# Thread: convergence of a sequence wrt a metric

1. ## convergence of a sequence wrt a metric

(1) Let [0,1]* be the set of all functions N-->[0,1] (ie of all sequences in [0,1]). Define a sequence (Fn) in [0,1]* by:

Fn(k)= 0 if k different from n; 1 otherwise.

Consider the following metric on [0,1]*: d1(f,g):=sup{|f(n)-g(n)|:n element of N}.
Determine whether (Fn) converges with respect to d1.

(2) same as in (1) but with respect to the metric d2(f,g):= (sum from i=0 to infinity)[(2^-i)|f(i)-g(i)|.

Note that N is the set of all natural numbers including zero.

I do not know how to check convergence with respect to a metric.

2. Convergence of a sequence $\displaystyle (x_n)$ implies that it satisfies the Cauchy condition, namely $\displaystyle \forall \epsilon>0\ \exists N = N(\epsilon) \cdot m,n>N \Rightarrow d(x_m,x_n) < \epsilon$. (Note that the Cauchy criterion does not imply convergence unless the metric space is complete.)

Now consider (1). We have $\displaystyle d(f_m,f_n) = 1$ whenever $\displaystyle m \neq n$. Hence the Cauchy criterion with $\displaystyle \epsilon = 1/2$ does not hold.

On the other hand, convergence of a sequence $\displaystyle x_n$ to $\displaystyle y$ is defined as saying that the sequence of real numbers $\displaystyle d(x_n,y)$ converges to zero.

For (2), you have to decide what the sequence converges to, and I suggest you consider the sequence $\displaystyle z_n = 0$. Then consider $\displaystyle d(f_n,z) = 2^{-n}$ to see that $\displaystyle f_n \rightarrow z$.