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\subsubsection*{Remarks}

\begin{enumerate}[(1)]
    \item Intuition: A set is open if for each
        point in the set it contains all points
        nearby as well. A set is a neighbourhood
        of $a$ if it contains all points near $a$.
    \item The open ball $B_\eps(a)$ is open. Why?
        If $x \in B_\eps(a)$ then $d(x, a) =
        \delta < \eps$, say, so by triangle
        inequality $B_{\eps - \delta}(x) \subset
        B_\eps(a)$.
    \item If $\mathcal{N}$ is an open set and $a
        \in \mathcal{N}$ then certainly
        $\mathcal{N}$ is a neighbourhood of $a$:
        \[ a \in \mathcal{N} \subset \mathcal{N} .\]
        However, a neighbourhood of $a$ need not
        be open. For example in $\RR$ with the
        usual metric then $[-1, 1]$ is a
        neighbourhood of 0:
        \[ 0 \in (-1, 1) \subset [-1, 1] \]
        And $[-1, 1] \cup \{396\}$ is a
        neighbourhood of 0 (throwing in extra
        stuff doesn't stop something from being a
        neighbourhood).
    \item $\mathcal{N}$ is a neighbourhood of $a$
        if and only if $\exists \eps > 0$ with
        $B_\eps(a) \subset \mathcal{N}$:
        \begin{enumerate}
            \item[$\Leftarrow$] $a \in B)\eps(a)
                \subset \mathcal{N}$.
            \item[$\Rightarrow$] $a \in G \subset
                \mathcal{N}$ and $\exists \eps >
                0$ such that $B_\eps(a) \subset
                G$.
        \end{enumerate}
    \item A set $G$ is open if and only if it's a
        neighbourhood of each of its points.
\end{enumerate}

\begin{flashcard}[NbdCts]
\begin{proposition}[Generalising continuity]
    Let $(X, d)$, $(Y, e)$ be metric spaces and let
    $f : X \to Y$.
    \begin{enumerate}[(i)]
        \item $f$ is continuous at $a \in X$ if
            and only if \cloze{whenever $\mathcal{N} \in
            Y$ is a neighbourhood of $f(a)$ we
            have $f^{-1}(\mathcal{N}) \subset X$ a
            neighbourhood of $a$};
        \item $f$ is a continuous function \cloze{if and
            only if whenever $G \subset Y$ open we
            have $f^{-1}(G) \subset X$ open.}
    \end{enumerate}
\end{proposition}
\end{flashcard}

\begin{proof}
    \begin{enumerate}[(i)]
        \item \begin{enumerate}
            \item[$\Rightarrow$] Suppose $f$ is
                continuous at $a \in X$. Let
                $\mathcal{N}$ be a neighbourhood
                of $f(a)$. Then $\exists \eps > 0$
                such that $B_\eps(f(a)) \subset
                \mathcal{N}$. But $f$ continuous
                at $a$ so $\exists \delta > 0$
                such that $B_\delta(a) \subset
                f^{-1}(B_\eps(f(a))) \subset
                f^{-1}(\mathcal{N})$. So
                $f^{-1}(\mathcal{N})$ is a
                neighbourhood of $a$.
            \item[$\Leftarrow$] Suppose
                $f^{-1}(\mathcal{N})$ is a
                neighbourhood of $a$ for every
                neighbourhood $\mathcal{N}$ of
                $f(a)$. Let $\eps > 0$. In
                particular, $B_\eps(f(a))$ is a
                neighbourhood of $f(a)$ so
                $f^{-1}(B_\eps(f(a)))$ is a
                neighbourhood of $a$ so $\exists
                \delta > 0$ such that $B_\delta(a)
                \subset f^{-1}(B_\eps(f(a)))$. So
                $f$ is continuous at $a$.
        \end{enumerate}
        \item \begin{enumerate}
            \item[$\Rightarrow$] Suppose $f$ is a continuous
                function. Let $G \subset Y$ be open.
                Let $a \in f^{-1}(G)$. Then $f(a) \in
                G$ and $G$ open so $G$ is a
                neighbourhood of $f(a)$. Moreover, $f$
                is continuous at $a$ so by (i) we have
                $f^{-1}(G)$ a neighbourhood of $a$.
                Hence $\exists \delta > 0$ such that
                $B_\delta(a) \subset f^{-1}(G)$. So
                $f^{-1}(G)$ is open.
            \item[$\Leftarrow$] Suppose $f^{-1}(G)$ open
                whenever $G$ is open in $Y$. Let
                $a \in X$. Let $\mathcal{N}
                \subset Y$ be a neighbourhood of
                $f(a)$. Then $\exists G \subset Y$
                open such that $f(a) \in G \subset
                \mathcal{N}$. By assumption
                $f^{-1}(G) \subset X$ open. Now $a
                \in f^{-1}(G) \subset
                f^{-1}(\mathcal{N})$ with
                $f^{-1}(G)$ open so
                $f^{-1}(\mathcal{N})$ is a
                neighbourhood of $a$. So by (i),
                $f$ is continuous at $a$. So $f$
                is a continuous function.
        \end{enumerate}
    \end{enumerate}
\end{proof}

