Compact implies Sequentially Compact

[boopbeep]

Homework Statement [/b]
I need help proving that if X is a metric space and E a subset of X is compact, then E is sequentially compact.


I know I need to consider a sequence x_n in E, and I want to say that there is a point a in E and a radius r > 0 so that Br(a) [the ball of radius r with center a] contains x_k for infinitely many k's. If I show this, then I think I can conclude that any subsequence of x_n converges to a. Can someone please help?

 

[Hurkyl]

So... your conjecture is that every sequence in a compact metric space is actally convergent?

 

[boopbeep]

No, that is not one of the assumptions...my second paragraph in the problem description is a hint in the back of the textbook as to how to go about the problem.

 

[Office_Shredder]

If any subsequence of x_n converges to a, then x_n converges to a too. Rather, what you can conclude is a single subsequence of x_n converges to a (which is what sequential compactness is about)

 

[cfgauss36]

If any subsequence of x_n converges to a, then x_n converges to a too. Rather, what you can conclude is a single subsequence of x_n converges to a (which is what sequential compactness is about)

This is not true- consider a_n=(-1)^n=1,-1,1,-1,1,-1,...

Then a_{2n}=1 is a subsequence that converges to 1, but a_n does not converge at all- nonetheless to 1.

 

[HallsofIvy]

If any subsequence of x_n converges to a, then x_n converges to a too. Rather, what you can conclude is a single subsequence of x_n converges to a (which is what sequential compactness is about)

This is not true- consider a_n=(-1)^n=1,-1,1,-1,1,-1,...

Then a_{2n}=1 is a subsequence that converges to 1, but a_n does not converge at all- nonetheless to 1.

Office Shredder meant "if every subsequence". He chose the wrong word.