Anand Deopurkar

Posted on Wednesday, 04 February, 2004 - 02:51 pm:

I know that if

S \subseteq R^{n}

and if

S

is closed and bounded then

S

is compact.
By compactness I mean that every infinite open cover has a finite subcover.

I know that the result is not true in every metric space for example in Q. But is it it true in any complete metric space ?

Michael Doré

Posted on Wednesday, 04 February, 2004 - 03:07 pm:

No. The problem is that in a general metric space, a set being bounded really tells you very little. This is for the following reason - if you have a metric

d

then if you define a new metric

d' = min (d, 1)

(the ''truncation metric'') then things like completeness and compactness are obviously the same for

(X, d')

and

(X, d)

, yet any subset of the metric space

(X, d')

is trivially bounded! So a counterexample to the question you asked is the metric space

(R, d)

where

d (x, y) = min (| x - y |, 1)

. Then the subset

R

itself is closed and bounded but not compact.

In fact the crucial property is not boundedness but what is called ''total boundedness''. A metric space $(X, d)$ is totally bounded iff for every $e > 0$ you can write $X$ as a finite union of balls of radius $ε$ . (It is then easy to see that total boundedness implies boundedness. The converse holds for any subset of $R^{n}$ , but not in general.)

It is then easy to check that a metric space is compact iff it is complete and totally bounded.

If you want to think about subsets of a metric space, then you use the definition: a subset $S$ of the metric space $(X, d)$ is totally bounded iff for every $ε > 0$ $S$ can be _covered_ by a finite union of balls of radius $ε$ . Then you can show that a subset of a complete metric space is compact iff it is closed and totally bounded.

Michael

Anand Deopurkar

Posted on Wednesday, 04 February, 2004 - 03:12 pm:

Thanks... I am thinking over it.