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

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  • $\begingroup$ @Asaf: I don't think that this really is a duplicate, even if the question is briefly addressed there. $\endgroup$ – t.b. May 31 '11 at 11:33
  • $\begingroup$ @Theo: You are correct. :-) $\endgroup$ – Asaf Karagila May 31 '11 at 11:41
  • $\begingroup$ What about such map from a non compact set to compact set in nice topologies? $\endgroup$ – Alex May 31 '11 at 12:14
  • $\begingroup$ @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. $\endgroup$ – t.b. May 31 '11 at 12:16
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    $\begingroup$ 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. $\endgroup$ – Alex May 31 '11 at 12:27
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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.

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    $\begingroup$ 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 $a, b \lt 1$. $\endgroup$ – Tom Collinge Mar 27 '16 at 9:06
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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.

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    $\begingroup$ I like this answer a lot :) Very clever! $\endgroup$ – Prism May 5 '13 at 22:10
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    $\begingroup$ 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? $\endgroup$ – Liam Cooney Oct 29 '16 at 5:15
  • $\begingroup$ @LiamCooney, perhaps you could ask a separate question. $\endgroup$ – lhf Oct 29 '16 at 10:59
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    $\begingroup$ I did not understand the very last line. Can anybody explain ? $\endgroup$ – user9026 Jun 7 at 15:16
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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.

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

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    $\begingroup$ On the contrary: when mentioning subsets of the reals, assume the standard topology unless otherwise stated. $\endgroup$ – GEdgar May 31 '11 at 13:24
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    $\begingroup$ @GEdgar: When taking a course in real analysis? Sure. When taking a course in point-set topology? Not if you want to be accurate. $\endgroup$ – Asaf Karagila May 31 '11 at 14:00
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    $\begingroup$ @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. $\endgroup$ – GEdgar Jun 1 '11 at 13:25
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    $\begingroup$ @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. $\endgroup$ – Asaf Karagila Jun 1 '11 at 13:33
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    $\begingroup$ @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). $\endgroup$ – robjohn Nov 4 '12 at 15:28
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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.

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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.

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    $\begingroup$ How do you know that $f^{-1}$ would also be continuous? $\endgroup$ – Daniel Fischer Aug 14 '14 at 17:20
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    $\begingroup$ Each continuous one-to-one mapping(equivalently, bijection) always is continuous. $\endgroup$ – George Aug 15 '14 at 4:57
  • $\begingroup$ 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. $\endgroup$ – George Aug 15 '14 at 9:26
  • $\begingroup$ That argument should be made in the answer (although invariance of domain is serious overkill for the one-dimensional setting). $\endgroup$ – Daniel Fischer Aug 15 '14 at 12:37

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