Need help in a real analysis question

[zztc2004]

I am trying to prove a question :
Assume K$$\inR^{m}$$ is compact and {xn} (n from 1 to infinite) is a sequence of points in K that does not converge . Prove that there are 2 subsequences that converge to different points in K .
Hint : Let yi=x$$_{ni}$$ be one subsequence that converges to a point in K. what is the consequence of the fact that the whole sequence goes not converge to y .

 

[Bacle]

Hi, zz:

You may be able to argue that if two subsequences {x_ni} and {x_nj} of {x_n}
converge to the same value x, then the original sequence {x_n} must itself converge
to x . It seems clear this would be true if X is metric/metrizable, when you can just use the triangle inequality, but let's see the more general case:

So assume that. Then for every 'hood U_x of x, ( 'hood :=neighborhood) there
is an N with : x_ni is in U_x for i>n, and x_nj is also in U_x.

I think that is almost it.

 

[dimitri151]

Look up Heine-Borel Theorem. See if that and the given condition of compactness tells you anything about K. Also look up Bolzano-Weierstrass Theorem. See if that helps you show there is a convergent subsequence. Also, if you can show there is one converging subsequence why is it that the whole given sequence doesn't converge? Can you do a subsequence-ectomy, then B-W Th again on what's left? I won't say more.