Analysis Question, metric spaces and subsequences?

Question

Hi I'm having a lot of trouble with this question, any help or hints on either or both would be greatly appreciated! These two parts are part of one question so they should go together.
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1) Suppose that {x_n} is a sequence in a metric space X with some distance function d(x,y), and that z is a point in X s.t. for each subsequence {x_n_k} of {x_n} there exists at least one sub-subsequence {x_n_k_l} which is convergent to z. Prove that {x_n} is convergent to z.

2) Then consider the function f:X*X->R (f is a mapping of the cartesian product of X with X into the real numbers)
and let f(x,y)=d(x,y)/(1+d(x,y)) for all x,y in X.

Show that f defines a distance function on X and that the identity mapping from X to X is uniformly continuous when viewed as a mapping between metric spaces in either way: (X,d)->(X,f) or (X,f)->(X,d)
(X,d) means metric space X with distance function d(x,y), likewise (X,f) means metric space X with distance function f(x,y).


Thank you so much!

Answer 1

1) maybe this works
assume {x_n} is not convergent to z, so it has a subsequence p(1) witch doesn't converge to z. But this subequence, by hypothesis, has a sub-subsequence that converges to z. Eliminating this sub-subequence from p(1) we obtain either
a finite set( in this case is a contradiction) or other sub-subsequence p(2) which is not convergent to z.
But p(2) itself is a subsequence of {x_n} and we can apply the same procedure to find p(3), p(4) and so on.
But we notice that we eliminate succesively from p(1) an infinity of cardinality N of disjoint sets of cardinality N.
That is, p(1), a set of cardinality N=aleph includes a set of cardinality N^N, contradiction.