Why is there no continuous bijection from (0,1] to R?

I read somewhere that the reason is due to the fact that (0,1] is a connected set and the image of a connected set must be connected, so why can't we have a connected image?


The continuous image of $(0,1)$ must be connected, but by bijectivity this is $\mathbb R$ except the image of $1$. Removing any point from $\mathbb R$ makes it disconnected, a contradiction. Specifically, the two connected components of $\mathbb R\backslash\{a\}$ are $(-\infty,a)$ and $(a,+\infty)$.


Tesla Daybreak’s answer is good. And this can be viewed from order structure.

Note that any injective continuous function from $(0,1]$ must be strict increase(or decrease). Hence $f(1)$ is either the maximal value or minimal value, but $\mathbb{R}$ don’t have maximal element or minimal element. This implies $f$ can’t be onto, if it is injective.

  • $\begingroup$ why would this argument work for (0,1)? is it because (0,1) does not have a minimal or maximal element? $\endgroup$
    – Bill
    Apr 10 '21 at 10:09
  • 1
    $\begingroup$ @William: Yes. If you want to understand this properly you will need to look at IVT (intermediate value theorem), which guarantees the property Landau states. More specifically, given any continuous $f$ on any real interval $I$, there are no $a,b,c$ such that $a,b<c$ and $f(c)$ is strictly between $f(a)$ and $f(b)$. Once you prove this, you can apply it to $c=1$. $\endgroup$
    – user21820
    Apr 11 '21 at 3:08

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.