# Does there exists a onto continuos function $f\colon (0,1) \to[0,1]$ which is one to one . [duplicate]

Iam trying to solve this problem , i can visualize graphically it is not possible. iam trying to do the proof by contradiction. if f:(0,1) to [0,1] is a continuos onto function then i have to prove that f can never be 1-1 . so iam letting that if f is 1-1 then either f is strictly increasing or strictly decreasing function. so case (1) if f is strictly increasing function then there exits some t in (0,1) such that f(t)=0 , so for x lesser than t f(x)k , f(x) <0 which is also not possible .so f can't be 1-1 . My method is right ??? is there is any alternative approach then plz tell

• Hi, and welcome to MSE! For future reference, mathjax can be used to format equations. – Air Conditioner Feb 22 '18 at 20:28
• Your reasoning is correct. – Umberto P. Feb 22 '18 at 20:30
• Hint: Let $f\colon (0,1)\to [0,1]$ be continuous and surjective. Take $a=f^{-1}(0)$ and $b=f^{-1}(1)$. Prove that $f([c,d])=[0,1]$ using the intermediate value theorem, where $c=min(a,b)$ and $d=max(a,b)$. Conclude that $f$ is not injective. – Júlio César Feb 22 '18 at 20:37
• I think you should post this as an answer. – H1ghfiv3 Feb 22 '18 at 20:42

This is merely an elaboration of the solution provided by Júlio Cesar under the original comment (all credit to him), which is in itself "only" making precise your original correct reasoning.

Assume $f:(0,1) \to [0,1]$ was continuous and bijective. Then we find unique $a := f^{-1}(0)$, $b := f^{-1}(1)$ and may assume wLog that $0 < a < b < 1$. Choose $x \in (0,a)$ and let $y := f(x) \in [0,1]$. By the intermediate value theorem, we find some $c \in (a,b)$ with $f(c) = y = f(x)$. Since $x < a < c$, $f$ is not injective, so we arrive at a contradiction.

$[0,1]$ is closed.

$f$ continuous and one to one implies $f^{-1}([0,1])$ is closed.

this is not the case for $(0,1)$

Using sequential characterisation of the limit, it is easy to prove the continuity of $f^{-1}$.

• Strictly speaking, You would need to know, however, that $f^{-1}$ is a continuous function (this is true but requires more work than the reasoning from the OP). – H1ghfiv3 Feb 22 '18 at 20:34
• But $(0,1)$ is closed in itself... – Sangchul Lee Feb 22 '18 at 20:34
• He probably means closed as a subset of $\mathbb R$ (not in its own subspace topology). – H1ghfiv3 Feb 22 '18 at 20:35
• That is what I am suspecting. Or perhaps he/she is really caring about compactness. – Sangchul Lee Feb 22 '18 at 20:37
• Any continuous, bijective function $f: I \to J$ between two subsets of $\mathbb R$ has a continuous inverse. But this is special to such subsets and cannot be generalised to functions between between arbitrary topological spaces. You usually proof it by showing first that such $f$ must be strictly monotone on its connected components. – H1ghfiv3 Feb 22 '18 at 20:58