\magnification 2200 \parindent 0pt \parskip 10pt \pageno=31 \hoffset -0.2truein \hsize 7.1truein \vsize 9.7truein \def\u{\vskip -10pt} \def\s{\vskip -5pt} \def\v{\vskip -9pt} 26. Compact Spaces Defn: A collection ${\cal C}$ of subsets of $X$ is said to have the {\bf finite intersection property} if for every finite subcollection $\{C_1, ..., C_n\}$ of ${\cal C}$, $C_1 \cup ...\cup C_n \not= \emptyset$ \end Thm 4: If $f:D \rightarrow R$ is continuous and if $D$ is compact, then $M = sup_{\{p \in X\}} f(p)$ and $m = inf_{\{p \in X\}} f(p)$ exist and there exist points in $p, q \in D$ such that $f(p) = M$ and $f(q) = m$. Defn: $f: D \rightarrow Y$. is {\bf uniformly continuous} if all $\epsilon > 0$, there exists a $\delta > 0$ such that \centerline{ $d_X(x, p) < \delta$ implies $d_Y(f(x), f(p)) < \epsilon$} \vskip -10pt for all $x \in D$. Defn: If $f:D \rightarrow Y$ is continuous and if $D$ is compact, then $f$ is uniformly continuous. \end Thm 3.12 (Heine-Borel Thm) Suppose that a family ${\cal F}$ of open intervals covers the closed interval $I = [a, b]$. Then a finite subfamily of ${\cal F}$ covers $I$. A family of sets, ${\cal F} = \{F_\alpha ~|~ \alpha \in A\}$, covers the set $S$ if \centerline{$S \subset \cup_{\alpha \in A}F_\alpha$} ${\cal F}$ is said to be a cover of $S$. Defn: In a metric space $X$, a set $A$ is compact if every sequence contains a subsequence which converges to a point in $A$. A set $A$ is bounded if there exists an open ball, $B(p, r)$ such that $A \subset B(p, r)$. Thm 6.13: Compact metric implies closed \& bounded. Thm 6.14: In ${\cal R}^n$ with standard metric, {closed \& bounded implies compact.} Thm: Convergence implies Cauchy. Thm: In a compact metric space, Cauchy implies convergence. Thm: In $R^n$, Cauchy implies convergence (even though $R^n$ is not compact). Thm 6.15 (Heine-Borel Theorem) \v Suppose $X$ is a metric space and $A \subset X$, then the following are equivalent: \v (1.) Every sequence contains a subsequence which converges to a point in $A$. \v (2.) Every open cover of $A$ contains a finite subcover. \end