# Continuous bijection from $\mathbb{R}^{2} \to \mathbb{R}$

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

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–  David Mitra May 29 '12 at 16:29

• 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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Please use $\LaTeX$ next time you want to copy your comment into an answer. –  Gigili May 29 '12 at 16:50
@Gigili: Are the cardinal numbers useful here at this problem? Thanks –  Babak S. May 29 '12 at 17:01

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 $\mathbb R^2\setminus\{P\}$.

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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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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. –  user20266 May 29 '12 at 16:46