| Anand
Deopurkar |
I know that if S Í Rn 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 ? |
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| Michael Doré |
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
(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
. 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 e. (It is then easy to see that total boundedness implies boundedness. The converse holds for any subset of Rn, 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 e > 0 S can be _covered_ by a finite union of balls of radius e. Then you can show that a subset of a complete metric space is compact iff it is closed and totally bounded. Michael |
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| Anand
Deopurkar |
Thanks... I am thinking over it. |