A description/proof of that for compact \(C^\infty\) manifold, sequence of points has convergent subsequence
Topics
About: \(C^\infty\) manifold
The table of contents of this article
Starting Context
- The reader knows a definition of \(C^\infty\) manifold.
- The reader knows a definition of compact topological space.
- The reader knows a definition of convergence of sequence of points on topological space.
Target Context
- The reader will have a description and a proof of the proposition that for any compact \(C^\infty\) manifold, any sequence of points has a convergent subsequence.
Orientation
There is a list of definitions discussed so far in this site.
There is a list of propositions discussed so far in this site.
Main Body
1: Description
For any compact \(C^\infty\) manifold, M, any sequence of points, \(S = (p_1, p_2, . . .)\), has a convergent subsequence.
2: Proof
If S has only finite elements, S itself is a convergent subsequence with the last element as the convergent point.
Suppose S has infinite elements. There is a point, \(p \in M\), such that every neighborhood of p contains infinite elements of S, because otherwise, every point on M would have a neighborhood that contains only finite elements of S, and the set of all such neighborhoods would constitute an open cover, which would have a finite sub cover, but that is impossible because S would have only finite elements then. Now, take any chart around p, \((U_p, \phi)\), and for each natural number, i, recursively in ascending order, we can take an open set, \(U_{p-i} \subseteq U_p\), that corresponds to an open ball around \(\phi (p)\) on \(\phi (U_p)\) whose diameter is smaller than both \(i^{-1}\) and the diameter for i - 1, which is possible because \(\phi (U_p)\) is open on \(\mathbb{R}^n\). \(U_{p-i}\) contains infinite elements of S, of which we take the first not-yet-taken element as the i-th element of our subsequence. Thus, the i-th and all the subsequent elements of the subsequence are contained in \(U_{p-i}\). The issue is for any neighborhood of p, \(U'_p\), to choose a \(U_{p-j} \subseteq U'_p\). Take \(\phi (U'_p \cap U_p)\), and there is an open ball, \(B_{\phi (p)-j^{-1}} \subseteq \phi (U'_p \cap U_p)\) for a j, and \(U_{p-j} \subseteq \phi^{-1} (B_{\phi (p)-j^{-1}}) \subseteq U'_p \cap U_p \subseteq U'_p\). Thus for any neighborhood of p, the j-th and all the subsequent elements of the subsequence are contained in the neighborhood, which means that the subsequence converges to p.
3: Note
The \(C^\infty\) manifold is not required to have any metric, as convergence is defined for any topological space without the concept of metric, but nevertheless, the proposition requires the topological space being a \(C^\infty\) manifold as it uses a chart.
It can be any metric space instead of \(C^\infty\) manifold, as it has open balls directly in the space.
