% vim: tw=50
% 10/11/2022 11AM

\subsection{Products}

Have $\RR$ with the usual topology. Would like
$\RR \times \RR$ to be $\RR^2$ with the Euclidean
topology. In general, if $(X, \tau)$ and $(Y,
\sigma)$ are topological spaces, what sensible
topology can we put on $X \times Y$? \\
In general, $\tau \times \sigma$ is not going to
be a topology. For example, $\RR \times \RR$
\begin{center}
    \includegraphics[width=0.6\linewidth]
    {images/58d40c5660e911ed.png}
\end{center}
Open ball not in $\tau \times \sigma$, but each
point in ball \emph{is} in some set in $\tau
\times \sigma$ confined in the ball. But open ball
\emph{is} union of some sets in $\tau \times
\sigma$. \myskip
In general, the product topology from $\tau,
\sigma$ is going to be collection of all unions of
sets in $\tau \times \sigma$.

\begin{definition*}
    A \emph{$\pi$-system} on a set $X$ is a
    non-empty subset $\Pi \subset \mathcal{P} X$
    such that $A, B \in \Pi \implies A \cap B
    \in \Pi$.
\end{definition*}

\begin{proposition}
    Let $\Pi$ be a $\pi$-system on a set $X$. Then
    \[ \tau = \left\{ \bigcup_{A \in \Sigma} A
    \mid \Sigma \subset \Pi \right\} \cup \{\emptyset,
    X\} \]
    is a topology on $X$.
\end{proposition}

\begin{proof}
    Clearly $\emptyset, X \in \tau$ and it's
    closed under arbitrary unions. Now suppose $G,
    H \in \tau$. If $G = \emptyset, X$ or $H =
    \emptyset, X$, then $G \cap H \in \tau$
    trivially. Otherwise, $G = \bigcup_{A \in
    \Phi} A$, $H = \bigcup_{B \in \Theta} B$ for
    some $\Phi, \Theta \subset \Pi$. Then
    \[ G \cap H = \bigcup_{\substack{A \in \Phi\\B
    \in \Theta}} (A \cap B) = \bigcup_{C \in
    \Sigma} C \]
    where $\Sigma = \{A \cap B \mid A \in \Phi, B
    \in \Theta\} \subset \Pi$.
\end{proof}

\myskip
We call $\tau$ the topology \emph{generated by}
$\Pi$.

\begin{proposition}
    Let $(X, \tau), (Y, \sigma)$ be topological
    spaces. Then $\tau \times \sigma$ is a
    $\pi$-system on $X \times Y$.
\end{proposition}

\begin{proof}
    $\emptyset = \emptyset \times \emptyset \in
    \tau \times \sigma$. So $\tau \times \sigma
    \neq \emptyset$. Now suppose $A, B \in \tau
    \times \sigma$. Then $A = G \times H$, $B = K
    \times L$ for some $G, K \in \tau$ and some
    $H, L \in \sigma$. So
    \[ A \cap B = (G \cap K) \times (H \cap L) \in
    \tau \times \sigma \qedhere \]
\end{proof}

\begin{flashcard}[product-topology-defn]
\begin{definition*}
    Let $(X, \tau), (Y, \sigma)$ be topological
    spaces. The \emph{product topology} on $X
    \times Y$ is the topology generated be the
    $\pi$-system
    \[ \cloze{\{U \times V : U \in \tau, V \in
    \sigma\}} \]
\end{definition*}
\end{flashcard}

\noindent
Exercise: If $X = Y = \RR$, $\tau = \sigma =
\text{usual topology}$. Then the product topology
on $\RR^2$ is the Euclidean topology. (Example
sheet 3 with guidance).

\begin{flashcard}[product-hausdorff-compact]
\begin{theorem}
    \begin{enumerate}[(a)]
        \item A product of \cloze{Hausdorff} spaces is
            \cloze{Hausdorff}.
        \item A product of \cloze{compact} spaces is
            \cloze{compact}.
    \end{enumerate}
\end{theorem}
\end{flashcard}

\begin{hiddenflashcard}[product-hausdorff-is-hausdorff-proof]
Proof that product of Hausdorff spaces is
Hausdorff? \\
\cloze{
Let $(x, y), (z, w) \in X \times Y$ with $(x, y)
\neq (z, w)$. WLOG $x \neq z$. Pick $G, H$ open in
$X$ with $x \in G$, $z \in H$, $G \cap H = \emptyset$.
Then $G \times Y, H \times Y$ are non-intersecting
open sets in $X \times Y$ containing $(x, y)$ and
$(z, w)$ respectively.
}
\end{hiddenflashcard}

\begin{hiddenflashcard}[product-compact-is-compact-proof]
Proof that product of compact spaces is compact?
\\
\cloze{
Let $\mathcal{C}$ be an open cover of $X \times
Y$. For each $x \in X$ define the following:
\begin{itemize}
    \item For each $y$ pick some $G_{(x, y)} \in
        \mathcal{C}$ containing $(x, y)$. Then
        pick $U_{(x, y)}$, $V_{(x, y)}$ open in
        $X$ and $Y$ respectively with $(x, y) \in
        U_{(x, y)} \times V_{(x, y)} \subset
        G_{(x, y)}$.
    \item Note $V_{(x, y)}$ is an open cover of
        $Y$. Take a finite subcover, $V_{(x,
        y_i)}$. Let $U_x = \bigcap_{i=1}^n
        U_{(x, y_i)}$. Then $U_x$ open in $X$ and
        \[ U_x \times Y \subset \bigcup_{i = 1}^n
        (U_{(x, y_i)} \times V_{(x, y_i)}) \subset
        \bigcup_{i = 1}^n G_{(x, y_i)} \]
\end{itemize}
Now $U_x$ forms an open cover of $X$, so take a
finite subcover, $U_{x_i}$. Then
\[ X \times Y \subset \bigcup_{i = 1}^n (U_{x_i}
\times Y) \subset \bigcup_{i = 1}^n \bigcup_{j =
1}^n G_{(x_i, y_j)} \]
so we have found a finite subcover.
}
\end{hiddenflashcard}

