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While revising, I came across this question(s):

A) Is there a continuous function from $(0,1)$ onto $[0,1]$?

B) Is there a continuous one-to-one function from $(0,1)$ onto $[0,1]$?

(clarification: one-to-one is taken as a synonym for injective)

I figured the answer to A is yes, with $\frac{1}{2}\sin(4\pi x)+\frac{1}{2}$ as an example.

The answer to part B is no, but what is the reason?

Sincere thanks for any help.

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1  
Related. – Rudy the Reindeer Jul 7 '12 at 9:35
2  
You can avoid confusion by using "surjective" and "injective" instead of "onto" and "one-to-one" respectively. – Rudy the Reindeer Jul 7 '12 at 10:09
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@MattN.: the answer to that question does not answer this question. – robjohn Jul 7 '12 at 10:15
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@Asaf: True, but the OP has already gotten to a continuous surjection, so the question is really only (B). – Cameron Buie Jul 7 '12 at 11:58
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@Asaf: Fair point. If it is closed, I will vote to reopen. – Cameron Buie Jul 7 '12 at 12:13
up vote 6 down vote accepted

There isn't such a function. If $f$ is such a function, then $f$ is monotonic, and $f^{-1}$ too. But a monotonic function defined on an interval is continuous iff its range is an interval. So we have that $f$ is an homeomorphism from $(0,1)$ to $[0,1]$, which is impossible since one is compact and the other is non compact.

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A slightly simpler solution, perhaps, for the first part would be as follows:

Consider the function from $\left[\frac14,\frac34\right]$ to $[0,1]$ defined by $x\mapsto 2\left(x-\frac14\right)$. This is certainly a continuous function, and it is certainly onto $[0,1]$.

Define now:

$$f(x)=\begin{cases}0 & x<\frac14\\ 2\left(x-\tfrac14\right) & x\in\left[\tfrac14,\tfrac34\right]\\ 1 & x>\frac34\end{cases}$$

Of course this is not an injective function, but it is continuous and onto, as required.

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1  
Was not the OP asking about a $1-1$ function? – Giovanni De Gaetano Jul 7 '12 at 9:37
    
@GiovanniDeGaetano: The first part reads differently. – Asaf Karagila Jul 7 '12 at 9:38
    
I'm sorry, I must have misread the question then. – Giovanni De Gaetano Jul 7 '12 at 11:48

Your example for A is fine.
For B "for a continuous function on an interval, being one-to-one is equivalent to being increasing throughout or decreasing throughout" so that no such function exists. (see (2.5)-(2.6) of this paper)

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The link was broken. – Raymond Manzoni Jun 10 '15 at 6:45

If such a function exists then (0,1) and [0,1] are homeomorphic. But if you delete 0 from [0,1] you get (0,1] which is connected. Deleting the point x in (0,1) which corresponds to 0 in [0,1] leaves you with (0,x) u (x,1) which is not connected ---> contradiction.

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2  
Any reason why you're posting a answer to a 3-year old question that already has an accepted answer? – Olorun Oct 2 '15 at 5:46
    
No particular reason. – Silent Duckling Oct 10 '15 at 12:27
    
Unfortunately this answer is incomplete. "Such a function" would be a continuous bijection from $(0,1)$ to $[0,1]$. The matter of the question is to show that any continuous bijection from $(0,1)$ to $[0,1]$ is an open mapping, and thus a homeomorphism. – Pete L. Clark Nov 29 '15 at 8:35

protected by Zev Chonoles Oct 2 '15 at 6:23

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