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Do you have an example of a path-connected non-hausdorff space on which two points can't be injectively path-connected? (that is, any path between them is not injective). I tried to figure out what such a space should look like, and what its topological properties should be, but I failed.

Thank you very much, AF

Eric Wofsey
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Arthur Filippi
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    This search may be of interest: https://topology.pi-base.org/spaces?q=Path%20Connected%2B~Arc%20connected – Steven Clontz Nov 23 '23 at 21:41
  • FYI: what you call "injectively path-connected" is usually called arc connected. – PatrickR Nov 23 '23 at 23:23
  • @PatrickR: That is not true. An arc is a homeomorphic path. Consider the interval with the trivial topology: it is clearly injectively path-connected, but not arc-connected. – tomasz Nov 24 '23 at 20:41
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    @tomasz That depends on how you define it. Arc connected is defined as "injectively path-connected" in Counterexamples in Topology (Steen & Seebach) p. 29. And I guess other sources use the stronger definition of "homeomorphic path". What is the most common one? – PatrickR Nov 25 '23 at 00:04
  • @PatrickR: You're right! There seems to be some disagreement when it comes to terminology, although it seems that the outlier is Steen & Seebach. At the very least, Wikipedia, Encyclopedia of Mathematics and Wolfram Mathworld do say that an arc is a homeomorphism from the interval or its image. As does Kuratowski, Engelking and Munkres. Bourbaki does not seem to say anything about it. – tomasz Nov 25 '23 at 01:00
  • @tomasz Wolfram is not a good source. But good to know about all the other ones. Pi-base will have to be expanded to give the two versions probably. – PatrickR Nov 25 '23 at 01:03

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Sure. It's not a very interesting example, but here goes: take the indiscrete (a.k.a trivial) topology on any countable set. It is trivially path connected, since every function to it is continuous... but there are no injections $[0,1]\to F$ where $F$ is countable.

FShrike
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  • Thanks @FShrike. Is it possible on a metric space? – Arthur Filippi Nov 23 '23 at 21:28
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    @ArthurFilippi No. It is a theorem that path-connected Hausdorff spaces (which includes all path connected metric spaces!) are arcwise connected, meaning in particular we can always choose an injective path. However this theorem is quite hard to prove – FShrike Nov 23 '23 at 21:29
  • Thanks. I guess Zorn's lemma is necessary to construct arcs from paths? – Arthur Filippi Nov 23 '23 at 21:36
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    @ArthurFilippi To be honest, I have never reviewed the proof (or even seen a reference to where it is proven). My guess is that it has nothing to do with Zorn's lemma but I could be wrong. The intuition for that guess is that Zorn is highly nonconstructive, highly arbitrary, highly "weird"; it is suspicious that we can expect a continuous function out of it, never mind a topological embedding [which is very 'regular']. – FShrike Nov 23 '23 at 21:40
  • Perhaps I'm missing something, but I'm pretty sure that an arc can be constructed from a path as the limit of maximal shortcuts. The main point is that if the codomain is Hausdorff, then the set of singular arguments is closed, so there is a well-defined "first shortcut". The more delicate part is that you then need to continue these shortcuts (possibly) transfinitely, but all of this is well- (and uniquely) defined thanks to compactness and the linear order on $[0,1]$, so no choice is necessary. – tomasz Nov 23 '23 at 23:30
  • @tomasz What do you mean by singular arguments? Elements $0\le t\le 1$ such that $f^{-1}{f(t)}$ has more than one element? I'm not convinced that these are closed. Of course, each fibre $f^{-1}{f(t)}$ is closed if the codomain is Hausdorff but the union of all these fibres need not be closed – FShrike Nov 23 '23 at 23:58
  • @FShrike: Yes. They are the projection of the preimage of the diagonal via $\gamma\times \gamma$. – tomasz Nov 24 '23 at 00:03
  • @tomasz ah right, that's a nice use of compactness. Interesting. How exactly does this shortcutting work, though? Consider the very first singular argument $t_0$. There need not be a first $t>t_0$ such that $t$ is also a singular argument, so for any subsegment I consider $[t_0, t_0+\delta)$ I might cross potentially infinitely many singular arguments, perhaps infinitely many singular arguments having the same value. – FShrike Nov 24 '23 at 00:07
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    No, but there is the last $t>t_0$ such that $\gamma(t)=\gamma(t_0)$. This gives you the first shortcut. Yes, you may need to repeat this infinitely many times, passing to the limit (also maybe infinitely many times), but I think this checks out in the end. Did not check carefully though. – tomasz Nov 24 '23 at 00:09
  • FYI, it's Corollary 31.6 in Willard. – PatrickR Nov 24 '23 at 01:44
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Perhaps the most natural example is the interval with a doubled endpoint. It is a (non-Hausdorff) path-connected compact smooth manifold, so it basically has all the nice properties you can imagine a space which is path-connected but not arc-connected can have, but it is not hard to see that there is no arc connecting the twin endpoints.

A more pathological (but still not trivial, as in FShrike's answer) example is the Sierpiński two-point space, i.e. $X=\{x,y\}$ where the open sets are $\emptyset, X, \{x\}$. Then the function $\gamma\colon [0,1]\to X$ given by $\gamma(1)=y$, $\gamma(t)=x$ for $t<1$ is a path, but clearly there is no injective path, since $X$ is finite.

tomasz
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