1. ## Homeomorphism

So part a) is fine.

part b) I can show it's not an ismometry (with values x=4, y=1 as a counterexample). But showing it's a homeomorphism i'm a little confused, as it's the identity map, will I simply have to show that y=x is bijective, continuous, and it's inverse is continuous?

part c) I'm not really sure on how it affects this.

help would be appreciated, thanks.

2. Originally Posted by pkr

So part a) is fine.

part b) I can show it's not an ismometry (with values x=4, y=1 as a counterexample). But showing it's a homeomorphism i'm a little confused, as it's the identity map, will I simply have to show that y=x is bijective, continuous, and it's inverse is continuous?
Yes, and since its inverse is simply itself, the only "hard" part is showing that it is continuous. Actually, that's very easy because you can show that every set is open with both metrics!

part c) I'm not really sure on how it affects this.

help would be appreciated, thanks.
Show that {1} is an open set in metric d2 but not in metric d1.

3. Thanks alot for your help, but the main problem i'm having with this module is showing a set is open so i'm finding the rest of b) and c) slighty hard. I know the definition of an open set is if there's a ball of centre a, radius r that's a subest of U say, i.e all points are interior points.

But when it comes to applying this I don't have many ideas, and haven't seen nor done many examples of this myself....

4. You show that a set is open by showing that there exist a neighborhood about every member of the set that is itself contained in the set.

For example, if n is in Z then the "neighborhood of n with radius 1/2", {m||m-n|< 1/2} is just {n} since there is no other integer within distance 1/2 of n.

Similarly, with d2= 1 if $x\ne y$ and 0 if x=y, the neighborhood of n with radius 1/2 consists only of {x} because any other point has distance 1.

That is, in both of these cases singleton sets are neighborhoods themselves and so open. But every set, A, can be written as the union of the singleton sets containing the points in A and, therefore, since the union of any collection of open sets is open, every set is open. (And, since the complement of a set is still a set and so open, every set is closed.)

For Q, with d1= |x-y|, given any $\epsilon> 0$, there exist points other than x such that $|x-y|< \epsilon$ so singleton sets are NOT open. But we still have the same proof that Q, with d2, has the property that all sets are open so these are not homeomorphic.