First I'll state my current mathematical knowledge on the subject of homeomorphisme: very low. I'm doing a class of dynamical systems and I got the following question:

Let $f:[0,1] \rightarrow (2,8)$ and $f$ is one-to-one and continuous. Now show $f$ is not onto.

So I started of by stating that $f$ is either continuously increasing or continuously decreasing, that's because $f$ is one-to-one. Now intuitively it makes sense that $f$ cannot be onto since it maps a closed interval into a open one, but I don't know how to show/prove this.

The exercise has a hint that $[0,1]$ and $(2,8)$ are not homeomorphic but I have no idea how to use this.

  • $\begingroup$ We don't need $f$ to be one one. Just continuity is sufficient to conclude. $\endgroup$
    – Paramanand Singh
    Jan 10, 2018 at 11:11

1 Answer 1


The image of the compact space $[0,1] $ under the continuous function $f $ is compact, hence a proper subset of $(2,8) $ because this isn't compact.

  • $\begingroup$ I don't really understand your reasening, how does compactness relate to a function being onto? $\endgroup$ Jan 10, 2018 at 16:43
  • $\begingroup$ Since the image $\operatorname{Im}(f)$ of $f$ is compact and $(2,8)$ is not compact we cannot have $\operatorname{Im}(f)=(2,8)$, thus $f$ cannot be surjective. $\endgroup$ Jan 10, 2018 at 16:46
  • $\begingroup$ I know this to be true but do you have a prove, like a proof used in real analysis? $\endgroup$ Jan 12, 2018 at 19:11
  • $\begingroup$ See math.stackexchange.com/q/26514 and math.stackexchange.com/q/874044 $\endgroup$ Jan 12, 2018 at 19:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.