I know that the result is true for closed interval $[0,1]$ by using intermediate value property.

But in the case where we consider open interval $(0,1)$ does the solution change?

  • 1
    $\begingroup$ Similarly, it's also false for $(0,1]$ (though it won't be a bijection) by the function $x\mapsto\frac x2$. $\endgroup$ – Akiva Weinberger Dec 10 '15 at 14:41
  • $\begingroup$ That's nice. :) $\endgroup$ – Error 404 Dec 10 '15 at 14:48
  • $\begingroup$ Yes, I understood that. :) Hence the result will be true only for closed intervals. $\endgroup$ – Error 404 Dec 10 '15 at 14:50

Try $x\mapsto x^2$ for a counterexample.


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