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?


2 Answers 2


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, 2021 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, 2021 at 3:08

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .