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Can path connectedness be defined without using the unit interval or more generally the real numbers?

I.e., do we need Dedekind cuts or Cauchy convergence equivalence classes of the rational numbers (metric space completion) in order to define any object topologically equivalent to the unit interval?

Compared to the definition of connectedness, which only uses open and closed sets, having to use the unit interval to define path connectedness seems somewhat like using a sledgehammer.

I suspect that the answer might be no, since for every Hausdorff path connected space, the paths are homeomorphic to the unit interval (at least according to the relevant Wikipedia article). In particular, every locally path-connected Hausdorff space has a bunch of 1-manifolds as subsets.

Still it is unclear to me, since it seems like it should be able to specify all of the unit interval's topological properties without having to recourse to its analytic definition.

Your thoughts or help would both be greatly appreciated.

EDIT: this question probably has something to do with homotopy theory: https://en.wikipedia.org/wiki/Homotopy, with which I am rudimentarily familiar at best.

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    $\begingroup$ I think you will find that you need something topologically equivalent to the unit interval to do it. $\endgroup$ – Justin Benfield Jun 13 '16 at 4:05
  • $\begingroup$ @JustinBenfield Well then I guess what I mean is do we need to use Dedekind cuts or Cauchy convergence equivalence classes of the rational numbers to define something topologically equivalent to the unit interval? $\endgroup$ – Chill2Macht Jun 13 '16 at 4:17
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    $\begingroup$ It seems to me this question is probably unfalsifiable. $\endgroup$ – user98602 Jun 13 '16 at 4:29
  • $\begingroup$ @MikeMiller Why? I don't understand. Can you help me rewrite it please? $\endgroup$ – Chill2Macht Jun 13 '16 at 4:45
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    $\begingroup$ I think topologically characterizing the interval is usually done by using connectedness, which is probably too close to Dedekind cuts for your liking. There are other constructions of the reals BTW. (one of my favorites is the finite hyperrationals modulo infinitesimal hyperrationals) $\endgroup$ – Hurkyl Jun 13 '16 at 9:17
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There are really two separate questions here: can you define the unit interval space without talking about real numbers, and can you define path-connectedness without talking about the unit interval space? The answer to both is yes; let me address the second question first.

Let $P$ be a topological space and let $a,b\in P$ be two points. Say that a space $X$ is $(P,a,b)$-connected if for any $x,y\in X$, there is a continuous map $f:P\to X$ such that $f(a)=x$ and $f(b)=y$. Of course, for $(P,a,b)=([0,1],0,1)$, this is just the usual definition of path-connectedness.

However, there is a more "universal" characterization of path-connectedness that doesn't require you to know about the space $[0,1]$. Namely, a space $X$ is path-connected iff it is $(P,a,b)$-connected for all compact Hausdorff spaces $P$ with two distinct points $a,b\in P$.

To prove this, suppose $X$ is path-connected, $P$ is a compact Hausdorff space, $x,y\in X$, and $a,b\in P$ are distinct. Since $X$ is path-connected, there is a path $g:[0,1]\to X$ such that $g(0)=x$ and $g(1)=y$. By Urysohn's lemma, there is a continuous map $h:P\to [0,1]$ such that $h(a)=0$ and $h(b)=1$. The composition $gh:P\to X$ is then continuous and satisfies $g(a)=x$ and $g(b)=y$.

The idea here is that you could use any space $P$ with two chosen points $a$ and $b$ to define a notion of "paths" in a space. However, if you restrict to compact Hausdorff spaces $P$, then the ordinary interval $[0,1]$ is the "strongest possible kind of path" you can have in a space: if you have a $[0,1]$-path between two points, then you have a $P$-path for every other compact Hausdorff space $P$ as well.

(Of course, all we used about compact Hausdorff is that we know there is a map $P\to [0,1]$ separating $a$ and $b$. However, I phrased everything in terms of the compact Hausdorff condition since this is a natural condition you can define without already knowing about the space $[0,1]$.)


OK, now let me say a little about the first question. There are in fact many different ways to uniquely characterize the space $[0,1]$ up to homeomorphism without reference to the reals or anything that is essentially equivalent to constructing the reals. In fact, you can deduce one from the answer I gave to the second question above.

Namely, say that a compact Hausdorff space $(P,a,b)$ equipped with two distinct points is a universal path if it has the special property of $[0,1]$ noted above: whenever there is a $(P,a,b)$-path between two points $x$ and $y$ of an arbitrary space, there is also a $(Q,c,d)$-path from $x$ to $y$ for any compact Hausdorff space $Q$ with two distinct points. Say that a universal path $(P,a,b)$ is minimal if for any other universal path $(Q,c,d)$, there is an embedding $P\to Q$ sending $a$ to $c$ and $b$ to $d$.

