TITLE: Continuous injective map $f:\mathbb{R}^3 \to \mathbb{R}$?
QUESTION [24 upvotes]: How would you show that there is no continuous injective map  $f:\mathbb{R}^3 \to \mathbb{R}$?
I tried the approach where if $a \in \mathbb{R}$ then $\mathbb{R} \backslash \{a\} $ is not clearly connected so there can't exist a continuous function otherwise $\mathbb{R} \backslash \{a\} $ would be connected?

REPLY [2 votes]: We can prove there is no continuous injection from $\mathbb{R}^{2}$ to 
 $\mathbb{R}.$Then we confine the map from $\mathbb{R}^{3}$ to $\mathbb{R}$ on the $xy$ plane.The map is still continuous and injective so we have contradction.
We just need to find 2 different points and $f$ has same value on the two points.
So just consider 4 vertexs $\{O(0,0),A(1,0),B(0,1),C(1,1)\}$ of the unit cube.We can assume $f$ takes different value on the $4$ points or we just find what we want.Assume $f(O)$ is the smallest and $f(O)<f(A)<f(B)$.Using mean value theorem and we can find a point $D$ on the segment $OB$ and $f(D)= f(A).$
This method just needs continuity for each variable and can be generalized to the case of $\mathbb{R}^{n}.$