0

I'm asked to state why there cannot be a surjective and continuous function $$g: [0,1]\longrightarrow \text{Cantor Set} $$ I know that $g^{-1}$ exists and is continuous & surjective since the previous exercise was for me to show exactly that. Now I know that if $g$ were a bijection, then if $g$ were to be continuous, then it would be a homeomorphism and would have to preserve connectedness of $[0,1]$, which it doesn't, implying that $g$ cannot be continuous.

But this requires me to have a bijectivity of $g^{-1}$, where does surjectivity alone enter in this case?

Sumanta
  • 9,534
Taken Spark
  • 157

1 Answers1

0

If $g: [0,1] \to C$ is continuous then $g[[0,1]]$ is connected. The only connected non-empty subsets of $C$ are of the form $\{c\}$ ($C$ is totally disconnected), so $g$ is constant.

Henno Brandsma
  • 242,131
  • and so it isn't surjective because it is defined as a map $[0,1]\longrightarrow C$ but there are elements of $C$ that are not images of a set in $[0,1]$ by $g$ $\Leftrightarrow$ non-surjectivity. Thank you! This was very clear. – Taken Spark Jan 23 '21 at 16:34
  • @TakenSpark $C$ is uncountable and you only have one point in the image. It doesn't get less surjective than that. – Henno Brandsma Jan 23 '21 at 16:35