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is it possible to construct a continuous bijective map from $\mathbb{R}$ to $\mathbb{R^{2}}$. if it is, please give an example.If not, how to prove?

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the inverse would not be continuous of course, else it's a homeomorphism. –  Alex Jan 2 '13 at 21:07
    
Do you want the map to be simply 1-1? Or do you want it also to be onto? –  1015 Jan 2 '13 at 21:09
    
sorry, I meant it to be bijective, but the Peano curve is not. –  Alex Jan 2 '13 at 21:13
    
@Beni: A space-filling curve is never injective. –  Brian M. Scott Jan 2 '13 at 21:13
    
@Alex: you might also want to change your title. –  1015 Jan 2 '13 at 21:38

2 Answers 2

up vote 5 down vote accepted

The answer is no.

Suppose that $f:\Bbb R\to\Bbb R^2$ is a continuous bijection. For $n\in\Bbb Z^+$ let $I_n=[-n,n]$, and let $K_n=f[I_n]$. Then $\Bbb R^2=\bigcup_{n\in\Bbb Z^+}K_n$, so by the Baire category theorem some $K_m$ has non-empty interior. Choose $x$ and $r$ so that $B(x,r)\subseteq K_m$, and let $g=f\upharpoonright I_m$. Then $g$ is a continuous bijection from the compact set $I_m$ onto $K_m$, so $g$ is a homeomorphism. (In case this isn’t familiar, note that $g$ is closed: every closed subset of $I_m$ is compact, $g$ preserves compactness, and every compact subset of $K_m$ is closed, since $K_m$ is compact. Finally, a closed, continuous bijection is clearly a homeomorphism.)

Let $J=g^{-1}[B(x,r)]$; $B(x,r)$ is connected and open, so $J$ is an open interval in $I_m$ homeomorphic to $B(x,r)$. But this is impossible: $J$, being an interval, has cut points, and $B(x,r)$ has none.

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If $f:\mathbb{R}\to \mathbb{R}^2$ is continuous and injective, then $f$ cannot be surjective. The reason is as follows. For each $n\in \mathbb{N}$, since $[-n,n]$ is compact, it is homeomorphic to $f([-n,n])$. Then due to invariance of domain, $f([-n,n])$ is a nowhere dense closed subset of $\mathbb{R}^2$. Therefore, by Baire category theorem, $f(\mathbb{R})=\cup_{n=1}^\infty f([-n,n])$ is also nowhere dense in $\mathbb{R}^2$.

Remark: With an almost identical argument, the result can be generalized as follows. If $m\ne n$, then there is no continuous bijection from $\mathbb{R}^m$ to $\mathbb{R}^n$.

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@ThomasAndrews: Please refer to this. –  23rd Jan 2 '13 at 21:30
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@EduardoSiva: Please see the "Consequences" paragraph here, which implies that $f([-n,n])$ has no interior point in $\mathbb{R}^2$. –  23rd Jan 2 '13 at 21:43
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Invariance of domain isn’t actually necessary. –  Brian M. Scott Jan 2 '13 at 22:03
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@richard a interval $I\in R$ do not be homeomorphic to a ball $B\in \mathbb{R}$ because if you take a point of each, we $B-{p}$ have conected and $I-{q}$ is not conected –  user52188 Jan 2 '13 at 22:10
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@richard my point is that you could argue more simply, as has been shown, in my view you used a bazooka to kill an ant. –  user52188 Jan 2 '13 at 22:28

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