\documentclass[letterpaper,11pt]{article}

\title{18.00B Lecture Notes}
\renewcommand{\today}{February 16, 2007}

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

\begin{document}
\maketitle

\definition{A \dt{metric space} $X$ is a set $X$ with some distance
  function $d:X\times X\to[0,\infty)$ s.t.
\alphlist {
  \item $d(p,p) = 0$ and $d(p,q)>0$ if $p \ne q$
  \item $d(p,q) = d(q,p)$
  \item $d(p,q) \leq d(p,r) + d(r,q)$ (``Triangle inequality'')
}}

\example {
  $$X = {\mbox{rooms in MIT}}$$
  $$d(p,q) = \mbox{shortest walking distance between their centers}$$
}
\example{
  $$\R, d(x,y) = |x-y|$$
  \alphlist {
    \item trivial
    \item trivial
    \item
      $$\mbox{First,} |x+y| \leq |x| + |y|$$
      $$d(x,y) = |x - y + z - z| \leq |x - z| + |z - y| = d(x,z) + d(z,y)$$
      
  }
}
\example {
  $$C^n \mbox{, } d(\v z,\v w) = \|\v z - \v w\|$$
  \alphlist {
    \item Straightforward from definition
    \item Trivial
    \item As above, first show $\|\vz + \vw\| \leq \|\vz|| + \|\vw\| $
      \begin{eqnarray*}
        \|\vz+\vw\|^2 &=& <\vz+\vw,\vz+\vw> \\
        &=&<\vz,\vz> + <\vz,\vw> + <\vw,\vz> + <\vw,\vw>\\
        &=& \|\vz\|^2 + \|\vw\|^2 + 2\Re(<\vz,\vw>) \\
        2\Re(<\vz,\vw>) &\leq& 2\|<\vz,\vw>\| \leq 2\|\vz\|\|\vw\|\\
        \|\vz+\vw\|^2 &\leq& \|\vz\|^2 + \|\vw\|^2 + 2\|\vz\|\|\vw\| \\
        &\Rightarrow& \|\vz + \vw\| \leq \|\vz|| + \|\vw\|
      \end{eqnarray*}
      $$d(\vz,\vw) = \|\vz - \vw + \v s - \v s\| \leq \|\vz - \v s\|
      + \|\v s - \\vw\| = d(\vz,\v s) + d(\vw, \v s)$$
  }
}

\example{
  Euclidean space: $R^n, d(\v x,\v y) = ||\v x-\v y||$
  $$\left(||\v x|| = \sqrt{<\v x, \v x>}\right)$$
  Everything we said above about complex vector spaces still holds,
  since $\R$ is a subfield of $\C$
}

\claim{
  if $E \subset X$, and $(X,d)$ is a metric space, so is $(E,d)$
}

\example {
  $R^2$ with the ``taxi driver'' metric $d(\v x, \v y) = |x_1-y_1| + |x_2-y_2|$
}

\example {
  $X = $ a set of bounded functions on $R$, with
  $d(f,g) = \sup_{x \in \R}|f(x) - g(x)|$
}

\definition{
  Let $(X,d)$ be a metric space, and $p\in X$. The \dt{open ball}
  $B_r(p)$ of radius $r>0$ around $p$ is
  $$ B_r(p) = \{x\in X \st d(x,p) < r \}$$
  We call the ball ``open'' because it doesn't contain its boundary
  (We use $<r$ instead of $\leq r$)
  % Add some figures here?
}

\definition{
  $E \subset X$ is \dt{open} (in the topology given by a metric $d$)
  if
  $$\forall p \in E \mbox{ } \exists \mbox{ } r > 0 \st B_r(p) \subset E$$
}

\lemma{Open balls are open.
}
\proof{
  Given $q \in B_r(p)$, we want to find some $B_{\epsilon}(q)$
  contained in $X$
  \begin{eqnarray*}
    q \in B_r(p) &\Leftrightarrow& d(p,q) < r \\
    \mbox{Set } \epsilon &=& \frac{r - d(p,q)}{2} \\
    x \in B_\epsilon(q) &\Rightarrow& d(p,x) \leq d(p,q) + d(q,x) \\
    &\leq& d(p,q) + \epsilon < r \\
    &\Rightarrow& x \in B_r(p)
  \end{eqnarray*}
}



\definition{
  A \dt{limit point} $p\in X$ of a set $E \subset X$ satisfies \\
  
  $\forall r > 0 \exists q \ne p \subset (B_r(p) \isect E)$
}

\end{document}