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- Dec 15th 2008, 05:17 PMEricaMaeCompact Metric Spaces
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- Dec 15th 2008, 06:45 PMchabmgph
Hi EricaMae,

To prove A is disconnected, you will need two proper subsets in A being relatively open and closed other than the empty set. So you have the right idea there.

Now let $\displaystyle P'=P \cap A$ and $\displaystyle Q'=Q \cap A$. See if you can prove them to be relatively open and closed in A. - Dec 15th 2008, 06:56 PMEricaMae
Okay, I have P and Q open sets in A such that A is contained in (P intersect Q), P intersect A is not empty, Q intersect A not empty, and P intersect Q is empty. I know that P' and Q' are relatively open and closed, (because P' is A intersect P with A closed and P open?) same for Q'. which would imply that A is disconnected because more than the whole set and empty set are relatively open and closed, but does it matter that the question says in a compact metric space A is a subset?

- Dec 15th 2008, 07:15 PMchabmgph
P' is relatively open since A is relatively open and closed in A when you considered A as a subspace of Y, So a finite intersection of open sets is open. Similarly, Q' is relatively open.

Though you will have to showed that they are also relatively closed. (Hint: P' and Q' are complement of each other in A).

As far as the part of A being compact, I am not so sure. I suspect that it is to make sure the set is not "too big" as you may have problems in some cases. It will have to leave it to someone who is more knowledgeable to answer. Sorry. (Doh)