# Continuity in metric spaces

## Homework Statement

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).

## Homework 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

## 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

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MathematicalPhysicist
Gold Member
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.

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

We aren't given that f is continuous, which is why I'm stuck.
If f is not continuous, then the statement in the OP is false.