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Can anyone give an example of a continuous bijection from $\mathbb{R}^{2} \to \mathbb{R}$

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    A [similar question](http://math.stackexchange.com/questions/116350/continuous-injective-map-f-mathbbr3-to-mathbbr/)2012-05-29

3 Answers 3

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Ok, I will add my hint as an answer so that it's not unanswered.

  • Not possible is my guess. If you remove finitely many points from $\mathbb R^2$ it remains connected where as $\mathbb R$ does not. That should be a hint.
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    @Gigili: Are the cardinal numbers useful here at this problem? Thanks2012-05-29
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Suppose $f:\mathbb R^2\to\mathbb R$ is a function. If $f$ is injective then there exist points $a in the range of $f$. Let $P=f^{-1}(b)$. Then $\mathbb R^2\setminus\{P\}$ is the union of the disjoint nonempty sets $f^{-1}(-\infty,b)$ and $f^{-1}(b,\infty)$. If $f$ is also continuous, then these sets are open. This violates connectedness of $\mathbb R^2\setminus\{P\}$.

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It is a well known result from basic topology that a continuos injective (bijective) map $f:X\rightarrow Y$ from a compact space $X$ into a Hausdorff space $Y$ is closed (a homeomorphism).

If you apply this result in your case to any closed disc in $\mathbb{R}^2$ and it's image you see that your bijection is a local homeomorphism, hence a homeomorphism. Using, e.g., Chandrasekhar's reason you get a contradiction.

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    @N.I right you are, thanks.2012-05-29