anonymous
  • anonymous
X is a compact metric space, M subset X, show that M is compact
Mathematics
  • Stacey Warren - Expert brainly.com
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schrodinger
  • schrodinger
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watchmath
  • watchmath
Do you use open cover to define compactness?
mathteacher1729
  • mathteacher1729
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.
anonymous
  • anonymous
@mathteacher : Sorry I am not actually conversant with that

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watchmath
  • watchmath
By the way the statement is not true, we need M to be closed.
anonymous
  • anonymous
@watchman you are right M is closed, that is the question
watchmath
  • watchmath
How you define compactness in your class? is it every open cover has finite subcover?
anonymous
  • anonymous
he used sequences, but I have seen you definition too
watchmath
  • watchmath
Ok, so you use every bounded sequence is convergent ?
anonymous
  • anonymous
yes
watchmath
  • watchmath
Sorry, I think it should be this definition Every sequence in A has a convergent subsequence, whose limit lies in A
anonymous
  • anonymous
I just checked the course material he has the first definition also in it, so I think it can be used
watchmath
  • watchmath
You mean the open cover definition?
anonymous
  • anonymous
yes
watchmath
  • watchmath
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.
anonymous
  • anonymous
good
anonymous
  • anonymous
Can you help if the definition is "X is compact if every sequence of X has a convergent subsequence
watchmath
  • watchmath
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
anonymous
  • anonymous
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

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