Show that Z is totally bounded and perfect

[Demon117]

Homework Statement

For integers m and n, let d(m,n)=0 if m=n and d(m,n) = 1/5^k otherwise, where k is the highest power of 5 that divides m-n. Show that d is indeed a metric.

Show that, in this metric, the set Z of integers is totally bounded and perfect.

The Attempt at a Solution

I frankly do not know where to begin with this proof. I guess my first question is, what must I show precisely?

 

[micromass]

Well, what are the definitions?? That are the things you need to show!

 

[Demon117]

Well, what are the definitions?? That are the things you need to show!

I think that this is obvious, which means I should have asked a different question.

Totally Bounded - Given epsilon>0 I must show that the set Z has a finite covering by epsilon neighborhoods.

Perfect - Need to show that each point p in Z, is a cluster point. Z clusters at p, if each open ball is an infinite set.

Now, clearly in the metric each point p,q in Z is a cluster point because d(p,q) --->0 as k gets larger and larger, by the definition of the metric. Therefore Z is perfect in the metric.

I don't really understand how to show that it is totally bounded in the metric. How do I come up with some finite covering that fits this idea.

 

[micromass]

The way I always approach such questions is by giving a description of the open balls. Given n in Z and given a certain epsilon, what does B(n,epsilon) look like?? Of course, we can always choose $$\epsilon=5^k$$, this will make no difference...

 

[Demon117]

The way I always approach such questions is by giving a description of the open balls. Given n in Z and given a certain epsilon, what does B(n,epsilon) look like?? Of course, we can always choose $$\epsilon=5^k$$, this will make no difference...

Would this imply that Z is compact in the metric? or complete?

 

[micromass]

No, I haven't said anything about compactness or completeness. You only need to know the balls to form an idea on how to prove total boundedness...