TITLE: Is there a one-one onto continuous function $f:[0,1]\rightarrow[0,1]^2$?
QUESTION [7 upvotes]: Is there a one-one onto continuous function $f:[0,1]\rightarrow[0,1]^2$? 
I was trying to prove that there is no such function, but failed.
Any suggestions?

REPLY [3 votes]: Here's an alternate solution using properties of connected sets: 
Suppose for contradiction such a function $\varphi$ existed, then it would have a continuous inverse (cf. baby Rudin 4.17) $$\varphi^{-1} : [0,1]^2 \rightarrow [0,1]$$ Now, $[0,1]^2 \setminus\{\varphi(1/2)\}$ is connected (since $\varphi$ is a bijection we are just removing one point from the square) and by properties of continuous functions we know that $\varphi^{-1}$ maps connected sets to connected sets (cf. baby Rudin 4.22) . But $$\varphi^{-1}([0,1]^2 \setminus\{\varphi(1/2)\}) = [0, 1/2)\cup(1/2, 1]$$ which is not connected, so we have a contradiction and hence no such map can exist.