Constructing a sequence in a manifold

Click For Summary
SUMMARY

The discussion centers on proving that a submanifold S of a manifold M is an embedded submanifold, given that every smooth function on S can be extended to a neighborhood W of S in M. The argument begins by assuming S is not embedded, leading to a contradiction through the construction of a sequence x_n in S-U converging to a point p. The key insight is that the first countability of manifolds ensures the existence of such a sequence, solidifying the conclusion that S must indeed be an embedded submanifold.

PREREQUISITES
  • Understanding of submanifolds and embedded submanifolds in differential geometry
  • Familiarity with smooth functions and their extensions
  • Knowledge of first countability and second countability in topology
  • Basic concepts of convergence in metric spaces
NEXT STEPS
  • Study the properties of embedded submanifolds in differential geometry
  • Learn about the implications of first countability in manifold theory
  • Explore the construction of smooth functions and their extensions in topology
  • Investigate examples of sequences in manifolds and their convergence behavior
USEFUL FOR

Mathematicians, particularly those specializing in differential geometry and topology, as well as students seeking to deepen their understanding of manifold theory and embedded submanifolds.

JYM
Messages
14
Reaction score
0
Given S is a submanifold of M such that every smooth function on S can be extended to a smooth function to a neighborhood W of S in M. I want to show that S is embedded submanifold.
My attempt: Suppose S is not embedded. Then there is a point p that is not contained in any slice chart. Since a submanifold is locally embedded, let U be a neighborhood of p that is embedded. Consider a smooth function on S that is supported in U and equal to 1 at p. If there is a sequence x_n in S-U that converges to p, then since f can be extended to a smooth function F on W but then 0=F(x_n) converges to F(p)=1, which is contradiction. My difficulty is to justify such a sequence exists. Please provide me your help.
 
Physics news on Phys.org
No. We can construct such a sequence. Now I get the idea; the result follows from first countablity of manifolds ( as second countable is first countable).
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
709
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 4 ·
Replies
4
Views
4K
  • · Replies 4 ·
Replies
4
Views
1K
  • · Replies 13 ·
Replies
13
Views
5K
  • · Replies 20 ·
Replies
20
Views
5K
  • · Replies 15 ·
Replies
15
Views
2K
  • · Replies 10 ·
Replies
10
Views
3K
  • · Replies 10 ·
Replies
10
Views
3K
  • · Replies 36 ·
2
Replies
36
Views
5K