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

remmer
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3 Answers3

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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.
Gigili
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Suppose $f:\mathbb R^2\to\mathbb R$ is a function. If $f$ is injective then there exist points $a<b<c$ 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 $\mathbb R^2\setminus\{P\}$.

Jonas Meyer
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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.

  • This machinery is not really needed, as Chandrasekhar's approach could be formalized as "if such an $f$ exists, then $\mathbb{R}^2 - {(0,0)}$ is connected, therefore $f(\mathbb{R}^2 - {(0,0)}) = \mathbb{R} - {f(0,0)}$ is the image of a connected set by a continuous function and therefore is connected, but $\mathbb{R}$ minus a point is never connected. – Najib Idrissi May 29 '12 at 16:45
  • @N.I right you are, thanks. –  May 29 '12 at 16:46