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This question already has an answer here:

The problem : Give example of a continuous, onto function from $(0,1)$ to $[0,1]$. Is it possible for such a function to be one-one?

My partial solution : For the $1^{st}$ part of the question, I came up with this example -

$f : (0,1) \to [0,1]$ given by $f(x)=\sin\big({2\pi x}\big)$

This is continuous and onto, but not one-one.

What I'm asking : For the $2^{nd}$ part of the question, I feel that it should be provable that there cannot exist a continuous, onto, one-one function from $(0,1)$ to $[0,1]$ (If not, we need a counter-example). Any help regarding this proof (or what would really surprise me, a counter example)?

Thanks in advance.

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marked as duplicate by dantopa, kingW3, erfink, Namaste, Leucippus Jun 4 '17 at 0:49

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    $\begingroup$ Hint: think about $f^{-1}\left(0\right)$. $\endgroup$ – Guy Jun 2 '17 at 15:24
  • $\begingroup$ Wouldn't such a function have to be strictly monotone, so map open intervals to open intervals? $\endgroup$ – MPW Jun 2 '17 at 15:24
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    $\begingroup$ $[0,1]$ is compact and $(0,1)$ not. In addition which property must $f$ have? $\endgroup$ – user160069 Jun 2 '17 at 15:25
  • $\begingroup$ May be we the assumption for $f$ is wrong, Can we really asssume such $f$? $\endgroup$ – Hirakjyoti Das Jun 2 '17 at 15:36
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Suppose $f$ is bijective, then $f$ must be strictly increasing or decreasing function. W.l.g. take $f$ to be increasing. Then $\exists x\in (0,1)$ such that $f(x)=1$. Now take $x<y<1$, since $f$ is increasing, $f(y)>1$, which leads to contradiction.

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Bijective continuous map implies the function is monotone, now $f(x)=0$ for some $x\in (0,1)$.

We know there exists a real number $y\in (0,x)$ and a real number $z\in (x,1)$.

Now since $f(x)=0$ and the function is bijective then both $f(y)$ and $f(z)$ are greater than $0$ so the function was decreasing somewhere in $(y,x)$ and increasing somewhere in $(x,z)$, contradicting the monotonicity of $f$.

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  • $\begingroup$ Thanks. To make it more tight (possibly), we may take $w$ such that $0 < w < \min\{f(y),f(z)\}$. By the Intermediate value theorem, there exists $t_1 \in (y,x)$ and $t_2 \in (x,z)$ such that $f(t_1)=f(t_2)=w$, contradicting the fact that $f$ is a bijection. $\endgroup$ – Dragon Jun 2 '17 at 16:19
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Assume there exists $x \in (0,1)$ such that $f(x) = 0$. Then, by openness of $(0,1)$, for some $\epsilon > 0$, the interval $(x-\epsilon, x+\epsilon)$ is in the interval $(0,1)$.

If now $f$ is one-to-one and onto, either $f(x-\epsilon) < f(x)$ or $f(x+\epsilon) < f(x)$. This contradicts the fact that we want $f$ to map to the closed interval $[0,1]$.

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If $f: A\rightarrow B$ is a continuous map, then $f$ pulls closed sets in $B$ to closed sets in $A$. Given, $f$ is onto and $[0, 1]$ is closed. So its pre-image should be closed. But its pre-image $(0,1)$ is not cloed in $\mathbb{R}$. A contradiction. So we can't have such $f$

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  • $\begingroup$ not correct (0,1) is closed in (0,1) $\endgroup$ – Arpan1729 Jun 2 '17 at 17:48

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