that $[0,1]^2\setminus \{x\}$ is connected for any $x \in [0,1]^2$.

It is easier (for me) to work with $\phi = f^{-1}$. However I must show that $\phi$ is continuous. Suppose $y_n \to y$, then I must show that $\phi(y_n) \to \phi(y)$. Let $x_n = \phi(y_n), x = \phi(y)$. One slightly technical way is to show that every subsequence of $x_n$ contains a further subsequence that converges to $x$. From this we will conclude that $\phi$ is continuous.

Suppose $x_{n_k} \to z$. Since $f$ is continuous, we have $y_{n_k} = f(x_{n_k}) \to f(z) = y$. Hence $z = x$. (So, in fact, the entire sequence, not just a subsequence, converges to $x$.) Hence $\phi$ is continuous.

Now suppose $\phi:[0,1]^2 \to [0,1]$ is a continuous bijection. Let $x = \phi^{-1} (\frac{1}{2} )$, then $\phi([0,1]^2\setminus \{x\})$ is connected, however we see that $\phi([0,1]^2\setminus \{x\}) = [0,\frac{1}{2}) \cup (\frac{1}{2},1]$ which is not connected. Hence a contradiction.

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technically he asked for a continuous bijection the other way, so you'd have to show that the inverse function is also continuous. –  Deven Ware Oct 1 '13 at 0:44
    
@DevenWare: Thanks - I'm just processing that now! –  copper.hat Oct 1 '13 at 0:44
    
@copper.hat Your answer shows an other side of the die. Please let it live. –  Nicolas Essis-Breton Oct 1 '13 at 1:09
    
@NicolasEssis-Breton: :-). –  copper.hat Oct 1 '13 at 1:13

As the other answers state, there is no bijection. However, since you mention numerics, an approximation might be of interest:

The Lissajous curve $\begin{pmatrix}\sin(at+\delta)\\\cos(bt)\end{pmatrix}$ for an irrational ratio $a/b$, e.g. $a=1, b=\sqrt2$, is not closed and therefore maps $\mathbb R$ to a dense subset of $[0,1]^2$. Now take one of the usual $\mathbb [0,1]\to\mathbb R$ mappings, e.g. $t = \tan(\pi(u-\frac12))$ or $\operatorname{artanh}(2u-1)$, to obtain a map from $[0,1]$ to a dense subset of $[0,1]^2$. Now I wonder if there is an analysical formula to obtain the $t$ best approximating a given $(x,y)$...

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