hello :)

can anyone explain why for compactness of the set in finite and infinite dimensional metric spaces necessary and sufficient conditions are not the same ?

thanks for any help :D I'm desperate right now :D

- May 30th 2011, 09:42 AMsedam7Compact set in metric space (finite and infinite dimensional)
hello :)

can anyone explain why for compactness of the set in finite and infinite dimensional metric spaces necessary and sufficient conditions are not the same ?

thanks for any help :D I'm desperate right now :D - May 30th 2011, 09:52 AMPlato
- May 30th 2011, 10:01 AMsedam7
hmmm...

It's like in finite dimensional metric space if set i bounded and closed then that set is compact .... but in infinite dimensional metric spaces some "another story" (sorry but I can't express myself totally in English so I'm trying with at least google translate to get some badly needed words :D )

necessary condition for compactness of the set M subset metric space X is that for exists finite - net (or web... i don't know how to translate it) of the set M.

If metric space X is complete then that is sufficient condition .... :D - May 30th 2011, 12:29 PMPlato
Thank you for the clarification.

You see, the definition for compactness is the same in any metric space:*Any open cover has a finite subcover.*

But you are quite right that conditions that imply compactness do differ depending on the space and completeness.

In any finite Euclidean space a closed and bounded set is compact. That is necessary and sufficient.

But seems that you are confused about the notion of*totally bounded*sets, sets that have an

That is also known as precompact and in a metric space that implies compactness.

That is a bit broad, but I hope it helps. - May 30th 2011, 12:38 PMsedam7
:D thank you very very much :D

I lost it in my head somewhere that every finite dimensional metric space is complete.... :D

and that implies compactness :D

thank you very much :D again :D