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

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

• @Asaf: I don't think that this really is a duplicate, even if the question is briefly addressed there. – t.b. May 31 '11 at 11:33
• @Theo: You are correct. :-) – Asaf Karagila May 31 '11 at 11:41
• 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

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.

• 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 9:06

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 there are lots of points in $(0,1)\setminus I$.

• I like this answer a lot :) Very clever! – Prism May 5 '13 at 22:10
• 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 5:15
• @LiamCooney, perhaps you could ask a separate question. – lhf Oct 29 '16 at 10:59

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.

• 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
• 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

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.

• On the contrary: when mentioning subsets of the reals, assume the standard topology unless otherwise stated. – GEdgar May 31 '11 at 13:24
• @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
• @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 1 '11 at 13:25
• @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 1 '11 at 13:33
• @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 4 '12 at 15:28

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.

• How do you know that $f^{-1}$ would also be continuous? – Daniel Fischer Aug 14 '14 at 17:20
• Each continuous one-to-one mapping(equivalently, bijection) always is continuous. – George Aug 15 '14 at 4:57
• In order to prove that $f^{-1}$ is continuous under our assumption, I use only one argument of the invariance of domain at 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 9:26
• 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

protected by Asaf Karagila♦Jan 3 '16 at 19:20

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).

Would you like to answer one of these unanswered questions instead?