# convergence of a sequence wrt a metric

• Sep 2nd 2006, 07:39 AM
TexasGirl
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.
• Sep 2nd 2006, 09:26 AM
rgep
Convergence of a sequence $(x_n)$ implies that it satisfies the Cauchy condition, namely $\forall \epsilon>0\ \exists N = N(\epsilon) \cdot m,n>N \Rightarrow d(x_m,x_n) < \epsilon$. (Note that the Cauchy criterion does not imply convergence unless the metric space is complete.)

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

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

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