## anonymous 5 years ago X is a compact metric space, M subset X, show that M is compact

1. watchmath

Do you use open cover to define compactness?

2. 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.

3. anonymous

@mathteacher : Sorry I am not actually conversant with that

4. watchmath

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

5. anonymous

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

6. watchmath

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

7. anonymous

he used sequences, but I have seen you definition too

8. watchmath

Ok, so you use every bounded sequence is convergent ?

9. anonymous

yes

10. watchmath

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

11. anonymous

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

12. watchmath

You mean the open cover definition?

13. anonymous

yes

14. anonymous
15. 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.

16. anonymous

good

17. anonymous

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

18. 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

19. 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