Topology of R

**kezman** · July 1st 2008, 07:13 PM

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.

**ThePerfectHacker** · July 1st 2008, 07:30 PM

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.

**squarerootof2** · July 1st 2008, 08:37 PM

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.

**ThePerfectHacker** · July 1st 2008, 08:53 PM

Originally Posted by squarerootof2

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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