# Continuity in metric spaces

1. Sep 18, 2012

### gotmilk04

1. The problem statement, all variables and given/known data
Show that if (x$_{n}$) is a sequence in a metric space (E,d) which converges to some x$\in$E, then (f(x$_{n}$)) is a convergent sequence in the reals (for its usual metric).

2. Relevant equations
Since (x$_{n}$) converges to x, for all ε>0, there exists N such that for all n$\geq$N, d(x$_{n}$,x)<ε.
So |x-x$_{n}$|<ε

3. The attempt at a solution
I understand that this will prove continuity, but I'm not sure how to get from d(x$_{n}$,x)<ε to what we want: d(f(x$_{n}$_,f(x))<ε

2. Sep 18, 2012

### MathematicalPhysicist

If f is continuous then from $d(x,x_n)\rightarrow 0$ you can deduce $d(f(x_n),f(x))\rightarrow 0$ (by definition).

Anyway for f general this might not be the case, take f(x)=1/x, x_n =1/n. x_n ->0 but f(0) is not defined.

3. Sep 18, 2012

### gotmilk04

We aren't given that f is continuous, which is why I'm stuck.

4. Sep 18, 2012

### micromass

Staff Emeritus
If f is not continuous, then the statement in the OP is false.