Continuous bijection from $(0,1)$ to $[0,1]$

Question

Does there exist a continuous bijection from $(0,1)$ to $[0,1]$? Of course the map should not be a proper map.

What about such map from a non compact set to compact set in nice topologies? — Alex, May 31 '11 at 12:14
@Alex: Take $f: [0,1) \to S^1 = {z \in \mathbb{C},:,|z| = 1}$ with $f(x) = e^{2\pi i x}$ This is continuous and bijective, but has no continuous inverse. — t.b., May 31 '11 at 12:16
what exactly is the reason for non existence? In the (0,1) case some sort of local compactness is the reason.Can the proof generalized to non existence of a map from open ball in R^n to a closed ball. — Alex, May 31 '11 at 12:27
@Alex: I should have said noncontinuous inverse. The point is that the inverse $g$ is already determined by $f$. Now if $z_{n} = e^{2\pi i x_{n}} = f(x_n)$ with $x_n \nearrow 1$ hen $z_n \to 1$ while $g(z_n) = x_{n}$ and and $g(1) = 0$, so $g(z_n) = x_n$ doesn't converge to $g(1)$ and thus $g$ isn't continuous. As for the generalization to open and closed balls in $\mathbb{R}^n$, I think you should ask this as a separate question, because clearly other techniques are required than the three (very similar if not identical) arguments you received here. — t.b., May 31 '11 at 12:37
@theo:please look at ncmathsadist 's answer to this question.What do you say on montonocity of 1-1 continuous function.Do nowhere monotonous functions exist? — Alex, May 31 '11 at 12:44
@Alex: A continuous $1-1$-function from an interval to an interval is strictly monotonous, that's true. This follows from the intermediate value theorem and the fact that intervals and points are precisely the connected subsets of an interval in $\mathbb{R}$. — t.b., May 31 '11 at 12:51
Alex, if you have any question on t.b.'s answer (or any of the ones you got) you may use the comment below the answer - I am sure he/she/they would be happy to assist! — AD - Stop Putin -, Dec 13 '11 at 21:52

t.b. · Answer 1 · 2011-06-01T08:04:53.557

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No. If $f:(0,1) \to [0,1]$ were continuous and bijective, there would be a unique point $x \in (0,1)$ such that $f(x) = 1$. However, since $f$ is continuous, the intervals $[x - \varepsilon, x]$ and $[x, x + \varepsilon]$ would be mapped to intervals $[a,1]$ and $[b,1]$, say. By bijectivity we'd have $a, b \lt 1$. Thus every value strictly between $\max{\{a,b\}}$ and $1$ would be assumed at least twice, contradicting bijectivity.

edited Jun 01 '11 at 08:04

answered May 31 '11 at 11:20

t.b.

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Interesting how this proof relies on $(0, 1)$ being open so that the intervals $[x- \epsilon, x] $ and $[x, x + \epsilon]$ both exist, and (0, 1] being "closed at 1" so that $a, b \lt 1$. – Tom Collinge Mar 27 '16 at 09:06

score 102 · Answer 2 · edited Oct 29 '19 at 00:46

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Let $f:(0,1) \rightarrow [0,1]$ be continuous and surjective. (Actually, we just need to suppose that $0$ and $1$ are in the image of $f$.) Let $a,b \in (0,1)$ such that $f(a)=0$ and $f(b)=1$. Let $I=[a,b]$ if $a<b$ or $I=[b,a]$ if $b<a$. Then, by the intermediate value theorem, $f(I)$ is an interval that contains $0$ and $1$ and so $f(I)$ contains $[0,1]$, which implies $f(I)=[0,1]$. But then $f$ cannot be injective because $(0,1)\setminus I$ is nonempty.

edited Oct 29 '19 at 00:46

Kenta S

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answered May 31 '11 at 12:08

lhf

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I like this answer a lot :) Very clever! – Prism May 05 '13 at 22:10
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I like this a lot too...is there any way one could extend this to showing that (0, 1) and [0, 1) are not homeomorphic? – Liam Cooney Oct 29 '16 at 05:15
@LiamCooney, perhaps you could ask a separate question. – lhf Oct 29 '16 at 10:59
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I did not understand the very last line. Can anybody explain ? – user9026 Jun 07 '20 at 15:16
Brilliant Proof awesome – Ekaveera Gouribhatla Feb 05 '21 at 06:31
@user9026 The closed interval preimage which covers the image cannot contain all points from the domain, because the domain is open. Any two points you select from an open interval will still have points to the left and to the right, which is a problem if you want continuous and not overlapping function. – Trixie Wolf Jun 17 '22 at 05:25

score 18 · Answer 3 · answered May 31 '11 at 12:01

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Suppose that $f:(0,1) \rightarrow [0,1]$ is 1-1 and continuous. By the intermediate value theorem, the image of any interval under $f$ is an interval. Since $f$ is 1-1, it is either (strictly) monotone increasing or decreasing. Hence, $f(0,1)$ is an interval. Without loss of generality, assume $f$ is increasing; were it not this analysis would apply to $1 - f$.

