1
$\begingroup$

Find a continuous bijection from $(0,1]$ to $(0,1)$.

I have solved the question for "Find a continuous bijection from $(0,1)$ to $(0,1]$". Moreover there is no such map. But cannot solve if there is a continuous bijection from $(0,1]$ to $(0,1)$.

Please help.

$\endgroup$
6
  • $\begingroup$ There is no such function. $\endgroup$
    – geetha290krm
    Dec 15, 2022 at 9:44
  • $\begingroup$ What you ask or claim is unclear. $\endgroup$
    – Andrea Mori
    Dec 15, 2022 at 9:44
  • 2
    $\begingroup$ You can pretty easily prove that if $f:(0,1]\to(0,1)$ is a continuous bijection, then so is its inverse $f^{-1}:(0,1)\to(0,1]$, but you already know such cannot exist $\endgroup$
    – Lorago
    Dec 15, 2022 at 9:45
  • $\begingroup$ @AndreaMori I have edited. $\endgroup$
    – user1116521
    Dec 15, 2022 at 9:48
  • $\begingroup$ You'd have to pick what to map 1 to, and then there'd be problems when looking at both sides of that image. $\endgroup$
    – coffeemath
    Dec 15, 2022 at 9:52

1 Answer 1

Reset to default
3
$\begingroup$

Note that if $f:(0,1]\to (0,1)$ is bijective and continuous, then $f|_{(0,1)}:(0,1)\to (0,1)\setminus\{f(1)\}$ is continuous. However, continuous functions map connected spaces to connected spaces, ergo since $(0,1)$ is connected, $f((0,1))=(0,1)\setminus\{f(1)\}$ is connected - which is obviously not (it decomposes as $(0,f(1))\cup (f(1),1)$). Therefore there cannot exist such $f$.

$\endgroup$

You must log in to answer this question.