\subsubsection*{Remarks}

\begin{enumerate}[(1)]
    \item This says that we can define continuity
        entirely in terms of open sets without
        mentioning the metric.
    \item We saw previously that homeomorphisms
        preserve convergence and continuity.
        Proposition 23(ii) says homeomorphisms also
        preserve open sets: to be precise, if $f :
        X \to Y$ is a homeomorphism then $G
        \subset X$ is open if and only if $f(G)
        \subset Y$ is open. (Why? $G =
        f^{-1}(f(G))$ with $f$ continuous and
        $f(a) = (f^{-1})^{-1}(G)$ with $f^{-1}$
        continuous.)
\end{enumerate}

\myskip
What else is preserved by homeomorphisms? \myskip
Suppose $f : X \to Y$ is a homeomorphism and $X$
is sequentially compact. Let $(y_n)$ be a sequence
in $Y$. Then $f^{-1}(y_n))$ is a sequence in $X$
and so has a convergent subsequence
$f^{-1}(y_{n_j}) \to x \in X$, say. But
convergence of sequences is preserved by
homeomorphisms. Hence $y_{n_j} =
f(f^{-1}(y_{n_j})) \to f(x) \in Y$. So $Y$ is
sequentially compact. \\
So if $X, Y$ homoeomorphic spaces, then
\[ \text{$X$ sequentially compact} \iff \text{$Y$
sequentially compact} \]
`Sequential compactness is a \emph{topological
property}' \myskip
If $X, Y$ are homeomorphic, and one of them has a
particular topological property, then so does the
other. \myskip
What about completeness? Not so good.

\begin{example*}
    We saw $(0, 1)$ and $\RR$ with the usual
    metric in each case are homeomorphic. But
    $\RR$ is complete and $(0, 1)$ is not. So
    completeness is not a topological property.
    \begin{hiddenflashcard}[completeness-topological-property]
        Proof that if $X$ is complete and $f : X
        \to Y$ is a homeomorphism then $Y$ is
        complete? \\
        \cloze{
        This is \fcemph{false}. For example, take
        $X = \RR$ and $Y = (0, 1)$ with usual
        metrics. (This shows that completeness is
        not a property that can be defined nicely
        for topological spaces).
        }
    \end{hiddenflashcard}
\end{example*}

\noindent
What went wrong? Property of being a Cauchy
sequence is not preserved by homeomorphisms.

\begin{remark*}
    Suppose $(x_n)$ is a sequence in a metric
    space $X$ and $x \in X$. Then
    \begin{align*}
        x_n \to x
        &\iff \forall \eps > 0 ~ \exists N ~
        \forall n \ge N \quad d(x_n, x) < \eps \\
        &\iff \forall \eps > 0 ~ \exists N ~
        \forall n \ge N \quad x_n \in B_\eps(x) \\
        &\iff \text{for all neighbourhoods
        $\mathcal{N}$ of $x$, $\exists N$ such
        that $\forall n \ge N$, $x_n \in \mathcal{N}$.}
    \end{align*}
    This defines convergence solely in terms of
    neighbourhoods. Can't do something similar for
    Cauchy sequences using neighbourhoods / open
    sets.
\end{remark*}

