Vrftsjtryk

This was part of an exercise given to the students in my class last week which I wasn't sure how to do. The question starts by studying the metric $d_1$ defined on $mathbb{R}^infty={x=(x_i)mid iinmathbb{N}}$ given by

$$d_1(x,y)=sum^infty_{i=1}frac{1}{2^i}frac{|x_i-y_i|}{1+|x_i-y_i|},qquad x,yinmathbb{R}^infty.$$

The students are asked to show that this is indeed a metric and compare it with another metric defined on the same space, none of which is difficult. Finally the question asks to show that if $(x^{(n)})_{ninmathbb{N}}subseteqmathbb{R}^infty$ is a sequence, then its convergence to a point $xinmathbb{R}^infty$ with respect to $d_1$ is equivalent to its pointwise convergence to the same point.

As I understand it, pointwise convergence should mean that for each $iinmathbb{N}$, and each real $epsilon>0$ there exists $N=N(epsilon,i)inmathbb{N}$ such that $|x^{(n)}_i-x_i|<epsilon$ whenever $n>N$.

Now one direction is clear to me, but proving that pointwise convergence implies $d_1$-convergence is not. Can anyone help me out with this direction?

What I need to show is that given any real $delta>0$ there exists $M=M(delta)inmathbb{N}$ such that

$$d_1(x^{(n)},x)=sum^infty_{i=1}frac{1}{2^i}frac{|x^{(n)}_i-x_i|}{1+|x^{(n)}_i-x_i|}<delta$$

whenever $n>M$, using the assumption of pointwise convergence which I believe to have interpreted correctly just above. The problem seems to be that whilst a suitable $M=M(epsilon,i)$ exists for each $i$, it is by no means clear to me that this collection of $M$s is bounded above.

Any help would be greatly appreciated.

So this is the answer I ended up giving my students. I'll use the notation introduced in the question itself.

We assume that $(x^{(n)})_{ninmathbb{n}}$ is a sequence converging pointwise to a point $xinmathbb{R}^infty$. We also assume given a real $delta>0$. Now, since for any $n$, the value $d_1(x^{(n)},x)$ is bounded above by the absolutely convergent series $sum^infty_{n=1}frac{1}{2^n}=1$ we can find for the given sequence $(x^{(n)})_{ninmathbb{n}}$ an integer $K$ such that the inequality

$$sup_{ninmathbb{N}}left{sum^infty_{i=K+1}frac{1}{2^i}frac{|x_i^{(n)}-x_i|}{1+|x_i^{(n)}-x_i|}right}<frac{delta}{2}$$

is satisfied. Notably $K$ is independent of $n$.

Now for each $i=1,dots,K$ we use the assumption of pointwise convergence to get integers $N_i$, $i=1,dots,K$, such that

$$|x_i^{(n)}-x_i|<frac{delta}{frac{2K}{2^i}-delta}.$$

Note that if necessary we are free to replace $K$ be any larger (but finite) integer so as to guarantee that $frac{2K}{2^i}-deltaneq 0$ for each $i$.

After a little rearranging we get from this that

$$sum^K_{i=1}frac{1}{2^i}frac{|x_i^{(n)}-x_i|}{1+|x_i^{(n)}-x_i|}<sum^K_{i=1}frac{delta}{2K}=Kcdot frac{delta}{2K}=frac{delta}{2}$$

whenever $n>N:=max_{i}n_i$, and we use this to get

$$d_1(x^{(n)},x)=sum^infty_{i=1}frac{1}{2^i}frac{|x_i^{(n)}-x_i|}{1+|x_i^{(n)}-x_i|}=sum^K_{i=1}frac{1}{2^i}frac{|x_i^{(n)}-x_i|}{1+|x_i^{(n)}-x_i|}+sum^infty_{i=K+1}frac{1}{2^i}frac{|x_i^{(n)}-x_i|}{1+|x_i^{(n)}-x_i|}<frac{delta}{2}+frac{delta}{2}=delta$$

which is enough to conclude that the sequence $(x^{(n)})_{ninmathbb{N}}$ converges to $x$ with respect to the metric $d_1$. This is exactly what we needed to show, so we are done.