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Math Help - Topology of R

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    Member kezman's Avatar
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    Topology of R

    Let x_n a secuence of real numbers so that \lim_{n \to \infty}{x_n} = x. Prove that A = ({x_n} : n \in \mathbb{N}) \cup (x) is compact.
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    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 \{ x_n\} \cup \{ x\} 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.
    Last edited by ThePerfectHacker; July 1st 2008 at 07:53 PM.
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    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.
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    Quote Originally Posted by squarerootof2 View Post
    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.
    For \sqrt{2} you gave a rational argument , 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 \{ S_i| i\in I\} be an open covering of \{x_n\} \cup \{x\}. Then x\in S_{a} for some a\in I. Since this set is open there is an \epsilon > 0 such that if |y-x|<\epsilon then y\in S_a for all y in \mathbb{R}. However, the sequence \{x_n\} is convergent and so there is N\in \mathbb{N} such that if n>N then |x_n-x|<\epsilon and by above it means x_n\in S_a. Now look at the points x_1,x_2,...,x_N. The point x_1 lies in some open set in \{S_i\}, say, S_1, and x_2 lies in say S_2, ..., and x_N lies in say S_N. Then \{ S_1,S_2,...,S_N,S_a\} is an open covering for \{x_n\}\cup \{x\}. We found a subcovering. Thus, \{x_n\}\cup \{x\} is a compact set.
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