# Given two points in a manifold, can one always find a topological ball that contains both?

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?

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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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Oh many thanks for the explanation. It is becoming clear now –  El Moro Apr 29 '11 at 6:48
I am not really sure if this works so you should take it with a grain of salt. In particular I am not certain one can always choose the balls to be small enough so that only the obvious intersections exist. –  Qiaochu Yuan Apr 29 '11 at 6:51
@Qiaochu: perhaps we could modify the argument by covering the path with charts, reducing to finitely many, and then checking that this property of paths is true in $\mathbb{R}^n$? (not that it's that clear to me that it's true for $\mathbb{R}^n$ though) –  Zev Chonoles Apr 29 '11 at 7:16
If the manifold is path-connected there is an embedded path between any two points. To prove that requires a few special cases (depending on the dimension of the ambient manifold) then the neighbourhood Qiaochu wants is given to you by the tubular neighbourhood theorem, well, really, the proof of it. –  Ryan Budney Apr 29 '11 at 7:32
In 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. –  Ryan Budney Apr 29 '11 at 7:47

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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Thank you Zev. Notably for reminding me of the equivalence above –  El Moro Apr 29 '11 at 7:09

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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Oh many thanks Ryan. I conclude then that a compact connected manifold always do have this feature. –  El Moro Apr 29 '11 at 17:32
I 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? –  yasmar Apr 29 '11 at 17:58