Questions involving a set under the usual real metric

[GridironCPJ]

Homework Statement

Let S={1/k : k=1, 2, 3, ...} and furnish S with the usual real metric. Answer the following questions about this metric space:

(a) Which points are isolated in S?

(b) Which sets are open and closed in S?

(c) Which sets have a nonempty boundary?

(d) Which sets are dense in S?

(e) Which sets are nowhere dense in S?

Homework Equations

Use any variations of the definitions of the terms involved, no restrictions.

The Attempt at a Solution

I was able to show (a). I just took an open ball around each point 1/k in S s.t. the radius of each ball is 1/2k, so each of these balls contains only the point they center. Thus, all points of S are isolated. If you see a problem with this, please let me know.

For the rest, I'm confusing myself. Sets in S can only contain the points 1/k for k=1, 2, 3, ..., so no set can be open, right? If I take an open ball centered at a 1/k, then I cannot find an open ball inside of that open ball s.t. the smaller open ball is contained in S. However, none of these open balls are even sets in S since they have infinitely many points not in S, and S is our metric space, so..? It's amazing how elementary set theory can be so mind-twisting at times. I would appreciate help on as many of these as possible.

 

[Dick]

You are off to a good start in finding that all points are isolated. You found that every point in S is itself an open set, right? This is an example of a discrete topological space. The sets in S that are open are the intersection of open balls in the reals with S.

 

[GridironCPJ]

You are off to a good start in finding that all points are isolated. You found that every point in S is itself an open set, right? This is an example of a discrete topological space. The sets in S that are open are the intersection of open balls in the reals with S.

I havn't convinced myself that every point in S is an open set. My textbook has a definition for open sets as the following:

A set G in S is is called open if for each x in G, there is an r>0 s.t. B(x, r) is in G.

This is clearly not the case for single points of S, as any open ball is not contained in the corresponding subset, which is just a point of S.

 

[algebrat]

But isn't B(1/k,1/[k(k+1)])={1/k}. Since:
1/k-1/(k+1)=(k+1-k)/[k(k+1)]
Aren't we in what's called a (topological) subspace? What's the definition of ball there?

 

[GridironCPJ]

But isn't B(1/k,1/[k(k+1)])={1/k}.


Since:
1/k-1/(k+1)=(k+1-k)/[k(k+1)]



Aren't we in what's called a (topological) subspace? What's the definition of ball there?

A ball is under the regular metric of the reals. If each singleton is open, wouldn't that make them all closed as well since the complement of each singleton is just an arbitrary union of open sets?

I'm convinced that no set in this space is dense by the way, unless anyone would like to correct me if I'm wrong.

 

[algebrat]

A ball is under the regular metric of the reals.

[STRIKE]Are we to treat S as a subspace with it's own topology? Is your books definition of ball for subspaces, or just R^n?[/STRIKE]

What does "furnish" mean at top. What is a ball in this "furnishing"? Is the ball I found above work with the books definitions? Ignore my question if it's resolved, I might be beating a dead horse.

If each singleton is open, wouldn't that make them all closed as well since the complement of each singleton is just an arbitrary union of open sets?

I agree.

I'm convinced that no set in this space is dense by the way, unless anyone would like to correct me if I'm wrong.

My first guess is you are right. Not obvious to me how to prove it though, looks like fun.

 

[GridironCPJ]

[STRIKE]Are we to treat S as a subspace with it's own topology? Is your books definition of ball for subspaces, or just R^n?[/STRIKE]

What does "furnish" mean at top. What is a ball in this "furnishing"? Is the ball I found above work with the books definitions? Ignore my question if it's resolved, I might be beating a dead horse.


I agree.

My first guess is you are right. Not obvious to me how to prove it though, looks like fun.

I think it just means that the space has that specific metric. The balls are the same as they are in the reals under the usual metric.

 

[Dick]

A ball is under the regular metric of the reals. If each singleton is open, wouldn't that make them all closed as well since the complement of each singleton is just an arbitrary union of open sets?

I'm convinced that no set in this space is dense by the way, unless anyone would like to correct me if I'm wrong.

That's right. Every singleton is open and closed. And, yes, the only set that is dense in S is S itself. Can you prove it?