\noindent
Just seen sequential compactness is a topological
property. Can define sequential compactness just
in terms of neighbourhoods / open sets:
\[ \text{sequentially compact} \leftarrow
\text{convergence of sequences} \leftarrow
\text{neighbourhoods} \]
Is there a `nicer' way to do this?

\begin{definition*}
    Let $X$ be a metric space. An \emph{open
    cover} of $X$ is a collection $\mathcal{C}$ of open
    subsets of $X$ such that
    \[ X = \bigcup_{G \in \mathcal{C}} G \]
    A \emph{subcover} of $\mathcal{C}$ is an open
    cover $\mathcal{B}$ of $X$ with $\mathcal{B}
    \subset \mathcal{C}$. We say $X$ is
    \emph{compact} if every open cover of $X$ has
    a finite subcover.
\end{definition*}

\begin{example*}[The Heine-Borel theorem]
    $[0, 1]$ with the usual metric is compact.
    \begin{proof}
        Let $\mathcal{C}$ be an open cover of $[0,
        1]$. Let
        \[ A = \{x \in [0, 1] \mid \exists
            \mathcal{B} \subset \mathcal{C}
        \text{finite with $[0, x] \subset
        \bigcup_{G \in \mathcal{B}} G$} \]
        We know $\exists G \in \mathcal{C}$ with
        $0 \in G$. So $0 \in A$ so $A \neq
        \emptyset$. Clearly $A$ bounded above by
        1. So $A$ has a supremum, $\sigma = \sup
        A$, say. \\
        As $G$ is open, $\exists \eps > 0$ such
        that $[0, \eps) = B_\eps(0) \subset G$. So
        $\frac{\eps}{2} \in A$ so $\sigma > 0$. \\
        Suppose $\sigma < 1$. Can find $G' \in
        \mathcal{C}$ with $\sigma \in G$. As
        $\sigma = \sup A$, we can find $x \in A$
        with $x \in G'$. So have $\mathcal{B}
        \subset \mathcal{C}$ finite with $[0, x]
        \subset \bigcup_{G \in \mathcal{B}} G$.
        But $\exists \eps > 0$ such that $(\sigma
        - \eps, \sigma + \eps) = B_\eps(\sigma
        \subset G$. So
        \[ \left[0, \sigma + \frac{\eps}{2}
        \right] \subset \bigcup_{G \in \mathcal{B}
        \cup \{G'\}} G \]
        So $\sigma + \frac{\eps}{2} \in A$.
        Contradiction. Hence $\sigma = 1$. Can
        find $G'' \in \mathcal{C}$ such that $1
        \in G''$. As $G''$ open, can find $\eps >
        0$ such that $(1 - \eps, 1] = B_\eps(1)
        \subset G''$. As $1 = \sup A$ can find $x
        \in A \cap (1 - \eps, 1]$. That says we
        have finite $\mathcal{B} \subset
        \mathcal{C}$ with $[0, x] \subset
        \bigcup_{G \in \mathcal{B}} G$. Then
        $\mathcal{B} \cup \{G''\}$ is an open
        cover of $[0, 1]$ and so a subcover of
        $\mathcal{C}$. So $[0, 1]$ is compact.
    \end{proof}
\end{example*}

\begin{hiddenflashcard}[heine-borel-theorem]
\begin{theorem*}[Heine-Borel Theorem]
    \cloze{
    $[0, 1]$ with the usual metric is compact.
    }
\end{theorem*}

\begin{proof}
    \cloze{
    Let $\mathcal{C}$ be an open cover of $[0,
    1]$. Let
    \[ A = \left\{x \in [0, 1] ~\Bigg|~ \exists
    \mathcal{B} \subset \mathcal{C}
    \text{ finite with $[0, x] \subset
    \bigcup_{G \in \mathcal{B}} G$}\right\} \]
    Show $0 \in A$ hence non-empty. Take $\sigma
    = \sup A$. Get a contradiction unless $\sigma
    = 1$. \fcemph{Note that we are not yet done}.
    Finally, construct a finite subcover
    using the fact that $\sigma = 1$.
    }
\end{proof}
\end{hiddenflashcard}