Continuous function from $(0,1)$ onto $[0,1]$

Question

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.

You can avoid confusion by using "surjective" and "injective" instead of "onto" and "one-to-one" respectively. — Rudy the Reindeer, Jul 07 '12 at 10:09
possible duplicate of Continuous bijection from $(0,1)$ to $[0,1]$ — Rudy the Reindeer, Jul 07 '12 at 10:10
@MattN.: the answer to that question does not answer this question. — robjohn, Jul 07 '12 at 10:15
@robjohn But "that question" asks "Does there exist a continuous bijection from $(0,1)$ to $[0,1]$"? and b) of this question here asks "Is there a continuous one-to-one function from $(0,1)$ onto $[0,1]$?". How are they not the same? — Rudy the Reindeer, Jul 07 '12 at 11:16
@Matt: Because this is not just (B) but rather (A) and (B). There is no answer on the suggested duplicate which answers (A). — Asaf Karagila, Jul 07 '12 at 11:50
@AsafKaragila a) is not part of the question because it's already answered in the OP. — Rudy the Reindeer, Jul 07 '12 at 11:55
@Matt: So all those questions appearing on the site which go like "This is a question, this is my answer, is this correct?" should be closed as not a real question? — Asaf Karagila, Jul 07 '12 at 11:56
@Asaf: True, but the OP has already gotten to a continuous surjection, so the question is really only (B). — Cameron Buie, Jul 07 '12 at 11:58
@Cameron: But the same can be applied on other "is this correct" questions because the OP already has a proof (and often a very good proof). We don't close those questions, do we? If the OP was sure of their example, why bring it up? Why even bother putting this in the question to begin with? Why not mention that as an example, or point out that it is true and only ask for the second part? — Asaf Karagila, Jul 07 '12 at 12:04
@Matt: The OP wrote that the answer for (B) is no, so there is an answer. Why duplicate? Why not simply NARQ instead, if there is no question in this post? (also, see my comment to Cameron.) — Asaf Karagila, Jul 07 '12 at 12:08

score 13 · Accepted Answer · edited Apr 20 '13 at 12:34

13

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.

edited Apr 20 '13 at 12:34

Thomas Andrews

177,126

answered Jul 07 '12 at 09:41

Ahriman

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2.$$ f(x) = \left{ \begin{array}{cl} 0 & x < \frac14 \ 2x-\frac12 & \frac14 \leq x \leq \frac34 \ 1 & x > \frac34 \end{array}\right. $$ – Sep 03 '18 at 01:25
Isn't a continuous function – Sep 03 '18 at 01:26
Yes, it absolutely is a continuous function. – Not Legato Oct 13 '20 at 18:57

Asaf Karagila · Answer 2 · 2012-07-07T09:41:02.970

12

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.

edited Jul 07 '12 at 09:41

answered Jul 07 '12 at 09:35

Asaf Karagila

393,674

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Was not the OP asking about a $1-1$ function? – Giovanni De Gaetano Jul 07 '12 at 09:37
@GiovanniDeGaetano: The first part reads differently. – Asaf Karagila Jul 07 '12 at 09:38
I'm sorry, I must have misread the question then. – Giovanni De Gaetano Jul 07 '12 at 11:48

Raymond Manzoni · Answer 3 · 2012-07-07T09:42:41.770

3

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)

edited Jul 07 '12 at 09:42

answered Jul 07 '12 at 09:37

Raymond Manzoni

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The link was broken. – Raymond Manzoni Jun 10 '15 at 06:45

Krishnan · Answer 4 · 2018-09-04T16:50:14.397

1

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))

edited Sep 04 '18 at 16:50

answered Sep 04 '18 at 16:44

Krishnan

29

This does not address part B of the question. – copper.hat Nov 19 '23 at 17:53

score -3 · Answer 5 · answered Oct 02 '15 at 05:45

-3

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.

answered Oct 02 '15 at 05:45

Silent Duckling

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Any reason why you're posting a answer to a 3-year old question that already has an accepted answer? – Olorun Oct 02 '15 at 05:46
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 08:35

Continuous function from $(0,1)$ onto $[0,1]$

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