Complete metric space
I am working on this problem, but I am not sure about one of my arguments. Could someone help me?
Let be a metric space. Define . Prove that is complete iff is complete.
Suppose is complete. Let be a Cauchy sequence in . Let . Then there exists N such that for all then
If , then implies that .
If , then I want to conclude that , but I can't justify this.
So is Cauchy in . Since is complete, converges to some . This implies there exists M such that if , then . Let . Then . So, if then . So, converges to .
I think the converse is analogous, if I can show that any Cauchy sequence in is Cauchy in , then the result would follow.
Right, the point is just that Cauchy sequences in one metric are Cauchy in the other, because d=d' when either one is small. Likewise, a sequence converges under d if and only if it converges under d'.
Thanks for your help, Tinyboss. But could you help me answer the part I can't justify?
Originally Posted by Tinyboss
You can always assume and forget the other case. Because if you find an N that works for , it always works for any larger .