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This question is related to question 92206 "What properties make $[0, 1]$ a good candidate for defining fundamental groups?" but is not exactly equivalent in my opinion. It is even suggested in one of the answers to 92206 that "there is nothing fundamental about the unit interval," but i would like to know what is fundamental about the unit interval. I have learned some answers from the answers to 92206, but i wonder if there is more. Another related question: "Topological Characterisation of the real line".

The question is:

What would be a "natural" topological characterization of the closed interval $[0,1]$?

Motivating questions (please do not answer them):

  • Why is all the algebraic topology built around it?
  • Why does it appear (in the guise of Lie groups) in the study of general locally compact topological groups (Gleason-Yamabe theorem)?
  • Is it really necessary to define it algebraically as a subset of a field ($\mathbb R$) just to use it as a topological space?

I am mostly interested in the significance of $[0, 1]$ for the study of Hausdorff compacts (like metrizability, Urysohn's lemma).

I have some ideas and have asked already on Math.StackExchange, but decided to duplicate here.

One purely topological way to define $[0, 1]$ up to homeomorphism would be to define path connectedness first: $x_1$ and $x_2$ are connected by a path in a topological space $X$ if for every Hausdorff compact $C$ and $a,b\in C$, there is a continuous $f\colon C\to X$ such that $f(a)=x_1$ and $f(b)=x_2$. Then it can be said that in every Hausdorff space $X$ with $x_1$ and $x_2$ connected by a path, every minimal subspace in which $x_1$ and $x_2$ are still connected by a path is homeomorphic to $[0, 1]$.

Maybe in some sense it can be said that $[0, 1]$ is the "minimal" Hausdorff space $X$ such that every Hausdorff compact embeds into $X^N$ for some $N$, but i do not know if this can be made precise.

In one of the answers to 92206 it was stated that $([0, 1], 0, 1)$ is a terminal object in the category of bipointed spaces equipped with the operation of "concatenation." This is the kind of answers i am interested in. As 92206 was concerned mostly with the fundamental group and tagged only with [at.algebraic-topology] and [homotopy-theory], i am asking this general topological question separately.

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This has already been discussed various times on mathoverflow. You should make precise why you are not satisfied with the answers, or/and in how far your question is different. –  Martin Brandenburg Mar 6 '13 at 15:51
    
@Martin can you give links please? I have looked for this question, but haven't found. It does not exist on MathOverflow in this form. I will edit it if i find a better way to state it, but if i just keep adding precisions, i am afraid it will become unreadable. –  Alexey Muranov Mar 6 '13 at 16:12
    
I am looking not just for any characterization, but for one that would show that "in some sense" $[0, 1]$ is the best topological space. –  Alexey Muranov Mar 6 '13 at 16:22
    
My answer mathoverflow.net/questions/76134/… also discusses $[0,1]$ from the point of view of its coalgebraic structure, as you mention in your last paragraph. This is similar also to Liviu Nicolaescu's answer below. But I disagree with your comment that "in some sense $[0,1]$ is the best topological space". –  Theo Johnson-Freyd Mar 6 '13 at 17:59

6 Answers 6

Not too long ago (2005) Harvey Friedman announced an attractive, novel characterization of the unit interval that seems to be little known, and might be the kind of answer you are looking for:

Up to isomorphism, the unit interval is the only complete totally ordered set (with end points) that has a continuous "betweenness function".

[Since arithmetical operations are continuous, it is clear that there are lots of continuous betweenness functions on the unit interval].

Here is the official characterization:

Theorem (H. Friedman). Let $X$ be a linearly ordered set with left and right endpoints and the least upper bound property. The following two statements are equivalent:

(a) $X$ is isomorphic to the usual closed unit interval.

(b) There is some $f:X^{2}\rightarrow X$ that is continuous relative to the order-topology on $X$, and $x < f(x,y) < y$ whenever $x < y$.

Friedman's proof appears in this FOM posting.

PS: Friedman's proof, when coupled with usual techniques of imposing an order on a continuum with at most two non-cut points (the "separation order") yields the following purely topological characterization [the main new idea being: in the classical characterization of the unit interval as the unique second countable continuum with exactly two non-cut points, "second countable" can be swapped with "supports a continuous betweenness function relative to the separation order"].

Theorem. Up to homeomorphism, the unit interval is the only continuum $X$ (Hausdorff, connected, and compact) in which all but two points of $X$ are cuts points, and which additionally has the property that $X^2$ supports a continuous "betweenness" function (relative to the separation order).

