# Metric spaces and convergent sequences

1. Feb 26, 2012

### 80past2

1. The problem statement, all variables and given/known data
let {xi} be a sequence of distinct elements in a metric space, and suppose that xi→x. Let f be a one-to-one map of the set of xis into itself. prove that f(xi)→x

2. Relevant equations
by convergence of xi, i know that for all ε>0, there exists some n0 such that if i≥n0, then d(xi,x)<ε.
by one to one, i know that if f(x)=f(y), then x=y.
xis are distinct (which implies that there are an infinite number of points?)
f(xi) ⊆ {xi} (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(xi),x)<ε.

3. 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(xa)=xb 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(xi) to converge, not create a new sequence by reordering my f(xi) such that it converges), and we already knew that sequence exists.

2. Feb 26, 2012

### 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)?

Last edited: Feb 26, 2012