Dear Cenap, Can you give proof? Will it hold if n points are chosen ,instead of only 2?
Finite or infinite dimensional manifolds no matter if manifold us connected we can find a chart containing n points... think sphere S^2
Think one U containing p and another V containing q so U \cap V is again a chart if intersection is smooth transversal by analtical continuation intersection is also a chart... infinire dimensional case manifold id diffeomorphiz to one open set in Hilbert space..
Here is an elementary argument. As Stampfli points out, we must assume that M is connected.
For any two points p,q of M there is a sequence B_1, … , B_k of closed n-balls in M such that p is in int(B_1) — int(B_2), q is in int(B_k) — int(B_{k-1}), and int(B_1), … , int(B_k) is a chain; i.e., int(B_i)\cap int(B_j) is non-empty iff | i — j | ≤ 1. Prove by induction on k that there is a single n-ball B whose interior U contains p and q and lies in int(B_1)\cup … \cup int(B_k).
The case k = 1 is obvious (B = B_1), so assume true for some k≥1. Suppose B_1, … , B_k, B_{k+1} is a sequence as above for points p and q. Let r be a point of int(B_k) \cap int(B_{k+1}). Then B_1, … , B_k is a sequence as above for p and r; hence, the interior V of some n-ball C lies in int(B_1), \cup … \cup int(B_k) and contains p and r. Since r and q lie in int(B_{k+1}), there is a homeomorphism g:B_{k+1} —> B_{k+1} such that g(r) = q and g is the identity bd(B_{k+1}). Let h:M —> M be the extension of g by the identity on M — B_{k+1}. Then h(p) = p, and, so U = h(V) is the interior of an n-ball B = h(C) with the desired properties.
With care one can extend this kind of argument to a finite set p_1, … , p_m, using an induction on m and sequences as above whose first term is an n-ball whose interior contains p_1, … , p_{m-1}.
Choose a smooth injective regular curve joining the two points p and q. Then a small tubular neighbourhood arond the curve is the the desired domain of the chart (the proof applies also for a finite number of points).
I will try to sum up a little bit. We must first suppose that there is an n such that every point of the manifold has a neighborhood diffeomorphic with an open set in R^n (for a fixed n). So we exclude situations like a manifold consisting of a sphere disjoint union with a line, and also situations like two intersecting planes, because the points on the line of their intersection have no such neighborhoods in the topology induced by R^3. Also, all kinds of manifold with bord are excluded.
Then there is an easier problem, which is to find a map containing p and q. For this one take small disjoint balls of radius \epsilon around the points, and their union is a map with disconnected domain. This can be always done.
A second problem is to find a connected map containing p and q. We suppose the manifold to be connected, so the question makes sense. The best idea presented so far is to construct a compact path connecting p and q, consisting only of regular points. This compact path have always a tubular neighborhood which is a map. (Just take a neighborhood of every point of the path, which is a map, and extract a finite cover.)
So the problem reduces to the existence of this compact path consisting of regular points only. The path can be only continuous. But we have supposed at the beginning that every point has a neighborhood diffeomorphic with an open set in R^n, so we take any continuous path connecting p and q and we are done!
Reciprocally, if x and y are in the same map, then there is such a path.
When does this construction not work? An example:
1. M is the conical surface obtained by rotating the line x = y around the y-axis. Let p have coordinate y < 0 and let q have coordinate y > 0. Any path from p to q must go through 0, which is singular. So there is no connected map containing p and q. But the manifold does not fulfill the condition put from the beginning, that any point has a neighborhood diffeomorphic with an open set in R^2. Exactly the point 0 does not fulfill this condition.
Since no one has as yet commented on the proof I gave earlier, perhaps I should have used the word "chart" instead of "open n-ball.” This argument, which proves that any two points in a connected n-manifold lie in a single chart, works in the smooth, PL, or TOP category, since the homeomorphism in the inductive step can be chosen accordingly. This approach avoids reliance on arcs and properties required of them to make an argument work. Or, perhaps I missed the point of the question, in which case I apologize.
@ J. L. Bryant
YES, the word "chart" instead of "open n-ball" makes it really more clear. But which axiom is applied in order to assure a finite sequence of charts whose chain of domains link p with q? (If you have a path, then you take a finite cover with charts for this compact set - but if you do not consider a path, then what do you apply?) Is the second countability axiom enough for this?
On the other hand, if a FINITE sequence of open n-balls exists, then a compact path exists as well. Just take regular points p_i in interior B_i \cap interior B_{i+1}. All points p_i, p_{i+1} can be connected by a path consisting of regular points, so we get a path consisting of regular points from p to q. So the question is just in virtue of which axiom does the finite chain of open balls exist?
Hi Mihai. We only need to know that X is a topological space: If U is an open cover of a topological space X and p is a point of X, then the set P of points q of X for which there exists a finite chain of elements of U from p to q is both open and closed in X. (Clearly P is open, and one easily shows that X - P is also open.) Thus, if X is connected, then P = X (regardless of the cover U). As I comment above, the proof I offer works in the smooth, PL, or topological category. It doesn't look to me like it works, however, in a more restricted category, such as, e.g., complex analytic. I wouldn't know what to do in such cases.
Notice that we may assume a chart in an n-manifold M is the interior of a closed n-ball, since every chart is covered by such sets.
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