Uniform Continuity of Sequences in Metric Space

[jdcasey9]

Homework Statement

Prove that f:(M,d) -> (N,p) is uniformly continuous if and only if p(f(x_n), f(y_n)) -> 0 for any pair of sequences (x_n) and (y_n) in M satisfying d(x_n, y_n) -> 0.

Homework Equations

The Attempt at a Solution

First, let f:(M,d)->(N,p) be uniformly continuous.

Let \epsilon=2\delta.

lf(x_n)-f(y_n)l \leq lf(x_n)-x_nl + lx_n-f(y_n)l \leq lf(x_n)-x_nl + lx_n-y_nl + ly_n-f(y_n)l < \delta + 0 + \delta= 2\delta =\epsilon
(because f is uniformly continuous)

Therefore, p(f(x_n), f(y_n))->0.

Second, let p(f(x_n), f(y_n)) -> 0 for (x_n), (y_n) in M such that d(x_n, y_n) ->0.

We can do this nearly the same way, except at the end we say:

lf(x_n)-f(y_n)l \leq lf(x_n)-x_nl + lx_n-y_nl + ly_n-f(y_n)l -> 0 so it must be uniformly continuous.

 

[micromass]

lf(x_n)-f(y_n)l \leq lf(x_n)-x_nl + lx_n-f(y_n)l \leq lf(x_n)-x_nl + lx_n-y_nl + ly_n-f(y_n)l < \delta + 0 + \delta= 2\delta =\epsilon

Firstly, how did you define the absolute value? Absolute value is only defined on R, but now you're working in an arbitrary metric space.
Secondly, how did you define f(x)-f(y). Again, you're working in an arbitrary metric space, thus it may be that there is no addition/substraction defined on that space.