Josia Pietsch
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90 lines
2.5 KiB
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\subsection{Additional Tutorial}


\tutorial{15}{20240131}{Additions}




The following is not relevant for the exam,


but aims to give 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}


\label{fact:hdifffilterlimit}


$X$ 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}


\label{fact:compactiffufconv}


$X$ 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 get 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 $(f,g)^{1}(\Delta) \supseteq A$.


The LHS is a dense closed set, i.e.~the entire space.


\end{proof}




We can uniquely extend a continuous $f\colon X \to Y$


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 $f^{1}(V)$.


% Then


% \[


% \{\cF \in \beta\N : \cF \ni f^{1}(V)\}


% \]


% is a basic open set.




\todo{I missed the last 5 minutes}
