Let a secuence of real numbers so that . Prove that A = is compact.

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- July 1st 2008, 08:13 PMkezmanTopology of R
Let a secuence of real numbers so that . Prove that A = is compact.

- July 1st 2008, 08:30 PMThePerfectHacker
A non-empty set in R is compact if and only if it is closed and bounded (by Heine-Borel). Thus, it is sufficient to show is closed and bounded. It is definitely bounded because convergent sequences are bounded. The last thing to show it is closed which takes more work.

- July 1st 2008, 09:37 PMsquarerootof2
i think to show that A is closed we can use the argument that for a set to be closed it must contain all of its limit points (can be proven by contradiction, ie assuming it is not, then there's a subsequence converging outside the set, which gives us a contradiction). since our set is the values of the sequence x_n, and it contains the points it converges to (which is just one point x in this case as limit is unique in the topology of R). thus A contains all of its limit points (which is just x), and must be closed. since we showed bounded already, this set must be compact by heine borel theorem.

- July 1st 2008, 09:53 PMThePerfectHacker
For you gave a rational argument (Rofl), but there is a problem with it. You are assuming that each sequence must go in order of the original sequence i.e. it is a subsequence. It does not have to.

Here is an argument that does not use Heine-Borel theorem, in fact, I think that doing this problem directly by definition is the way to go. Let be an open covering of . Then for some . Since this set is open there is an such that if then for all in . However, the sequence is convergent and so there is such that if then and by above it means . Now look at the points . The point lies in some open set in , say, , and lies in say , ..., and lies in say . Then is an open covering for . We found a subcovering. Thus, is a compact set.