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\begin{flashcard}[HausdorffSpace]
    \begin{definition*}[Hausdorff Space]
        A topological space $X$ is
        \emph{Hausdorff} if \cloze{whenever $x, y \in X$
        with $x \neq y$ then there are disjoint
        open $G, H \subset X$ with $x \in G$ and
        $y \in H$.}
    \end{definition*}
\end{flashcard}

\subsubsection*{Examples}

\begin{enumerate}[(1)]
    \item Metric spaces are Hausdorff. Indeed, if
        $(X, d)$ is a metric space and $x, y \in
        X$, $x \neq y$ then let $\delta = d(x, y)
        > 0$ and take $G =
        B_{\frac{\delta}{2}}(x)$ and $H =
        B_{\frac{\delta}{2}}(y)$.
    \item Indiscrete spaces are not Hausdorff
        (assuming $|X| \ge 2$).
    \item The cofinite topology is not Hausdorff.
        Let $X$ be an infinite set with the
        cofinite topology and let $x, y \in X$
        with $x \neq y$. Let $G, H \subset X$ be
        open with $x \in G$, $y \in H$. Clearly
        $G, h \neq \emptyset$ so $X \setminus G$,
        $X \setminus H$ are finite and so
        \[ X \setminus (G \cap H) = (X \setminus
        G) \cup (X \setminus H) \]
        is finite. In particular $G \cap H \neq
        \emptyset$. Similarly, the cocountable
        topology is not Hausdorff.
\end{enumerate}

\begin{hiddenflashcard}[non-Hausdorff-examples]
Examples of topological spaces that are not
Hausdorff? \\
\cloze{
Indiscrete space ($|X| \ge 2$), cofinite topology
($|X|$ infinite), cocountable topology ($|X|$
uncountable). \\
We could also take two copies of $\RR$ and
glue everywhere except at $0$.
}
\end{hiddenflashcard}

\begin{proposition}
    Limits of convergent sequences in Hausdorff
    spaces are unique.
\end{proposition}

\begin{proof}
    Let $X$ be Hausdorff, let $a, b \in X$, and
    let $(x_n)$ be a sequence in $X$ with $x_n \to
    a$ and $x_n \to b$. \\
    Suppose $a \neq b$. Take open $G, H$ with $a
    \in G$, $b \in H$ and $G \cap H = \emptyset$.
    Now $G$ is a neighbourhood of $a$ so there is
    some $N_1$ such that $\forall n \ge N_1$, $x_n
    \in G$. Similarly, there is some $N_2$ such
    that $\forall n \ge N_2$, $x_n \in H$. Take $n
    = \max\{N_1, N_2\}$. Then $x_n \in G \cap H =
    \emptyset$, contradiction. Hence $a = b$.
\end{proof}

\myskip
Relationship to continuity?

\begin{proposition}
    Let $X, Y$ be topological spaces and let $f :
    X \to Y$ be continuous at $a \in X$. Let
    $(x_n)$ be a sequence in $X$ with $x_n \to a$.
    Then $f(x_n) \to f(a)$.
\end{proposition}

\begin{hiddenflashcard}[generalising-metric-continuity-sequences-fact]
Recall that in a metric space, $f$ is continuous
at $x$ if and only if for all $x_n \to x$, $f(x_n)
\to f(x)$. How does this generalise to topological
spaces? \\
\cloze{
If $f$ is continuous, then if $x_n \to x$ then
$f(x_n) \to f(x)$. \\
However, the converse is not true in general.
(This is sort of because convergent sequences give
a lot less information in topological spaces than
in metric spaces).
}
\end{hiddenflashcard}

\begin{proof}
    Let $\mathcal{N} \subset Y$ be a
    neighbourhood of $f(a)$. As $f$ is continuous
    at $a$ we know $f^{-1}(\mathcal{N})$ is a
    neighbourhood of $a$. As $x_n \to a$ we can
    fined $\mathcal{N}$ such that $\forall n \ge
    N$, $x_n \in f^{-1}(\mathcal{N})$. Then for
    all $n \ge N$, $f(x_n) \in \mathcal{N}$. So
    $f(x_n) \to f(a)$.
\end{proof}

\begin{warning*}
    CONVERSE NOT TRUE IN GENERAL
\end{warning*}

\begin{example*}
    Let $X = Y = \RR$, $X$ with cocountable
    topology, $Y$ with usual topology and $f : X
    \to Y$ be the identity function. Suppose $x_n
    \to 0$ in $X$. Then for sufficiently large
    $n$, $x_n = 0$ and so for sufficiently large
    $n$, $f(x_n) = x_n = 0 = f(0)$ so $f(x_n) \to
    f(0)$ in $Y$. \\
    However, $(-1, 1) \subset Y$ is open and $0
    \in (-1, 1)$ so $(-1, 1)$ is a neighbourhood
    of 0 in $Y$. But $f^{-1}((-1, 1)) = (-1, 1)
    \subset X$ is not a neighbourhood of 0 in $X$.
    So $f$ is not continuous at 0.
\end{example*}

\noindent
Here, even imposing condition that the spaces are
Hausdorff is not enough.

