315: For Compact C^\infty Manifold, Sequence of Points Has Convergent Subsequence

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A description/proof of that for compact $C^{\mathrm{\infty }}$ manifold, sequence of points has convergent subsequence

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Topics

About: $C^{\mathrm{\infty }}$ manifold

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The table of contents of this article

• Starting Context
• Target Context
• Orientation
• Main Body
• 1: Description
• 2: Proof
• 3: Note

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Starting Context

• The reader knows a definition of $C^{\mathrm{\infty }}$ manifold.
• The reader knows a definition of compact topological space.
• The reader knows a definition of convergence of sequence of points on topological space.

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Target Context

• The reader will have a description and a proof of the proposition that for any compact $C^{\mathrm{\infty }}$ manifold, any sequence of points has a convergent subsequence.

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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.

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Main Body

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1: Description

For any compact $C^{\mathrm{\infty }}$ manifold, M, any sequence of points, $S=(p_1,p_2,...)$, has a convergent subsequence.

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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, $(U_p,\varphi )$, 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 $\varphi (p)$ on $\varphi (U_p)$ whose diameter is smaller than both $i^{-1}$ and the diameter for i - 1, which is possible because $\varphi (U_p)$ 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, $U_p^{\prime }$, to choose a $U_{p-j}\subseteq U_p^{\prime }$. Take $\varphi (U_p^{\prime }\cap U_p)$, and there is an open ball, $B_{\varphi (p)-j^{-1}}\subseteq \varphi (U_p^{\prime }\cap U_p)$ for a j, and $U_{p-j}\subseteq {\varphi }^{-1}(B_{\varphi (p)-j^{-1}})\subseteq U_p^{\prime }\cap U_p\subseteq U_p^{\prime }$. 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.

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3: Note

The $C^{\mathrm{\infty }}$ 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 $C^{\mathrm{\infty }}$ manifold as it uses a chart.

It can be any metric space instead of $C^{\mathrm{\infty }}$ manifold, as it has open balls directly in the space.

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References

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