Metric spaces and convergent sequences

[80past2]

Homework Statement

let {x_i} be a sequence of distinct elements in a metric space, and suppose that x_i→x. Let f be a one-to-one map of the set of x_is into itself. prove that f(x_i)→x

Homework Equations

by convergence of x_i, i know that for all ε>0, there exists some n₀ such that if i≥n₀, then d(x_i,x)<ε.
by one to one, i know that if f(x)=f(y), then x=y.
x_is are distinct (which implies that there are an infinite number of points?)
f(x_i) ⊆ {x_i} (with this and one to one, does that imply the function is onto? i think yes, since the cardinality of the two sets are equal, but I'm not sure this will help me)

i want to show: for all ε>0, there exists some n₀ such that if i≥n₀, then d(f(x_i),x)<ε.

The Attempt at a Solution

so I've been staring at this problem for hours, not getting farther than stating my assumptions. I'm having a hard time even convincing myself that it's true, which is usually my first step. I'm thinking that since there are an infinite amount of points, then no matter what function i have, any mapping will "fill up" the earlier holes, leaving me with a remaining series that lies in any epsilon neighborhood, but I'm having trouble expressing that mathematically, and I'm not even sure if it's true. i got my hopes up by trying use a contradiction and assuming the negation of what I'm trying to show, thinking that since f(x_a)=x_b for some a, b in N, but all that does is show that there is a sequence, not one necessarily related by the i's (am i making sense? i mean i need the f(x_i) to converge, not create a new sequence by reordering my f(x_i) such that it converges), and we already knew that sequence exists.

 

[Dick]

If d(xi,x)<ε for all i≥n0 then the i values where d(xi,x)>=ε are finite in number, since i<=n0. So those f(i) values must have maximum, yes? Call it n1. So for i>n1 what about d(x_f(i),x)?