# An example of a close and bounded set that is not compact

1. Aug 27, 2010

### michonamona

Take the discreet metric on an infinite set A.

I understand that its closed (because it contains all of its limit points), but I don't understand why its bounded and why its not compact.

Also, when they say "an infinite set A" do they mean a set that extends to infinite (say, [1,n] for all n in the natural numbers)? or is it a set in which there are infinitely many points in it?

Thank you,

M

2. Aug 27, 2010

### lanedance

that would depend on the context, however i would read it as infinite amount of elements, the natural numbers exmaple you gave is not bounded.

to understand why it is not compact consider the open cover comprosed of the union of a small neighbourhood (say e<1) of every point in the set

3. Aug 27, 2010

### michonamona

O.K., I now understand why its bounded.

As for its compactness, the open cover will be the neighborhoods $$V_{q}$$ for each $$q \in A$$. The reason why its not compact is because we can find an open cover that doesn't have a finite sub-cover. So the union of the $$V_{q}$$ doesn't have a finite sub-cover?

4. Aug 27, 2010

Regardless of the naturals, it's very simple to see why it's a bounded metric space.

A space is compact if, for every open cover, there is a finite subcover. As you said, if you take a cover consisting of all the singletons (indeed, a set with one element is open), and if there is an infinite number of them, can there be a finite subcover?

5. Aug 28, 2010

### michonamona

I imagine set A as a space that is contained, within it are infinitely many points that are distance of 1 apart from each other (since were using the discreet metric). If this picture of A is true, then we can just take the neighborhoods of enough points in A so that A will be covered by these neighborhoods.

The other image of A that I have in my head is that of a set that is continually expanding. It is expanding because of the infinitely many points within it that are distance of 1 away from each other. I can understand why this set will not have finitely many sub-covers.

6. Aug 28, 2010

### lanedance

however just to point out a commment on bounded sets...

any set with the discrete topology is bounded, in terms the metric generated by the topology, as any neighbourhood of a point with radius >1 contains the whole set

this wasn't really what I said in post 2 & doens't quite gel with your expanding anology, though I think i get what you were trying to describe