\documentclass[letterpaper,11pt]{article}

\title{18.00B Lecture Notes}
\renewcommand{\today}{February 21, 2007}

\input{../share/100B.tex}

\begin{document}
\maketitle

\definition{The \dt{closure} of a set $X$ is the intersection of all
  closed sets containing $X$}

\theorem{
  If $E \subset \R$ is nonempty and bounded above, then $\sup E \in
  \clos{E}$ (Thus, if $E$ is closed, $\sup E \in E$)
}

\proof{
  Let $x=\sup E$. If $x \notin \clos{E}$, then $x \in \comp{E}$, and $
  \exists r \st B_r(x) \isect \clos{E} = \emptyset $ (since $\comp{E}$
  is open). Therefore $x - \frac{r}{2}$ is an upper-bound for $E$, but
  is less than $\sup E$
}

\definition{$E \subset Y \subset X$ is \dt{open relative to} Y if it
  is an open subset of the metric space $Y$}

\example{
  $$\Q \subset \R$$
  $$ \{q\in\Q \st 0 < q < 1\}$$ is open relative to $\Q$, but not open
  as a subset of $R$
}

\theorem{
  Given metric spaces $Y \subset X$, $E \subset Y$ is open relative to
  $Y$ iff $E = G \isect Y$, where $G \subset X$ is open.
}
\proof{

  Note that $B_r(p) \subset Y$ = $(B_r(p) \subset X) \isect Y$,
  because they use the same distance relation.

  \begin{eqnarray*}
    \forall p \in E, \exists r_p>0 \st B_{r_p}(p) \isect Y
    \subset E \\
    G = \union_{p\in E} B_{r_p}(p) \\
    \mbox{G is a union of open sets, so is open} \\
    G \isect Y \supset E \mbox{, trivially} \\
    G \isect T \subset E \mbox{, because $G$ is a union of subsets of $E$}
    \\
    \Rightarrow G \isect Y = E
  \end{eqnarray*}
  $$\forall p \in G, \exists r > 0 B_r(p) \subset G$$

  $p \in G\isect Y \Rightarrow \exists r>0 \st B_r(p) \isect Y \subset
  G \isect Y$
  So $G \isect Y$ is relatively open in $Y$
}

\definition{
  An \dt{open cover} $\{u_\alpha\}$ of $E \subset X$ is a collection
  of open sets $u_\alpha \subset X \st \union_\alpha u_\alpha
  \supset E$ 

  An open cover of $X$ has $\union_\alpha u_\alpha = X$
}

\definition{
  $E \subset X$ is \dt{compact} if any open cover $\{u_\alpha\}$ of $E$
has a finite subcover. $\{u_{\alpha_1}, u_{\alpha_2},...,u_{\alpha_n}\}$
}

\example{
  $X$ = any finite set
}

\example{
  $E \subset R^n$ closed and bounded.
}

\theorem{
  $E \subset X$ compact $\Leftrightarrow E$ is a compact metric space
}
\proof{
  \mbox{}
  $\rightarrow$
  
  Suppose $E\subset X$ compact, $\{u_\alpha\}$ some open cover of $E$

  $\exists \{G_\alpha\}$ open cover of $E \subset X \st G_\alpha\isect
  E = U_\alpha$
  $\exists$ a finite sub-cover $\{G_{\alpha_1}, ..., G_{\alpha_n}\}
  \Rightarrow \{u_{\alpha_1}, ..., u_{\alpha_n}\}$ is a cover of E

  $\leftarrow$

  Suppose $E$ compact, $\{G_\alpha\}$ an open cover of $E \subset X$
  $\{G_\alpha\isect E\}$ is an open cover of E, and has a finite
  subcover, which covers $E\subset X$ if we consider the corresponding
  $G_\alpha$

}

\theorem{$E \subset X$ compact $\Rightarrow E$ closed}
\proof{

  Given $p \notin E$, let $u_r = B_r(p) -
  \clos{B}_{r/2}(p)$. \\ $\{u_r,r>0\}$ is an open cover of $E \subset
  X \Rightarrow \exists $ a finite subcover $ \{u_{r_1},...,u_{r_n}\}$ \\
  Take $r =
  \frac{\min\{r_1,r_2,...,r_n\}}{2}$. \\ $B_r(p) \isect u_{r_i} =
  \emptyset$

  Therefore, $B_r(p)\isect E = \emptyset \Rightarrow E$ is closed
}


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