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Is this a topological characterization? –  Martin Brandenburg Mar 6 '13 at 22:00
    
@Martin: admittedly this is not a purely topological characterization, but clearly topology plays a pivotal role in its formulation (continuity of $f$ with respect to the product topology on $X^2$ induced by the order topology on $X$). –  Ali Enayat Mar 7 '13 at 0:28
    
@Martin/Ali: I guess one could use the van Dalen-Wattel topological characterization of ordered spaces to get rid of $<$. –  Ramiro de la Vega Mar 7 '13 at 15:07
    
@Ramiro: thanks for the pointer; I noticed another way to couch the result in terms of purely topological notions (as in the edit). –  Ali Enayat Mar 7 '13 at 21:16
    
@Martin: the edit was prompted by your remark and that of Ramiro. –  Ali Enayat Mar 7 '13 at 21:16

The following topological characterization is close to that of the real line which is indicated in another MO thread, but does not seem to have been pointed out here.

Let $X$ be a topological space, let $a,b$ be points of $X$ such that $a\neq b$. Assume that $X$ is compact, connected and separable. The following conditions are equivalent:

  1. There exists a homeomorphism $f\colon [0,1]\to X$ such that $f(0)=a$ and $f(1)=b$.
  2. Every connected subset of $X$ containing $\lbrace a,b\rbrace$ is equal to $X$.
  3. The space $X$ is locally connected, and any connected and compact subset of $X$ which contains $\lbrace a,b\rbrace$ is equal to $X$.
  4. For every $x\in X\setminus\lbrace a,b\rbrace$, the space $X\setminus\lbrace x\rbrace$ is not connected.
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What definition of separable do you use (i have heard two)? In any case, i prefer definitions without cardinality assumptions. –  Alexey Muranov Mar 6 '13 at 18:52
    
Alexey, I'm perplexed: are the two definitions you of separable have heard not equivalent? –  François G. Dorais Mar 6 '13 at 22:22
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@Alexey: separable = having a countable dense subset. What other definition have you heard? –  ACL Mar 6 '13 at 22:52
    
P.R. Halmos calls a space separable if it has a countable base of open sets. I think it is now called "second countable". (Also Halmos uses a different definition of Borel sets that is now common: for him, they are elements of the sigma-ring generated by compact sets in a locally compact space.) My first impression is that his definitions could be more meaningful than the ones i've heard elsewhere –  Alexey Muranov Mar 7 '13 at 6:17

A space which is homeomorphic to the closed unit interval is called a simple arc in the monograph "Dynamic topology" by Whyburn and Duda and there is a characterisation of it on p. 70 of this book. This assumes that the given space is a metric space, a condition which can be avoided by using the Urysohn metrization theorem.

On request, the characterisation is as follows: a space is a simple arc if and only if it is a non-degenerate (i.e., with more than one point) compact, connected set which is second countable and such that each point (with the exception of two specified ones---the endpoints) is a cut point. (A point in a connected space is a cut point, if its complement is disconnected).

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Can you add this characterization to your answer? –  Martin Brandenburg Mar 6 '13 at 18:01

This is only a guess, I don't know whether is true, but I find it plausible.

Consider all the triplets $(X, x_0, x_1)$ where $X$ is a compact, second-countable, connected Hausdorff space, and $x_0,x_1$ are distinct points in $X$. We say that the triplet satisfies the property $P$ if the following hold.

  • For any $x\in X$, $x\neq x_0,x_1$ the complement $X\setminus \lbrace x\rbrace$ has exactly two connected components, each containing exactly one of the points $x_0,x_1$. Denote by $C_i$ the component containing $x_i$, $i=0,1$.
  • Set

$$\bar{C}_i:= C_i\cup\lbrace x\rbrace$$

Then each of the triplets $(\bar{C}_i, x_i,x)$ is homeomorphic to the triplet $(X, x_i, x_{1-i})$.

Here is my claim: a triplet $(X.x_0,x_1)$ satisfies property $P$ if and only if it is homeomorphic to the triplet $(\; [0,1],0,1\;)$

Acknowledgments. I want to thank all the commenters for the useful explanations. (I've added second-countability to my claim which turned out to be an old result.) Here's an odd simple observation which I find intriguing, and it may or may not be useful.

Observe that on the space $\newcommand{\eT}{\mathscr{T}}$ $\eT$ of (homeomorphisms types of) second countable, compact, connected, Hausdorff triplets $(X,x_0,x_1)$ there is a structure of associative semigroup

$$ (X, x_0,x_1)* (Y,y_0,y_1)= (Z,x_0,y_1), $$

where $Z$ is the space obtained by gluing $X$ to $Y$ by identifying $x_1$ with $y_0$. This operation is clearly involved in defining the fundamental group.

Denote by $\newcommand{\be}{\boldsymbol{e}}$ $\be$ the triplet $([0,1],0,1)$. Observe that $\be$ is an idempotent $\be\ast\be=\be$. This fact alone makes possible the definition of the fundamental group.

The results described in the comments show that $\be$ is characterized by the property

$$ \be =t_1\ast t_2,\;\;t_1,t_2\in \eT \Longleftrightarrow t_1=t_2=\be. $$

Here is an amusing question. Is it true that $\be$ is the unique idempotent of the semigroup $(\eT,\ast)$? I'm inclined to believe that the answer is positive.

