1. ## 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.

2. 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.

3. 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.

4. 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.