Continuous bijection from $\mathbb{R}^{2} \to \mathbb{R}$ Can anyone give an example of a continuous bijection from $\mathbb{R}^{2} \to \mathbb{R}$
 A: 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 $\mathbb R^2\setminus\{P\}$.
A: 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.
A: Ok, I will add my hint as an answer so that it's not unanswered.

  
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*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.
  

