Can a Totally Bounded Metric Space Be Non-Compact?

[hitmeoff]

Homework Statement

Give an example of a totally bounded metric space which is not compact

Homework Equations

Def: A metric space X is totally bounded if for each e > 0, there exists a finite number of open balls of radius e that cover X

Def: A metric space X is compact if every open cover has a finite sub cover. I other words, X is compact if, whenever {U_alpha}_{alpha in A} is an open cover of X, there are finitely many U_alpha's such that X is a subset of Ualpha¹ u U^alpha2 u ...u Ualpha^m

The Attempt at a Solution

Suppose we had a metric space X that obeyed the first def. The for every e > 0, there's a finite set of open balls with radius e that covers X.

Question is, if we take the union of these balls, is that not a finite, open cover of X? I guess, I am not not exactly seeing what the difference is between compact and totally bounded. I think (0,1) with the standard Euclidean metric works (according to the back of the book, any bounded subset of Rⁿ would work). But still not exactly seeing how it does.

In my class notes, I have it showing that (0,1) is contained in the U(1/n_i, 1-1/n_i) but is not contained in (1/N, 1-1/N) for any N, maybe I am not understanding my notes, but (1/N, 1-1/N) for any N sounds like one specific subset of U(1/n_i, 1-1/n_i), which is in a way a union of subsets of U(1/n_i, 1-1/n_i), but what about any other union of subsets of U(1/n_i, 1-1/n_i), namely U(1/n_i, 1-1/n_i) itself.

 

[hunt_mat]

Think about the unit ball in infinite dimensions...

 

[micromass]

Unit ball in infinite dimensions won't be totally bounded

In fact, a metric space is compact iff it is complete and totally bounded. So, can you come up with an easy space that is not complete?

 

[hunt_mat]

Really? The unit ball in infinite dimensions isn't bounded?

 

[micromass]

Really? The unit ball in infinite dimensions isn't bounded?

It is bounded for sure, but not totally bounded!

 

[hunt_mat]

Hmmm, interesting.

Metric spaces is a little rusty...

 

[Ratpigeon]

open, totally bounded sets aren't compact, because compactness is equivalent to sequential compactness, and open sets aren't sequentially compact, bounded or no (Very easy to prove). You can probably prove it using the open cover definition of compactness - something about how for every point in your set, there is an infinite number of points closer to the limit point (which is excluded by definition),
So for delta small enough, you can manage to get an infinite number of points between the delta neighbourhood of any point x, and the limit point, that aren't included in your cover...
(Very sketchy, missing a several important steps, but probably the gist of the proof...)

 

[Citan Uzuki]

open, totally bounded sets aren't compact, because compactness is equivalent to sequential compactness, and open sets aren't sequentially compact, bounded or no (Very easy to prove).

The empty set is sequentially compact. Also there are many metric spaces in which some nonempty open sets are compact (example: any compact metric space. More generally, any locally compact metric space with a compact connected component).

Edit: sorry, I forgot the local compactness condition the first time around.