\begin{proof}
    Let $(X, \tau), (Y, \sigma)$ be topological
    spaces and let $\rho$ be the product topology
    on $X \times Y$.
    \begin{enumerate}[(a)]
        \item Suppose $X, Y$ Hausdorff. Let $(x,
            y), (z, w) \in X \times Y$ with $(x,
            y) \neq (z, w)$. WLOG $x \neq z$. As
            $X$ is Hausdorff, can find $G, H \in
            \tau$ with $G \cap H = \emptyset$, $x
            \in G$, $z \in H$. Then $G \times Y, H
            \times Y \in \rho$ with $(G \times Y)
            \cap (H \times Y) = \emptyset$ and
            $(x, y) \in G \times Y$, $(z, w) \in H
            \times Y$. So $X \times Y$ is
            Hausdorff.
        \item Suppose $X, Y$ compact. Let
            $\mathcal{C} \subset \rho$ be an open
            cover of $X \times Y$. Fix $x \in X$.
            For each $y \in Y$, there is some $G_y
            \in \mathcal{C}$ such that $(x, y) \in
            G_y$. Hence we can find $U_y \in
            \sigma$ and $V_y \in \tau$ such that
            $(x, y) \in U)y \times V_y \subset
            G_y$. In particular, we have $x \in
            U_y$ and $y \in V_y$. Thus $\{V_y \mid
            y \in Y\} \subset \sigma\}$ is an open
            cover of $Y$. So, as $Y$ is compact,
            it has a finite subcover $\{V_{y_1},
            \dots, V_{y_n}\}$, say. Let $W =
            \bigcap_{i = 1}^n U_{y_i}$. Then $W$
            is open in $X$ and $x \in W$.
            Moreover,
            \[ W \times Y \subset \bigcup_{i =
            1}^n (U_{y_i} \times V_{y_i}) \subset
            \bigcup_{i = 1}^n G_{y_i} \]
            Now, do this for each $x \in X$ to
            obtain $W_x = W$, $n_x = n$ and
            \[ G_{y_i}^{(x)} = G_{y_i} \]
            ($1 \le i \le n_x$) as above. Then
            $\{W_x \mid x \in X\} \subset \tau$ is
            an open cover of $X$. So, as $X$ is
            compact, it has a finite subcover,
            $\{W_{x_1}, \dots, W_{x_m}\}$, say.
            Now $X = \bigcup_{j = 1}^m W_{x_j}$
            and, for each $j$,
            \[ W_{x_J} \times Y \subset \bigcup_{i
            = 1}^{n_{x_j}} G_{y_i} ^{(x_j)} \]
            Thus $\{G_{y_1}^{(x_j)} \mid 1 \le j
            \le m, 1 \le i \le n_{x_j}\}$ is an
            open cover of $X \times Y$ and hence a
            finite subcover of $\mathcal{C}$. Thus
            $X \times Y$ is compact.
    \end{enumerate}
\end{proof}

\subsection{Quotients}

Consider the surface of a torus in $\RR^3$:
\begin{center}
    \includegraphics[width=0.6\linewidth]
    {images/427a6a0a60ed11ed.png}
\end{center}
We might be interested in for example the set of
continuous functions $T \to T$. Analysis is likely
to be unpleasant, for example what is the equation
defining $T$? What do we care about? Continuity
and convergence. So if we replace $T$ by a space
homemorphic to $T$ then we're happy, particularly
if the new space is analytically easier to work
with. For example, take closed unit square $[0, 1]
\times [0, 1]$ with Euclidean topology. Glue $(x,
0)$ to $(x, 1)$ for each $x$ and glue $(0, y)$ to
$(1, y)$ for each $y$.
\begin{center}
    \includegraphics[width=0.6\linewidth]
    {images/c48a385860ee11ed.png}
\end{center}
This seems to give us $T$. More formally, defined
an equivalence relation on $[0, 1] \times [0, 1]$,
$\sim$ say, with equivalence classes:
\[ \{(x, y)\} \quad (0 < x, y < 1) \]
\[ \{(x, 0), (x, 1)\} \quad (0 < x < 1) \]
\[ \{(0, y), (1, y)\} \quad (0 < y < 1) \]
\[ \{(0,0),(0,1),(1,0),(1,1)\} \]
Essentially, we could define $T = [0, 1]^2 /
\sim$, the set of equivalence classes. Maybe
better way to do this?
\begin{center}
    \includegraphics[width=0.6\linewidth]
    {images/0423ad5a60ef11ed.png}
\end{center}
Instead we could define an equivalence relation
$\sim$ on $\RR^2$ by $(x, y) \sim (z, w) \iff x -
z \in \ZZ$ and $y - w \in \ZZ$. Again, hopefully
could define $T = \RR^2 / \sim$. \\
But: What is the topology?