Question

A question about metric spaces/topology: Let L be the set of all sequences of real numbers x={x1, x2,...,xn,...} with the property that the series |x1| + |x2| + ... is convergent. If we define d(x,y)=|x1-y1| + |x2-y2| + ... for all x,y in L, prove that (L,d) is a metric space.

Answer 1

1) |xn-yn| is non-negative
2) for any n, |xn-yn| = 0 for xn=yn
3) |xn-yn| = |yn-xn|
4) and finally for the last property
d(x,y) = |x1-y1| + |x2-y2| + ...
d(y,z) = |z1-y1| + |z2-y2| + ...
d(x,z) = |x1-z1| + |x2-z2| + ...

as we know, |xn-zn| <= |xn - yn| + |yn-zn| (i guess triangle inequality,

d(x, z) < = d(x, y) + d(z, y)

hence it is a metric space

Answer 2

Thanks experimentX! It's so easy and yet I didn't see it. :-)

Answer 3

@experimentX: Working more on this I have two additional questions:
1) Do I not also have to show that the series defining d(x,y) converges for all x,y or does it not matter if d(x,y) is infinite for some choice of x and y? Would it be ok for the triangle inequality to be inf <= inf + inf?
2) So far we know that x=y => d(x,y)=0, but I think the reverse must also be shown, right? There could be a series with x<>y for which d(x,y) converges to zero... So I'd still have to show that d(x,y) => xn = yn .

Answer 4

The last formula should be d(x,y)=0 => xn = yn

Answer 5

2) Can be shown like this: All terms in d(x,y) are positive. There is no way that two terms can cancel each other out. So if there is just one difference between any xn and yn then d(x,y) will be non-zero. Therefore d(x,y)=0 implies xn=yn.

Answer 6

1) I think this can be shown like this: We know that |xn| + |yn| >= |xn + yn|. This also implies |xn|+|yn| >= |xn - yn|. Therefore sum(|xn|)+sum(|yn|) >= sum(|xn - yn|). Both, sum(|xn|) and sum(|yn|) are finite values, therefore d(x,y) = sum(|xn - yn|) must be finite, too. Is this right?