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What are the most general criteria we can impose on a locally path connected Hausdorff space $X$ and a path connected subset $A$ such that $\overline{A}$ is path connected? Do more restrictions need to be imposed on $X$ or $A$?

For instance, I know that if $\overline{A}$ is locally path connected then $\overline{A}$ is path connected; for all $x \in \overline{A}$ and some neighborhood $U$ of $x$ that is open in $\overline{A}$, there must be some path connected neighborhood $U' \subseteq U$ of $x$ that is open in $\overline{A}$. That is, there is some open subset $V'$ of $X$ such that $U' = V' \cap \overline{A}$. Since $x$ is a point of closure of $A$, $V'$ must contain some point $x' \in A \subseteq \overline{A}$, i.e. $x' \in U'$, so $x$ is path connected to $x'$ and hence also to $A$. This holds for all $x \in \overline{A}$ so $\overline{A}$ is path connected.

However, the tough part is proving that $\overline{A}$ is locally path connected, because $\overline{A}$ is probably (?) not open in $X$. I'm a complete novice so the only useful thing I know from browsing definitions is that all open subsets of a locally path connected space inherit the local path connectivity. Are there more ways to prove that a subspace inherits local path connectedness?

This is more specific, but would it help if I knew that $A$ was the set difference of two closed sets (i.e. the intersection of a closed set and an open set)?

I've been looking at stronger restrictions such as $X$ being locally simply connected, but the online documentation is scarce. Would local simple connectivity be "inherited more easily" by subspaces?

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    I don't think you'll get far imposing conditions on $X$: $\mathbb{R}^2$ is about as nice a space as you could wish for, but $A=\{(x,\sin(1/x))| x \in (0,1) \}$ still gives a counterexample. Here $A$ is homeomorphic to $\mathbb{R}$, another very nice space, suggesting imposing conditions on the topology of $A$ won't help either. It looks like you really do need conditions specifically on how $A$ sits in $X$.2012-06-25
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    @ChrisEagle: IMO that example is decisive enough to be an answer... oh except that now the question has been edited. I wonder if this is not really just two questions, now.2012-06-25
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    Good point, I guess I really have to prove some restrictions on $\overline{A}$ to exclude pathological cases such as the Topologist's Sine Curve (in fact, I'd prefer working on $A \cup \{x\}$, where $x \in \overline{A}$), such as local path connectivity (which I can't prove easily) or local simple path connectivity (which I don't fully understand).2012-06-25
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    Considering the edit, the problem in Chris' counterexample is not that the function doesn't exist, but that it fails to be continuous, which sounds like it would be hard to prove in generality.2012-06-25
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    Sorry for the change, but the initial question was to give a sufficient condition to construct the $p$ I mention in my edits. The second half of the question now contains another attempt at the same construction.2012-06-25
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    If your space has a metric, you can consider the following quantitative version of path-connectedness: a space is *quasiconvex* if two points $p,q$ can be joined by a curve of length $L\le Cd(p,q)$ where $C$ is a constant independent of $p,q$. The closure of a quasiconvex set is quasiconvex. And I think one can relax this condition to $L\le \omega(d(p,q))$ where $\omega$ is a modulus of continuity.2012-06-25
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    "This is more specific, but would it help if I knew that A was the set difference of two closed sets (i.e. the intersection of a closed set and an open set)?" I'm fairly sure the topologists sine curve is the intersection of $\bar{A}$ and the right half-plane.2012-06-25
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    @Kovalev: I think that will be quite helpful, but hopefully I don't have to use the magic bullet of metric spaces and even further restrictions.2012-06-26
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    As another way of looking at things, it is known that $x \in \overline{A}$ if and only if there exists a net of points in $A$ that converges to $x$. However, under what conditions are the points of closure of $A$ precisely the terminal points of paths in $A$? That is, we require that the nets are paths.2012-06-30
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    @LeonidKovalev: Do you have some references that discuss quasiconvex metric spaces? All I can find on the internet are very technical papers...2012-06-30
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    The book *Metric structures...* by Gromov has a short appendix on quasiconvex domains written by Pansu. Quasiconvexity appears in the book *Topics on analysis in metric spaces* by Heinonen, but only briefly. The webpage of [David Herron](https://math.uc.edu/~herron/research.html) has Beamer slides of two talks on the subject, which should be easier to follow than the papers themselves.2012-06-30
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    @LeonidKovalev: Thanks for the references. I'll probably need more time to digest them, but at first glance quasiconvexity seems like a rather strong and specific requirement on a metric space. Is my rough understanding correct that a metric space is quasiconvex if the "holes" in the space are bounded in diameter? The size of the largest hole would supply the constant $C$. If so, then perhaps a lot more "usual" spaces would fit the requirement, and I'd be more willing to use it.2012-06-30
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    No, that's not right since $C$ is a multiplicative constant (hence unit-less). Quasiconvexity is about the shape of holes rather than their size. Any size is OK as long as the shape is round-ish (like a square, for example). I agree that this is a strong requirement. This is why I mentioned a more general concept of "uniformly path-connected"; for every $\epsilon>0$ there exists $\delta>0$ such that any pair of points at distance $<\delta$ can be connected by a curve of length $<\epsilon$. This property passes to the closure for the same reason as uniform continuity does.2012-06-30
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    Actually, the definition in my comment of Jun 25 at 15:17 is probably better. The "modulus of path-connectivity" function $\omega$ should be required to be *finite* on $(0,\infty)$, with $\omega(0+)=0$. Defined this way, uniformly path-connected implies path-connected. And I don't remember seeing "uniformly path-connected" in the literature at all.2012-06-30
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    let us [continue this discussion in chat](http://chat.stackexchange.com/rooms/3945/discussion-between-herng-yi-and-leonid-kovalev)2012-07-01
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    @HerngYi Sorry, I don't use the chat. Feel free to ask by email, leonidvkovalev@gmail.com2012-07-01
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    @LeonidKovalev: I've decided to use a formalization of connected sets that is removed from its topological context: [Connectivity Spaces](http://www.math.shimane-u.ac.jp/memoir/39/D.Buhagiar.pdf)2012-07-14

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