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% 08/11/2022 11AM

\begin{remark*}
    If $X$ is a topological space and $K \subset
    X$ we might want to say `$K$ is compact'.
    Clearly meaningful since $K$ is topological
    space with subspace topology. Think further:
    \\
    Let $\tau$ be the topology on $X$. Then $K$ is
    compact if and only if whenever $\mathcal{C}
    \subset \tau$ with $K = \bigcup_{G \in
    \mathcal{C}} G \cap K$ then there is a finite
    $\mathcal{D} \subset \mathcal{C}$ such that
    $K = \bigcup_{G \in \mathcal{D}} G \cap K$.
    \myskip
    Equivalently, $K$ is compact if and only if
    whenever $\mathcal{C} \subset \tau$ with $K
    \subset \bigcup_{G \in \mathcal{C}} G$ then
    there is a finite $\mathcal{D} \subset
    \mathcal{C}$ with $K \subset \bigcup_{G \in
    \mathcal{D}} G$. So sometimes refer to
    $\mathcal{C}$ as being open cover of $K$ (in
    $X$).
\end{remark*}

\subsubsection*{Examples}

\begin{enumerate}[(1)]
    \item $[0, 1]$ with the usual topology is
        compact. (section 2.4 ``Heine Borel
        Theorem'' - `creeping along proof'). More
        generally, $S \subset \RR^n$ is compact if
        and only if $S$ is closed and bounded.
    \item A metric space is compact if and only if
        it is complete and totally bounded.
    \item Suppose $X$ is a discrete topological
        space. Then $\{\{x\} \mid x \in X\}$ is an
        open cover. So $X$ is compact if and only
        if $X$ is finite. (Note any finite space
        is compact).
    \item Let $X$ be indiscrete. Then the only
        open covers of $X$ are $\{\emptyset, X\}$
        and $\{X\}$, both of which are finite. So
        $X$ is compact.
\end{enumerate}

\begin{flashcard}[cts-function-on-compact-space-is-bounded]
\begin{theorem}
    A continuous real-valued function on a compact
    topological space is bounded and attains its
    bounds.
\end{theorem}
\prompt{

\begin{proof}
\cloze{
    To prove bounded, consider $f^{-1}((-n, n))$,
    for all $n$. This is an open cover of $X$,
    so has finite subcover. Deduce $f$ bounded. \\
    To prove that it attains bounds, suppose $\sup
    f = \sigma$ is not attained. Define $g(x) =
    \frac{1}{\sigma - f(x)}$. Then $g$ is a
    well-defined
    real-valued continuous function on a compact
    space, hence bounded (by previous part). So $|\sigma - f(x)|$ is
    lower bounded by a positive constant,
    contradicting definition of $\sigma$.
}
\end{proof}
}
\end{flashcard}

\begin{proof}
    Let $X$ be compact and $f : X \to \RR$ be
    continuous. Let $G_n = f^{-1}((-n, n))$ ($n
    \in \NN$). Then $\{G_n \mid n \in \NN\}$ is an
    open cover of $X$ and so, as $X$ compact, have
    a finite subcover $\{G_{n_1}, \dots,
    G_{n_k}\}$.For all $x \in G_{n_i}$, $|f(x)| <
    n_i$. Hence for all $x \in X$, $|f(x)| <
    \max_{1 \le i \le k} n_i$. Hence $f$ is
    bounded. \myskip
    Let $\sigma = \sup_{x \in X} f(x)$, and
    suppose $\sigma$ not attained by $f$. Then can
    define $g : X \to \RR$ by
    \[ g(x) = \frac{1}{\sigma - f(x)} \]
    which is well-defined and continuous. Hence by
    previous part, $g$ is bounded. But as $\sigma
    = \sup_{x \in X} f(x)$, so given $\eps > 0$ we
    can find $x$ such that $\sigma - f(x) < \eps$
    so $g(x) > \frac{1}{\eps}$. Contradiction.
    Similarly, $\in_{x \in X} f(x)$ is attained.
\end{proof}

\begin{remark*}
    Think of compactness as a `smallness'
    condition - next best thing to finiteness. For
    example: a real-valued function on a finite
    space is bounded (obvious). Here we have a
    continuous function on a compact space - how
    do we show boundedness? Use compactness to
    show space not `too big' - we can cover it
    with finitely many sets on each of which $f$
    is bounded. (Then it becomes obvious).
\end{remark*}

