Theorems

Finite sets are Compact

Proof: Consider an open cover {$G_\alpha$} covering $x_1,x_2 \dots x_N$.

$\forall x_i$ choose one $G_\alpha$ such that $x_i \epsilon {G_{\alpha_i}}$. Then {$G_{\alpha_{i}}_1 ^ N$} covers the set.

Compact sets are Bounded

Bounded: A set K is bounded in X, if $ K \subset N_r(x)$ for some $x \ \epsilon \ X$

Proof: Since K is Compact, every cover of K has a finite sub cover.

Let $B_x = N_{1}(x)$ (Ball of radius 1 around x). {$\{B_{x} \ x\epsilon K\}$} is an open cover and has a finite sub cover. Let {$B_{x_1}$,$B_{x_2}$,..$B_{x_N}$} be the finite sub cover.

Let $ r = max_{j=1:N} d(x_i,x_j)$ (r exists, because the set is finite).

Using the $\Delta$ inequality, $N_{r+1}(x_i)$ contains all K. So K is bounded in X.

Compactness is a property of the set and not the metric.

Compact sets are Closed

Nested closed sets in R are not empty

Proof: Let $I_n = [a_n \ b_n]$.

Since the sets are nested, if $m > n$ then $a_n <= \ a_m <= \ b_m <= \ a_n$.

Because all the $a_i$ are bounded by $b_i$, there exists an $x = sup({a_i})$ such that $ x \ge a_i \forall i$.

Also $ x \le b_n$ $\forall n$ because $b_n$ is an upper bound for $a_i$. So $x$ is in the intersection and hence the intersection is not empty.

This theorem also generalizes to k-cells in $R^k$.

R is uncountable

Proof: Suppose $R = \{ .. x_1,x_2, ... x_n\} $ is countable.

Choose $I_1$ as an interval which misses $x_1$. Choose $I_2 \subset I_1$ as an interval which misses $x_1 \ \& \ x_2$. $I_3 \subset I_2$ misses $x_1 \ \& x_2 \ \& x_3$ and so on.

The $I_n$s defined above are a nested sequence.

$=>$ there must be a point x in all $I_n$s which is not listed.

$=>$ R cannot be listed.

Heine - Borel theorem

Any closed interval $[a \ b]$ in R is compact in R. The theorem also generalizes to k-cells in $R^k$.