TITLE: Prove or Disprove : There exists a continuous bijection from $\mathbb{ R}^2$ to $\mathbb{R} $
QUESTION [7 upvotes]: This question was asked to me by a mathematics undergraduate to me and I was not able to solve it. So, I am asking it here.

Prove or Disprove : There exists a continuous bijection from $\mathbb{ R}^2$ to $\mathbb{R} $ .

I have no idea on how this problem can be tackled. It seems it has something to do with set theory but I only know elementary set theory( bijection from naturals) and  I am unable to solve it.

REPLY [3 votes]: Suppose $f:\Bbb R^2\to \Bbb R$ is a continuous injection.  Let $S=[0,1]^2$ and $T=[1/3,2/3]^2.$ Each of $S,T$ is compact and  connected so their images $f[S], f[T]$ are compact and  connected.
$(\bullet)\,$ So $f[S]$ and $f[T]$ are closed bounded real intervals.
Now $f|_S: S\to f[S]$ is a continuous bijection from the compact Hausdorff space $S$ to the compact Hausdorff space $f[S]$ so $f|_S:S\to f[S]$ is a homeomorphism.
Therefore $f$ maps the boundary of $T$ in the space $S$ to the boundary of $f[T]$ in the space $f[S].$
But the boundary of  $f[T]$ in the space $f[S]$ contains just 2 points by $(\bullet)$ and the boundary of $T$ in the space $S$ is infinite, and $f$ is 1-to-1, which is absurd.