Find a continuous, bijective function from $(0,1)$ onto $(0,1]$.

My attempt: Any such maps exists. This is my guess.

Suppose if $f$ is a bijection from $(0,1)$ onto $(0,1]$, then there exist $x\in (0,1)$ such that $f(x) = 1$. Now choose $0<a<x<b<1$ such that $f(a)< f(x)=1$ and $f(b)<f(x) = 1$ otherwise if $f(a)= f(x)=1$, then it would contradict that $f$ is injective.

Now I cannot proceed further. Please help.

  • 1
    $\begingroup$ @LeeMosher Sir but in my question involves continuity. $\endgroup$
    – user1116521
    Dec 14, 2022 at 17:48
  • $\begingroup$ @LeeMosher That link if for a bijection, not a continuous bijection. $\endgroup$
    – jjagmath
    Dec 14, 2022 at 17:55

1 Answer 1


There is no such map. To show this, recall that all continuous injections $(0, 1) \to \mathbb{R}$ must be strictly increasing or strictly decreasing.

So if we had a continuous bijection $f: (0, 1) \to (0, 1]$, take $x$ such that $f(x) = 1$. Then take $y < x < z$. Then $f(y), f(z) < f(x)$, so $f$ is neither strictly increasing nor strictly decreasing; contradiction.


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