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.