I hope you will bear with me and excuse me if my question is kind of obvious to many of you. In an $n$-manifold say we have $x,y$ in it. Could we always find an open $n$ dimensional ball contained in the manifold such as both points belong to it?
Given two points in a manifold, can one always find a topological ball that contains both?
3 Answers
You need at least that the manifold is path-connected. I think the following works: given a path $p : I \to M$ with no self-intersections from $x$ to $y$ we can cover every point in the image of $p$ by a small open subset homeomorphic to the open $n$-ball. By compactness this cover has a finite subcover, and by choosing the balls small enough (not in the sense of a metric but in the sense that they only intersect each other when consecutive) their union should also be homeomorphic to an $n$-ball.
Edit: Ryan Budney's explanation in the comments shows I am being naive. It is not obvious that the existence of a path implies the existence of a particularly nice path without additional work.
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3In dimension 1 the argument is very similar to the classification of 1-manifolds. In dimension 2 you can approximate the path by an immersion with regular double points, but then you have to do double-point removal and smoothing. In dimension 3 and higher you can approximate by an embedding. So it's not terribly uniform. – 2011-04-29
To complement Qiaochu's answer, here is why this is impossible when $M$ is not path-connected (note that manifolds are connected if and only if path-connected, this is Prop 1.8 in Lee Smooth Manifolds).
If $x,y\in M$ are in different connected components of $M$, then any open set of $M$ that contains $x$ and $y$ would necessarily be disconnected, which an open $n$-ball is not.
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0Thank you Zev. Notably for reminding me of the equivalence above – 2011-04-29
Another argument would be to put a complete Riemann metric on the manifold -- this can be done fairly easily, for example if the manifold was a closed subset of Euclidean space (can be done via the Whitney embedding theorem) then the induced metric is complete by Hopf-Rinow: http://en.wikipedia.org/wiki/Hopf-Rinow_theorem
Okay, so now the manifold is geodesically complete, so given any two points $p,q \in M$ take a shortest geodesic segment connecting $p$ to $q$. This is an embedded arc by design. Now a little regular neighbourhood of that arc is what you're looking for.
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1I guess it wasn't specified that the manifold was smooth; I assumed that was the source of the complication discussed in Qiaochu's answer. If our manifold doesn't admit a differentiable structure, I guess this answer is no longer valid. What about Qiaochu's answer, augmented by your comments? – 2011-04-29