- #1
5kold
- 13
- 0
different metrics... same topology?
Let (X,d) be a metric space. Define d':X x X-->[0,infinity) by
... d(x,y) if d(x,y)=<1
d'(x,y)=
... 1 if d(x,y)>=1
Prove that d' is a metric a d that d a d d' define the same topology on X.
This is a weird seeming metric. I am not sure how they describe the same thing except I have one example that kind of makes sense.
Imagine an infinite cylinder radius 1/2 "endowed" with metric d. Then imagine an eye observing it from directly above. To the eye, it sees no depth to the cylinder, so it can at best describe the cylinder with metric d'.
Now let x be on the very edge of the top of the cylinder and y be some ways down the side of the cylinder directly across from x. It doesn't matter to the eye.. for all it can see, x and y are just 1 unit apart since they are on opposite sides of the cylinder.
Maybe this doesn't make sense. But how would I show that d' is a metric? And what about them describing the same topology? Please help? Thanks!
Let (X,d) be a metric space. Define d':X x X-->[0,infinity) by
... d(x,y) if d(x,y)=<1
d'(x,y)=
... 1 if d(x,y)>=1
Prove that d' is a metric a d that d a d d' define the same topology on X.
This is a weird seeming metric. I am not sure how they describe the same thing except I have one example that kind of makes sense.
Imagine an infinite cylinder radius 1/2 "endowed" with metric d. Then imagine an eye observing it from directly above. To the eye, it sees no depth to the cylinder, so it can at best describe the cylinder with metric d'.
Now let x be on the very edge of the top of the cylinder and y be some ways down the side of the cylinder directly across from x. It doesn't matter to the eye.. for all it can see, x and y are just 1 unit apart since they are on opposite sides of the cylinder.
Maybe this doesn't make sense. But how would I show that d' is a metric? And what about them describing the same topology? Please help? Thanks!