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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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    $\begingroup$ Related. $\endgroup$ – Rudy the Reindeer Jul 7 '12 at 9:35
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    $\begingroup$ You can avoid confusion by using "surjective" and "injective" instead of "onto" and "one-to-one" respectively. $\endgroup$ – Rudy the Reindeer Jul 7 '12 at 10:09
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    $\begingroup$ @MattN.: the answer to that question does not answer this question. $\endgroup$ – robjohn♦ Jul 7 '12 at 10:15
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    $\begingroup$ @Asaf: True, but the OP has already gotten to a continuous surjection, so the question is really only (B). $\endgroup$ – Cameron Buie Jul 7 '12 at 11:58
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    $\begingroup$ @Asaf: Fair point. If it is closed, I will vote to reopen. $\endgroup$ – Cameron Buie Jul 7 '12 at 12:13
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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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  • $\begingroup$ 2.$$ f(x) = \left\{ \begin{array}{cl} 0 & x < \frac14 \\ 2x-\frac12 & \frac14 \leq x \leq \frac34 \\ 1 & x > \frac34 \end{array}\right. $$ $\endgroup$ – user464147 Sep 3 '18 at 1:25
  • $\begingroup$ Isn't a continuous function $\endgroup$ – user464147 Sep 3 '18 at 1:26
  • $\begingroup$ Yes, it absolutely is a continuous function. $\endgroup$ – Not Legato Oct 13 '20 at 18:57
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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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    $\begingroup$ Was not the OP asking about a $1-1$ function? $\endgroup$ – Giovanni De Gaetano Jul 7 '12 at 9:37
  • $\begingroup$ @GiovanniDeGaetano: The first part reads differently. $\endgroup$ – Asaf Karagila♦ Jul 7 '12 at 9:38
  • $\begingroup$ I'm sorry, I must have misread the question then. $\endgroup$ – Giovanni De Gaetano Jul 7 '12 at 11:48
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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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Answer for (B): Continuous image of a compact set is compact. [see1]

Suppose $f: [0,1] \rightarrow (0,1)$ is a bijection and continuous. Then $f([0,1])=(0,1)$, which is compact by above theorem.

But in $\mathbb{R}$, the closed interval $[0,1]$ is compact; but the open interval $(0,1)$ is not compact.

Therefore if such $f$ does not exist, then $f^{-1}$ does not exist too, where $f^{-1}: (0,1) \rightarrow [0,1]$. (your question in (B))

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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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    $\begingroup$ Any reason why you're posting a answer to a 3-year old question that already has an accepted answer? $\endgroup$ – Olorun Oct 2 '15 at 5:46
  • $\begingroup$ No particular reason. $\endgroup$ – Silent Duckling Oct 10 '15 at 12:27
  • $\begingroup$ 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. $\endgroup$ – Pete L. Clark Nov 29 '15 at 8:35

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