Suppose now that $f$ is onto; then we must have some $t\in(0,1)$ with $f(t) = 1$. Because $f$ is strictly monotone increasing, we would have to have $f(s) > 1$, for $t \le s < 1$. This violates the premise that $f(0,1) \subseteq [0,1]$. Hence, $f$ cannot be onto.

answered May 31 '11 at 12:01

ncmathsadist

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You need the continuity to get this. 1-1 alone does not imply monotone. – ncmathsadist May 31 '11 at 12:20
I don't think a continuous 1-1 function should be monotone.There may be non differentiable kind of things with nowhere monotonocity – Alex May 31 '11 at 12:20
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Yes, continuous 1-1 function defined on an interval is monotone. Interesting application of intermediate value theorem several times. – GEdgar May 31 '11 at 13:22

score 8 · Answer 4 · edited Dec 30 '12 at 20:22

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Since Theo gave an answer I am going to be nitpicking and add one remark. When speaking about continuity (especially when tagging under [topology]) it is best to mention the topology you are working with. In this case, you mean in the standard topology.

Otherwise, consider the discrete topology, i.e. every set is open:

Let $f\colon [0,1]\to (0,1)$ be any bijection, it is continuous since all sets are open, the preimage of an open set is an open set, thus $f$ is continuous.

edited Dec 30 '12 at 20:22

Rudy the Reindeer

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answered May 31 '11 at 11:43

Asaf Karagila

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On the contrary: when mentioning subsets of the reals, assume the standard topology unless otherwise stated. – GEdgar May 31 '11 at 13:24
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@GEdgar: When taking a course in real analysis? Sure. When taking a course in point-set topology? Not if you want to be accurate. – Asaf Karagila May 31 '11 at 14:00
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@Asaf: When answering a question on this board? YES. Or, if you want to be super-accurate, add "assuming the usual topology" to your answer. – GEdgar Jun 01 '11 at 13:25
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@GEdgar: The notation $(0,1)$ is just for a set. I would think that "Proving there exists a continuous bijection between $(0,1)$ and $[0,1]$; prove that if for some $\tau$ a topology on $\mathbb R$ there exists such bijection then some condition." would be an excellent homework assignment in a general topology class (not in this exact form of course). When I gave my answer the question was only tagged under [topology]. When doing mathematics one should strive to be as general and accurate as the context allows, I did exactly that. – Asaf Karagila Jun 01 '11 at 13:33
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@AsafKaragila: it would be complete to give an example of one such bijection (e.g. $0\mapsto\frac12$, $\frac1n\mapsto\frac1{n+2}$ for $n=1,2,3,\dots$, and $x\mapsto x$, otherwise). – robjohn Nov 04 '12 at 15:28

score 3 · Answer 5 · answered Oct 29 '19 at 03:31

Recall the following result:

Proposition 1: Let $f: (a,b) \to (c,d)$ be an injective continuous mapping from one open interval to another. Then the image of $f$ is an open interval.

Now assume we have a continuous bijection $f: (0,1) \to [0,1]$. The function $f$ maps some point in $(0,1)$ to $0$ and another point to $1$. If we remove these points we can restrict $f$ and define an injective and surjective continuous mapping

$\tag 1 f: (0,a) \sqcup (a,b) \sqcup (b,1) \to (0,1)$

on three non-overlapping intervals.

By proposition 1 the image of each of these three intervals is an open interval of $(0,1)$. Since $f$ is also surjective, we have a partition of $(0,1)$ into three open subsets. But it is not possible to express a connected topological space in such a manner.

score -2 · Answer 6 · edited Aug 16 '14 at 12:35

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There does not exist a continuous bijection from (0,1) to [0,1]. Indeed, let $f$ be such a function. Let consider a sequence $x_n=1-1/n$. Then from the sequence $(f(x_n))$ we can choose a subsequence $(f(x_{n_k}))$ which is convergent. Let denote this limit by $y$. Obviously, $y \in [0,1]$. Since $f^{-1}$ also is continuous, we get $f^{-1}(y)=\lim_{k \to +\infty}f^{-1}(f(x_{n_k}))=\lim_{k \to \infty}x_{n_k}=1$. But $1 \notin (0,1)$.

Remark(Why $f^{-1}$ must be continuous under our assumption?) By our assumption $f:(0,1)\to [0,1]$ is continuous bijection. Then $f:(0,1)\to [0,1]$ must be injective and continuous which following invariance of domain (see, http://en.wikipedia.org/wiki/Invariance_of_domain is homeomorphism. Hence $f^{-1}: [0,1]\to (0,1)$ is continuous.

edited Aug 16 '14 at 12:35

Martin Sleziak

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answered Aug 14 '14 at 17:13

George

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How do you know that $f^{-1}$ would also be continuous? – Daniel Fischer Aug 14 '14 at 17:20
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Each continuous one-to-one mapping(equivalently, bijection) always is continuous. – George Aug 15 '14 at 04:57
In order to prove that $f^{-1}$ is continuous under our assumption, I use only one argument of the invariance of domain at http://en.wikipedia.org/wiki/Invariance_of_domain asserted that if $U$ is open subset of $R^n$ and $f:U\to R^n$ is injective and continuous that $f(U)$ is open and $f:U \to f(U)$ is homeomorphism. My previous comment assumes this situation. – George Aug 15 '14 at 09:26
That argument should be made in the answer (although invariance of domain is serious overkill for the one-dimensional setting). – Daniel Fischer Aug 15 '14 at 12:37

Continuous bijection from $(0,1)$ to $[0,1]$

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