\subsection{Additional Tutorial}
\tutorial{15}{2024-01-31}{Additions}

The following is not relevant for the exam,
but gives a more general picture.

Let $X$ be a topological space
and let $\cF$ be a filter on $ X$.

$x \in X$ is a limit point of $\cF$ iff the neighbourhood filter $\cN_x$,
all sets containing an open neighbourhood of $x$,
is contained in  $\cF$.

\begin{fact}
    $X$ is Hausdorff iff every filter has at most one limit point.
\end{fact}
\begin{proof}
    Neighbourhood filters are compatible
    iff the corresponding points
    can not be separated by open subsets.
\end{proof}

\begin{fact}
    $X$ is (quasi-) compact
    iff every ultrafilter converges.
\end{fact}
\begin{proof}
    Suppose that $X$ is compact.
    Let $\cU$ be an ultrafilter.
    Consider the family $\cV = \{\overline{A} : A \in \cU\}$
    of closed sets.
    By the FIP we geht that there exist
    $c \in X$ such that $c \in \overline{A}$ for all $A \in \cU$.
    Let $N$ be an open neighbourhood of $c$.
    If $N^c \in \cU$, then $c \in N^c \lightning$
    So we get that $N \in \cU$.

    Let $\{V_i : i \in I\} $ be a family of closed sets with the FIP.
    Consider the filter generated by this family.
    We extend this to an ultrafilter.
    The limit of this ultrafilter is contained in all the $V_i$.
\end{proof}

Let $X,Y$ be topological spaces,
$\cB$ a filter base on $X$,
$\cF$ the filter generated by $\cB$
and
$f\colon  X \to Y$.
Then $f(\cB)$ is a filter base on $Y$,
since $f(\bigcap A_i ) \subseteq  \bigcap f(A_i)$.
We say that $\lim_\cF f = y$,
if $f(\cF) \to  y$.

Equivalently $f^{-1}(N) \in \cF$
for all neighbourhoods $N$ of $y$.

In the lecture we only considered $X = \N$.
If $\cB$ is the base of an ultrafilter,
so is $f(\cB)$.

\begin{fact}
    Let $X$ be a topological space
    and let $Y$ be Hausdorff.
    Let $f,g \colon  X \to Y$
     be continuous.
     Let $A \subseteq  X$ be dense such that
     $f\defon{A} = g\defon{A} $.
     Then $f = g$.
\end{fact}
\begin{proof}
    Consider $(f,g)^{-1}(\Delta) \supseteq A$.
\end{proof}

We can uniquely extend $f\colon  X \to  Y$ continuous
to a continuous $\overline{f}\colon  \beta X \to  Y$
by setting $\overline{f}(\cU) \coloneqq  \lim_\cU f$.

Let $V$ be an open neighbourhood of $Y$ in $\overline{f}\left( U) \right) $.
Consider $f^{-1}(V)$.
Consider the basic open set
\[
\{\cF \in  \beta\N  : \cF \ni f^{-1}(V)\}.
\]

\todo{I missed the last 5 minutes}
