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What is a purely topological characterisation of the real line( standard topology)?

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1-dimensional, locally Euclidean, connected, non-compact space. –  Francesco Polizzi Sep 22 '11 at 16:35
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saying "locally Euclidean" seems like a bit of a cheat... –  Steven Gubkin Sep 22 '11 at 16:40
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Googling the text of the question turns up a perfectly good answer as the first result, so let me cast my nonexistent vote to close. –  Artie Prendergast-Smith Sep 22 '11 at 16:48
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To define a manifold i have to define the standard topology on $ \mathbb{R}^n$ first. –  Suryateja Sep 22 '11 at 16:50
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ok ok. I was actually thinking of a characterization of $R$ among topological manifolds, and not in terms of point-set topology. –  Francesco Polizzi Sep 22 '11 at 17:38

4 Answers 4

up vote 30 down vote accepted

Here are a few examples that came up in a first search. Ward in "The topological characterization of an open linear interval", Proc. London Math. Soc.(2) 41 (1936), 191-198 proved the characterization of the real line as a connected, locally connected separable metric space, such that every point is a strong cut point (removing it leaves precisely two connected components). Franklin and Krishnarao proved that in this characterization "metric space" can be relaxed to "regular space", "On the topological characterization of the real line", Department of Mathematics, Carnegie-Mellon University, Report #69-36, 1969.

On a different note, Thron and Zimmerman prove in "A characterization of order topologies by means of minimal T0-topologies", Proc. Amer. Math. Soc. 27, (1971), 161-167, that order topologies $\tau$ on a set $X$ can be characterized as the topologies for which $(X,\tau)$ is $T_1$ and $\tau$ is the least upper bound of two minimal $T_0$ topologies. (Minimal here means that the open sets form a nested family of sets and that the complements of the point closures form a base for the topology.) Similarly the reals can be characterized as a connected, separable, $T_1$ space, and $\tau$ is the least upper bound of two noncompact minimal $T_0$ topologies.

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I like the formulation with regular better because in order to define metrisable one needs the real line, or else we need to replace it with a purely topological characterisation of metrisability (which likely will involve regularity anyway). –  Henno Brandsma Sep 22 '11 at 18:55
    
I don’t understand the given characterization of order topologies. Order topologies cannot be characterized as the unique topology such that <insert any topological condition here>, since different orders on X may generate different (even non-homeomorphic) topologies. –  Emil Jeřábek Sep 23 '11 at 14:22
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Never mind, I get it. The uniqueness requirement simply should not be there, and then it works. –  Emil Jeřábek Sep 26 '11 at 11:22

Gjergi's answer is almost certainly the best possible. But I wanted to add another one, that's a bit more "modern" in its approach to characterization problems. Namely, consider the category whose objects consist of a topological space $X$ and a pair of (edit: distinct, closed) points $l_X,r_X \in X$. This category has a monoidal structure, given by $$ (X,l_X,r_X) \otimes (Y,l_Y,r_Y) = \bigl( (X \sqcup Y) / (r_X = l_Y), l_X, r_Y \bigr). $$ Now consider the endofunctor $(-)^{\otimes 2}$ of this category that sends each object to its "tensor square". Recall that coalgebra for an endofunctor $F$ is an object $\mathcal X$ and a morphism $\mathcal X \to F (\mathcal X)$; given any endofunctor, there is a category of coalgebras. Unpacking when $\mathcal X = (X,l_X,r_X)$ and $F = (-)^{\otimes 2}$, a coalgebra is precisely a topological space $X$ with two marked points $l_X,r_X$, along with a continuous map $X \to X \sqcup_{r_X = l_X} X$ fixing the remaining marked points.

The amazing result is that the category of coalgebras of this functor has a terminal object, and that terminal object is precisely the closed interval $[0,1]$, with $l = 0$ and $r = 1$. Then it's clear how to characterize $\mathbb R = $ the open interval: it is the final coalgebra for this functor, minus its two marked points.

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Interesting. Two questions: 1) Where a reference for this result can be found? 2) Does the same characterization of $\mathbb{R}$ hold for more "combinatorial" categories, such as the category of simplicial complexes? –  Qfwfq Sep 22 '11 at 20:48
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The result is due to Peter Freyd mta.ca/~cat-dist/catlist/1999/realcoalg. –  David Corfield Sep 22 '11 at 21:34
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Theo, the result is fine for topological spaces equipped with two distinct points. The business about posets is a red herring. I refer you to this paper by Tom Leinster: arxiv.org/PS_cache/arxiv/pdf/1010/1010.4474v2.pdf –  Todd Trimble Oct 2 '11 at 0:57
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I have edited the question in such a way that Todd's comment doesn't make sense. He was responding to a temporary edit in which I wrote "Edit: Something like what I wrote below is certainly true, but Martin Brandenburg pointed out to me in an email that the link in the comments proves a related result for posets, characterizing the closed interval as a totally ordered set, not for topological spaces. Well, certainly you can extract the topology from the ordering, but take what I wrote with a grain of salt — I was going on apparently faulty memory." Since Todd says it's fine, I've un-edited it. –  Theo Johnson-Freyd Oct 2 '11 at 2:59
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This is amazingly natural, with the appropriate dose of hindsight. –  Mariano Suárez-Alvarez Oct 2 '11 at 18:45

Gjergji's answer looks very exhaustive. Let me just add the paper http://www.jstor.org/pss/2308632 where they propose a topological characterization, also in this case using the order topology (the real line is the unique linear space which is separable, connected and having neither maximum nor minimum). Personally, I would avoid the order and I really like Ward's characterization.

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Take the category of completely regular spaces (subspaces of compact Hausdorff spaces or uniformisable spaces, you do not need the real numbers) with continuous maps. The cogenerators in this category are precisely all spaces containing the real numbers as a subspace. Now you can choose all connected cogenerators which are contained in every cogenerator. You get (up to homeomorphisms) $[0,1]$, $[0,1)$ and $\mathbb{R}$. Only one of these objects can be embedded densely into the other two objects: $\mathbb{R}$. Notice that you can state this definition without refering to anything else than the morphisms in the category of completely regular spaces. This definition is quite amusing in my opinion, although it might not be very practical.

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