Question

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

Answer 1

Do you use open cover to define compactness?

Answer 2

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.

Answer 3

@mathteacher : Sorry I am not actually conversant with that

Answer 4

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

Answer 5

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

Answer 6

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

Answer 7

he used sequences, but I have seen you definition too

Answer 8

Ok, so you use every bounded sequence is convergent ?

Answer 9

yes

Answer 10

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

Answer 11

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

Answer 12

You mean the open cover definition?

Answer 13

yes

Answer 14

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Answer 15

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.
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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.

Answer 16

good

Answer 17

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

Answer 18

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

Answer 19

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