Give an example of this strange condition in metric spaces.

Asked Aug 20 '15 at 16:20

Active Aug 20 '15 at 16:48

Viewed 48 times

The condition is:

There exists homeomorphic metric spaces $(X,d), (X,d')$ such that only one of them is complete.

But is this possible?, If it is can someone give the example please?

Thanks a lot for the improvements, but the only thing that is changing is the metric not the space. And I think they are changing both in the answers :)

Thanks a lot in advance.

edited Aug 20 '15 at 16:48

asked Aug 20 '15 at 16:20

user162343

3,383

Thanks a lot for the improvements, but the only thing that is changing is the metric not the space. And I think they are changing both in the answers :) – user162343 Aug 20 '15 at 16:47
1

They're not, really - if $(X, d)$ is homeomorphic to $(X', d')$ via $f$, the function $d^: X^2\rightarrow X: (x, y)\mapsto d'(f(x), f(y))$ is a metric on $X$, and we have $(X, d^)$ is isometric to $(X', d)$. So we can just "port over" the metric on $X'$ to one on $X$. – Noah Schweber Aug 20 '15 at 16:52
See also this question. – user642796 Aug 20 '15 at 16:54
Ok thanks a lot give a second to review :) that question. – user162343 Aug 20 '15 at 18:08
Right, the thing is that I don't know topology stuff :) – user162343 Aug 20 '15 at 18:09
Thanks Noah, but I haven't seen that :( – user162343 Aug 20 '15 at 18:10

Give an example of this strange condition in metric spaces.

0 Answers0