A community for students.

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

anonymous

  • 5 years ago

X is a compact metric space, M subset X, show that M is compact

  • This Question is Closed
  1. watchmath
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Do you use open cover to define compactness?

  2. mathteacher1729
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Oh man, topology! :D or D:, depending. Hehe. Seriously though, for problems where you are asked to show that a subset or subspace of something has the same properties as that which it is a subspace/subset of... Just go down through the list of properties for a compact metric space, and show that M has them all.

  3. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    @mathteacher : Sorry I am not actually conversant with that

  4. watchmath
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    By the way the statement is not true, we need M to be closed.

  5. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    @watchman you are right M is closed, that is the question

  6. watchmath
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    How you define compactness in your class? is it every open cover has finite subcover?

  7. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    he used sequences, but I have seen you definition too

  8. watchmath
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Ok, so you use every bounded sequence is convergent ?

  9. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    yes

  10. watchmath
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Sorry, I think it should be this definition Every sequence in A has a convergent subsequence, whose limit lies in A

  11. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    I just checked the course material he has the first definition also in it, so I think it can be used

  12. watchmath
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    You mean the open cover definition?

  13. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    yes

  14. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    help me watchmath..--> http://openstudy.com/groups/mathematics/updates/4de3991e4c0e8b0b0c29abd8#/groups/mathematics/updates/4de3a0794c0e8b0ba139abd8

  15. watchmath
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Consider an open cover \(M\subset \bigcup U_\alpha\) of M. Since M is closed, then \(X\backslash M\) is open in X. Now \((X\backslash M)\cup \bigcup U_\alpha\) is an open cover of \(X\). Since \(X\) is compact , then it has a finite subcover \((X\backslash M)\cup \bigcup_{i=1}^n U_i\). But now \(\bigcup_{i=1}^n U_i\) is a finite subcover of \(\bigcup U_\alpha\). Hence every open cover of M has a finite subcover, i.e., M is compact. ============== Notice that here we don't use any property of metric space at all. It means that the statement is true not only in a metric space.

  16. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    good

  17. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Can you help if the definition is "X is compact if every sequence of X has a convergent subsequence

  18. watchmath
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Ok, take a sequence \((x_n)\) in M. Since \(M\subset X\) then we consider \((x_n)\) as a sequence in X. By compacness of X then this sequence has a convergent subsequence \((x_{n_k})\). But since M is closed then the limit of this sequence is in M too. Done! :D

  19. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Can you Help me with this \[T:X \rightarrow Y \] be a linear operator and \[dim X = dim Y = n < \infty \] Show that range \[R(T) = Y\] if and only if \[T^{-1}\] exist

  20. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.