\begin{example*}
    Take example as above but replace topology on
    $X$ by
    \[ \sigma = \{G \subset \RR \mid (X \setminus G)
    \text{ countable or $0 \not\in G$}\} \]
    This is a topology (check-exercise). And it is
    Hausdorff: suppose $x, y \in X$ with $x \neq
    y$. If $x, y \neq 0$ then $\{x\}, \{y\} \in
    \sigma$. While if $x = 0$, say, then $\RR
    \setminus \{y\}, \{y\} \in \sigma$. \\
    Now, neighbourhoods of 0 in $\sigma$ are
    exactly the same as in the cocountable
    topology. So exactly as before $x_n \to 0$ in
    $X$ implies $x_n \to 0$ in $Y$, but $f$ is not
    continuous at 0.
\end{example*}

\begin{remark*}
    In a metric space, the topology is completely
    determined by convergence of sequences. Not
    true for a general topological space. Hence
    we'll tend to concentrate more on continuity
    than convergence of sequences.
\end{remark*}

\subsection{Subspaces}

\begin{flashcard}[SubspaceTopology]
    \begin{definition*}[Subspace Topology]
        Let $(X, \tau)$ be a topological space and let
        $Y \subset X$. The \emph{subspace topology} on
        $Y$ is
        \[ \sigma = \cloze{\{G \cap Y \mid G \in
        \tau\}} \]
    \end{definition*}
\end{flashcard}

\noindent
Easy to check that this is a topology. Need to
check backward compatibility with metric space
definition:

\begin{proposition}
    Let $(X, d)$ be a metric space with topology
    $\tau$ induced by $d$. Let $Y$ be a subspace
    of the metric space $X$. Then $Y$ has the
    subspace topology.
\end{proposition}

\begin{proof}
    Let $\sigma$ be the topology on $Y$ induced by
    the metric $d \mid_{Y^2}$. Suppose $G \in
    \tau$. Let $y \in G \cap Y$. As $y \in G$ and
    $G$ open in $X$ we can find $\delta > 0$ such
    that $\forall x \in X$, $d(x, y) < \delta$
    implies $x \in G$. Then $\forall x \in Y$,
    $d(x, y) < \delta \implies x \in G \cap Y$. So
    $G \cap Y$ is a neighbourhood of $y$. So $G
    \cap Y \in \sigma$. \myskip
    Conversely, suppose $H \in \sigma$. For each
    $y \in H$ can find $\delta_y > 0$ such that
    $\forall x \in Y$, $d(x, y) < \delta_y
    \implies x \in H$. Consider the open balls
    \[ B_{\delta_y}(y) = \{x \in X \mid d(x, y) <
    \delta_y\} \]
    ($y \in H$). Each $B_{\delta_y}(y)$ is open ,
    for each $y \in H$, $y \in B_{\delta_y(y)}$
    and $B_{\delta_y}(y) \cap Y \subset H$. Let $G
    = \bigcup_{y \in H} B_{\delta_y}(y)$. Then $G$
    is open and $G \cap Y = H$. That is, we've
    found $G \in \tau$ such that $G \cap Y = H$.
\end{proof}

\begin{proposition}
    A subspace of a Hausdorff space is Hausdorff.
\end{proposition}

\begin{proof}
    Let $(X, \tau)$ be Hausdorff, $Y \subset X$,
    $\sigma$ the subspace topology on $Y$. Let $x,
    y \in Y$ with $x \neq y$. As $X$ Hausdorff can
    find $G, H \in \tau$ with $x \in G$, $y \in
    H$, $G \cap H = \emptyset$. Then $G \cap Y, H
    \cap Y \in \sigma$ with $x \in G \cap Y$, $y
    \in H \cap Y$, and $(G \cap Y) \cap (H \cap Y)
    = \emptyset$.
\end{proof}

\subsection{Completeness}

\begin{flashcard}[TopOpenCover]
    \begin{definition*}
        Let $(X, \tau)$ be a topological space. An
        \emph{open over} of $X$ is \cloze{a subset
        $\mathcal{C} \subset \tau$ such that $X =
        \bigcup_{G \in \mathcal{C}} G$.}
    \end{definition*}
\end{flashcard}

\begin{flashcard}[TopSubCover]
    \begin{definition*}
        Let $(X, \tau)$ be a topological space and
        $\mathcal{C}$ and open cover of $X$. A
        \emph{subcover} of $\mathcal{C}$ is \cloze{a
        subset $\mathcal{D}$ of $\mathcal{C}$ which is
        itself an open cover.}
    \end{definition*}
\end{flashcard}

\begin{flashcard}[TopCompact]
    \begin{definition*}
        We say that a topological space $X$ is
        \emph{compact} if \cloze{every open cover
        of $X$ has a finite subcover.}
    \end{definition*}
\end{flashcard}

\begin{flashcard}[TopSeqCompact]
    \begin{definition*}
        We say that a topological space $X$ is
        \emph{sequentially compact} if \cloze{every
        sequence in $X$ has a convergent
        subsequence.}
    \end{definition*}
\end{flashcard}

\noindent
A continuous real-valued function on a
sequentially compact topological space is bounded
and attains its bounds.

\begin{remark*}
    Traditional wording: here and elsewhere, if no
    topology is specified $\RR$ is generally
    assumed to have the usual topology. Proof
    similar to metric case.
\end{remark*}

\noindent
We've seen for a metric space that compact and
sequentially compact are equivalent.

\begin{warning*}
    This equivalence is not true for a general
    topological space.
\end{warning*}

\noindent
There exists a compact space that isn't
sequentially compact, and a sequentially compact
space that isn't compact, but both examples are
beyond the scope of this course. \myskip
Observe compactness and sequential compactness are
both topological properties, since they both use
only open sets and convergence of sequences.
\myskip
Given we don't want to think too much about
sequences in a general topological space, we'll be
concentrating primarily on compactness rather than
sequential compactness.