Are compact sets in an arbitrary metric space always bounded?

[PingPong]

Homework Statement

Prove that every compact set is bounded.

Homework Equations

The usual compactness stuff - a compact set in a metric space X is one that, for every open cover, there is a finite subcover.

The Attempt at a Solution

I'm really hesitant about this question because my professor kept repeating that there is much more to compactness in general metric spaces than there was in real analysis (where compact sets are closed and bounded). A proof that I've come up with is essentially the same proof used for the Heine-Borel theorem, and I don't think it works.

Suppose a set E is compact, and consider a neighborhood around a point p, I_n=N_n(p); of course, \{I_n\} serves as a cover for E because there exists an n such that every q in X is also in N_n(p)*. But E is compact, so there a finite subcover \{I_{n_k}\}, which implies that E is bounded.

My problem with that is with the part marked *. Isn't this already assuming that it's bounded? It works for real numbers because of the Archimedian property, but a general metric space doesn't have this property.

Of course, I could always try the opposite and show that if a set E is not bounded, then it is not compact, right? For example, if I used the metric d(x,y)=\infty if x\ne y, then this metric on any set makes the set unbounded. Then it boils down to finding a cover that has no finite subcover...such as, perhaps, I_n=\{N_n(p):n\in N\} for some p\in E.

Am I anywhere near close or am I just off my rocker?

 

[Dick]

Assume E is not bounded. Then E is not contained in a ball of finite radius. Pick a point x in E and consider the set of open balls B(x,n) for all integers n>0. Does the union of that set cover E? Does it have a finite subcover?

 

[PingPong]

Right, so I sort of started that in my second attempt. My problem is, I'm not sure if this covers the entire set E - I would say yes, because the ball (or I used a neighborhood) contains all points that are infinitely separated. If the answer is yes, then obviously there is no finite subcover (if there was, then E would be bounded) and so E is not compact.

So was I on the right track with my second attempt?

 

[HallsofIvy]

The answer is yes. I'm not at all clear on what you mean by "infinitely separated", nor what ball (you say "the" ball) you are talking about. Any two points, p, q, in A have distance d(p,q) which is, by definition of "metric", a finite number. There exist an integer n larger than that distance.

 

[quasar987]

For your information, the caracterisation is as follows:

"A subset S of a metric space M is compact iff it is complete (as a subspace) and totally bounded."

In the case where M itself is complete, we have that a subset of M is complete as a subspace iff it is closed. Therefor we have the following characterizations when M is complete:

"A subset S of a complete metric space M is compact iff it is closed and totally bounded."

Now this looks somewhat more like the caracterisation we know for compact sets in R! The only difference is the "totally bounded" part.

"A set S in a metric space is said to be totally bounded if for any epsilon>0, we can find a finite cover of S by epsilon-balls."

So it is a stronger requirement then just boundedness, but in R^n the two notions coincide.