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The long line as I know it doesn’t, as one of the components $C_i$ will be a short line. Anyway, one could add “second-countable” to the list to be safe against similar examples. –  Emil Jeřábek Mar 6 '13 at 15:21
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Compactness of $X$ guarantees 2nd countability. By the way, what is "the long line"? I have never encountered this term. –  Liviu Nicolaescu Mar 6 '13 at 15:31
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Compactness implies 2nd countability for metric spaces; there are plenty of compact spaces that aren't second-countable. A compact, connected, homogeneous Suslin line would give a nice counterexample, if there happens to be such a thing. –  François G. Dorais Mar 6 '13 at 16:20
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Actually, according to math.stackexchange.com/a/322519/22772, $([0,1],0,1)$ is characterized as the unique compact, Hausdorff, second-countable space $X$ with points $x_0\ne x_1$ such that $X\smallsetminus\{x_i\}$ is connected for $i=0,1$, and $X\smallsetminus\{x\}$ is disconnected for every $x\ne x_0,x_1$. This seems to imply that Liviu’s conditions with second countability added also characterize $[0,1]$. –  Emil Jeřábek Mar 6 '13 at 16:56
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@Liviu: Maybe you want to add "connected" to the definition of $eT$. Otherwise the Cantor space with any two points gives another idempotent. –  Ramiro de la Vega Mar 7 '13 at 15:23
up vote 3 down vote accepted

Consider the class of all Hausdorff compacts with distinct points (i.e. which have more than $1$ point) that are absolute retracts in the class of Hausdorff compacts. Then $[0,1]$ is up to homeomorphism the only member of this class that embeds into every other.

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In other words, in some sense, $[0,1]$ is the "simplest" non-trivial absolute retract. –  Alexey Muranov May 25 at 15:30

I'm not sure if this counts as a characterization, but it's a cool fact worth noting. If you start with $I$ and take the reflective subcategory it generates then you get the subcategory of completely regular spaces (aka Tychonoff spaces or $T_{3.5}$ spaces). These are very important for general topology and have good categorical properties, e.g. closure under subspaces, products, etc. The class of completely regular spaces coincides with the class of uniformizable spaces, and these have also been of historical importance.

On a related note, the coreflective subcategory generated by $I$ is the subcategory of $\Delta$-generated spaces. This forms a combinatorial model category, and has gained much interest in recent years as a place to do homotopy theory without having to worry about smallness issues. Now, as far as I know this doesn't characterize $I$, because maybe something else could generate the same subcategories. If someone knows whether or not such a generator has to be unique (or unique up to ---) I'd be interested. This is CW, so feel free to edit if you know this story and want to. I learned this from my advisor, Mark Hovey. It's somewhat far away from the work I do.

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Thanks for the answer, i wish i understood what a reflexive subcategory generated by something is and how it looks like (i do not even remember the definition of the adjoint functor). However, this seems to me not very relevant to my question, as i understand it. The unit interval already appears in the definition of completely regular spaces, so my question would be rather about the other part of your answer: why are completely regular spaces important for general topology? –  Alexey Muranov Aug 7 '13 at 19:05
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My advisor is fond of saying that point-set topology likes to study metric spaces and compact Hausdorff spaces. It's natural to consider completely regular spaces because they are \emph{exactly} subspaces of compact Hausdorff spaces. See for example the Imbedding Theorem of Chapter 34 of Munkres. Furthermore, they pop up when you want to do metrization, e.g in the Urysohn Metrization Theorem (completely regular plus second countable implies metrizable), or in uniformization. As I understand it, point-set topologists in the 60s-90s did a lot of work in this subcategory –  David White Aug 7 '13 at 19:14
    
Thanks for pointing out that completely regular spaces are exactly subspaces of compact Hausdorff spaces, i need to think about it. This is more like what i was asking (can you maybe add this to your answer?). About metrization, i was in particular wondering why topologists are so fond of metrizing everything, that is mapping pairs of points to $\mathbb{R}$ (and not to some other space or ordered abelian group). –  Alexey Muranov Aug 7 '13 at 19:24
    
@AlexeyMuranov: topologists do map pairs to ordered abelian (semi)groups, see e.g. R. Kopperman, All topologies come from generalized metrics, Amer. Math. Monthly 95 (1988) 89–97. –  Francois Ziegler Aug 7 '13 at 19:45
    
@FrancoisZiegler: thanks for the reference. I understand that you can map anything to anything else, but i had an impression that, when studying Hausdorff compacts, people prefer to map to $\mathbb{R}$ or $[0,1]$. I suspected they had a good reason to choose it, and i wanted to understand it. –  Alexey Muranov Aug 7 '13 at 19:48

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