2022-07-10

96: For Compact \(C^\infty\) Manifold with Boundary, Sequence of Points Has Convergent Subsequence

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description/proof of that for compact \(C^\infty\) manifold with boundary, sequence of points has convergent subsequence

Topics


About: \(C^\infty\) manifold

The table of contents of this article


Starting Context



Target Context


  • The reader will have a description and a proof of the proposition that for any compact \(C^\infty\) manifold with boundary, 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: Structured Description


Here is the rules of Structured Description.

Entities:
\(M\): \(\in \{\text{ the } C^\infty \text{ manifolds with boundary }\} \cap \{\text{ the compact topological spaces }\}\)
\(s\): \(\mathbb{N} \to M\)
//

Statements:
\(\exists s^` \in \{\text{ the subsequences of } s\}, \exists m \in M (s^` \text{ converges to } m)\)
//


2: Proof


Whole Strategy: Step 1: see that \(M\) is a 1st-countable topological space; Step 2: apply the proposition that any 1st-countable topological space is sequentially compact if the space is countably compact.

Step 1:

\(M\) is a 2nd-countable topological space, by the definition of \(C^\infty\) manifold with boundary.

\(M\) is a 1st-countable topological space, by the proposition that any 2nd-countable topological space is 1st-countable.

Step 2:

\(M\) is countably compact, because \(M\) is compact, obviously.

\(M\) is sequentially compact, by the proposition that any 1st-countable topological space is sequentially compact if the space is countably compact.

That means that any sequence has a convergent subsequence.


References


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