\noindent
More generally:

\begin{flashcard}[continuous-image-of-compact-space-is-compact]
\begin{theorem}
    A continuous image of a compact space is
    compact.
\end{theorem}

\begin{proof}
    \cloze{
    Let $f : X \to Y$ be continuous and $X$
    compact. Let $K = f(X) \subset Y$. Let
    $\mathcal{C}$ be an open cover of $K$ in $Y$.
    Then $\{f^{-1}(G) \mid G \in \mathcal{C}\}$ is
    an open cover of $X$, so by compactness, there
    is a finite $\mathcal{D} \subset \mathcal{C}$
    such that $\{f^{-1}(G) \mid G \in
    \mathcal{D}\}$ is an open cover of $X$. Then
    $\mathcal{D}$ is an open cover of $K$ in $Y$.
    So $K$ is compact.
}
\end{proof}
\end{flashcard}

\begin{remark*}
    This together with the fact that compact
    subsets of $\RR$ are closed and bounded gives
    an alternative proof of Theorem 33.
\end{remark*}

\begin{flashcard}[closed-subset-facts]
\begin{lemma*}
    \begin{enumerate}[(a)]
        \item A closed subsets of a
            \cloze{compact} space
            is compact.
        \item A compact subset of a
            \cloze{Hausdorff}
            space is closed.
    \end{enumerate}
\end{lemma*}
\end{flashcard}

\begin{proof}
    \begin{enumerate}[(a)]
        \item Let $X$ be a compact topological
            space and let $F \subset X$ be closed.
            Let $\mathcal{C}$ be an open cover of
            $F$ in $X$. Then $X \setminus F$ is
            open so $\mathcal{C}' = \mathcal{C}
            \cup \{X \setminus F\}$ then
            $\mathcal{C}'$ is an open cover of
            $X$. $X$ is compact so $\mathcal{C}'$
            has a finite subcover $\mathcal{D}'$.
            Let $\mathcal{D} = \mathcal{D}'
            \setminus \{X \setminus F\}$ if $X
            \setminus F \in \mathcal{D}'$ and
            $\mathcal{D} = \mathcal{D}'$
            otherwise. Then $\mathcal{D}$ is a
            finite subcover of $\mathcal{C}$. So
            $F$ is compact.
        \item Let $X$ be a Hausdorff space and let
            $K \subset X$ be compact. We want to
            show $K$ is closed, i.e. $X \setminus
            K$ is open, i.e. $X \setminus K$ is a
            neighbourhood of each of its points.
            \\
            Let $y \in X \setminus K$. Given $x
            \in K$, $x \neq y$ so as $X$
            Hausdorff, can find disjoint open
            $U_x, V_x \subset X$ with $x \in U_x$
            and $y \in V_x$. Then $\{U_x \mid x
            \in X\}$ is an open cover of $K$ in
            $X$ so it has a finite subcover
            $\{U_{x_1}, U_{x_2}, \dots,
            U_{x_n}\}$. Let
            \[ U = \bigcup_{i = 1}^n U_{x_i} \quad
            V = \bigcap_{i = 1}^n V_{x_i} \]
            We have $U, V$ open, $K \subset U$, $y
            \in V$ and $U \cap V = \emptyset$. In
            particular, we have found an open set
            $V$ such that $y \in V \subset X
            \setminus K$. So $X \setminus K$ is a
            neighbourhood of each of its points,
            so it's open, so $K$ is closed.
    \end{enumerate}
\end{proof}

\begin{flashcard}[continuous-bijection-compact-hausdorff-homeomorphism]
\begin{theorem}
    A continuous bijection from a \cloze{compact} space to
    a \cloze{Hausdorff} space is a homeomorphism.
\end{theorem}

\begin{proof}
    \cloze{
    Let $f : X \to Y$ be a continuous bijection,
    $X$ compact, $Y$ Hausdorff. Aim is to show
    $f^{-1} : Y \to X$ is continuous. \myskip
    Let $G \subset X$ open. Then $X \setminus G$
    is closed, so by Lemma 35(a), $X \setminus G$
    is compact. Hence by Theorem 34, $f(X
    \setminus G)$ is compact and so by Lemma
    35(b), $f(X \setminus G)$ is closed. That is,
    $Y \setminus f(G)$ is closed, i.e. $f(G)$ is
    open. But $f$ is a bijection, so
    $(f^{-1})^{-1}(G) = f(G)$ is open. So $f^{-1}$
    is continuous.
}
\end{proof}
\end{flashcard}