I now claim that $([0,1],0,1)$ is the unique minimal universal path (up to homeomorphism). We know it is universal. To show that it is minimal, let $(Q,c,d)$ be any universal path. Since the identity map $Q\to Q$ is a $(Q,c,d)$-path from $c$ to $d$ in $Q$, universality implies there is a $([0,1],0,1)$-path from $c$ to $d$ in $Q$. But if there is a path between two points of a Hausdorff space, there is also a path which is an embedding (see Does path-connected imply simple path-connected?). Thus there is an embedding $[0,1]\to Q$ sending $0$ to $c$ and $1$ to $d$.

Now suppose $(P,a,b)$ is any minimal universal path. The previous paragraph shows that there is a $([0,1],0,1)$-path from $a$ to $b$ in $P$. Now since $([0,1],0,1)$ is a universal path, minimality of $P$ says that $P$ embeds in $[0,1]$ sending $a$ to $0$ and $b$ to $1$. But since $P$ contains a path from $a$ to $b$, the image of this embedding contains a path from $0$ to $1$, and thus contains all of $[0,1]$. Thus the embedding is actually a homeomorphism $P\to [0,1]$.

As I mentioned, this is just one of many ways of characterizing $[0,1]$. For another characterization that also relates closely to the intuitive notion of "paths", see this answer by Tom Leinster on MO.

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  • $\begingroup$ This is amazing thank you so much! So basically we can use Urysohn's Lemma to characterize path connectedness in terms of compact Hausdorff spaces in general? (and the converse direction is trivial since the unit interval is compact Hausdorff.) I am pleasantly surprised that there is such a simple and satisfying amswer to the question. If you have the free time, I would also love to hear what you have to say about defining the unit interval space without talking about real numbers -- it would really change my whole understanding of the subject! $\endgroup$ – Chill2Macht Jun 13 '16 at 11:35
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    $\begingroup$ This is such a great answer. Eric, you really do need to go back to grad school so you can, you know, eventually get paid to kick butt, rather than doing it for free. $\endgroup$ – goblin Jun 13 '16 at 17:00
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    $\begingroup$ In some sense, this answer explains why the closed interval $[0,1]$ is important in topology. Definitions like the definition of a path connected space or a homotopy are important because they have the universal characterization that Eric has outlined above. The amazing universal property satisfied by $[0,1]$ means that we can reformulate all of these definitions in terms of one familiar space. That is an extremely powerful tool to have. $\endgroup$ – John Gowers Jul 29 '16 at 17:16
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It's precisely this question which motivates the definition of an arc-connected space,which is a slight but important generalization of path connectedness. An arc is a homeomorphism f between [0,1] and f([0,1]). What you'd like to give is a generalization that doesn't involve the real numbers in any way. It's a good question and I don't really have the answer.

I suppose one could use a partially ordered set as the indexing set and establish a continuous function that is analogous to path connectedness. What's not clear is whether or not the completeness property is needed for the indexing set to produce a generalization of path connecteness. If it is, then the answer is no since only the elements of a complete ordered field would satisfy the definition and that means the real numbers come in at some point.

Without going too deep into this, there is an area of point set topology called ordered topology, which studies topological spaces with an order relation. This area has been particularly useful in operator theory, where it's been found that operators can be defined on spaces where they induce an order which in turn induces specific topologies. To answer your question,I don't think we need to go that deeply into it, but there are several excellent introductory articles which you can consult. This one by Eilenberg I think you'll find quite helpful to get you started. I think you've got an excellent starting point for an undergraduate or beginning graduate level research topic here and this looks like a good place to begin.

Addendum: I did a little research and found out something quite interesting that's probably relevant to your question. It turns out every topological space with a partial order on it is Hausdorff. Since the interval [0,1] is clearly a partially ordered set (in fact,it's totally ordered,which is a stronger condition).Since partial ordering is the weakest order type,this strongly suggests that any ordered topological space which generalizes path connectedness may be Hausdorff. You can begin your research with this conjecture,free of charge. : )

Good luck!

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    $\begingroup$ I don't think "arc-connected space" is a generalisation of "path-connected space". It is the other way around. $\endgroup$ – Marc van Leeuwen Jun 13 '16 at 11:13
  • $\begingroup$ @MarcvanLeeuwen On further thought,you're probably right. My bad. $\endgroup$ – Mathemagician1234 Jun 13 '16 at 16:06

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