Josia Pietsch
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90 lines
2.5 KiB
TeX
90 lines
2.5 KiB
TeX
\subsection{Additional Tutorial}
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\tutorial{15}{2024-01-31}{Additions}
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The following is not relevant for the exam,
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but aims to give a more general picture.
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Let $X$ be a topological space
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and let $\cF$ be a filter on $ X$.
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$x \in X$ is a limit point of $\cF$ iff the neighbourhood filter $\cN_x$,
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all sets containing an open neighbourhood of $x$,
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is contained in $\cF$.
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\begin{fact}
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\label{fact:hdifffilterlimit}
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$X$ is Hausdorff iff every filter has at most one limit point.
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\end{fact}
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\begin{proof}
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Neighbourhood filters are compatible
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iff the corresponding points
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can not be separated by open subsets.
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\end{proof}
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\begin{fact}
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\label{fact:compactiffufconv}
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$X$ is (quasi-) compact
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iff every ultrafilter converges.
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\end{fact}
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\begin{proof}
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Suppose that $X$ is compact.
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Let $\cU$ be an ultrafilter.
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Consider the family $\cV = \{\overline{A} : A \in \cU\}$
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of closed sets.
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By the FIP we get that there exist
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$c \in X$ such that $c \in \overline{A}$ for all $A \in \cU$.
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Let $N$ be an open neighbourhood of $c$.
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If $N^c \in \cU$, then $c \in N^c \lightning$
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So we get that $N \in \cU$.
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Let $\{V_i : i \in I\} $ be a family of closed sets with the FIP.
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Consider the filter generated by this family.
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We extend this to an ultrafilter.
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The limit of this ultrafilter is contained in all the $V_i$.
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\end{proof}
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Let $X,Y$ be topological spaces,
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$\cB$ a filter base on $X$,
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$\cF$ the filter generated by $\cB$
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and
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$f\colon X \to Y$.
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Then $f(\cB)$ is a filter base on $Y$,
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since $f(\bigcap A_i ) \subseteq \bigcap f(A_i)$.
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We say that $\lim_\cF f = y$,
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if $f(\cF) \to y$.
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Equivalently $f^{-1}(N) \in \cF$
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for all neighbourhoods $N$ of $y$.
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In the lecture we only considered $X = \N$.
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If $\cB$ is the base of an ultrafilter,
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so is $f(\cB)$.
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\begin{fact}
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Let $X$ be a topological space
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and let $Y$ be Hausdorff.
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Let $f,g \colon X \to Y$
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be continuous.
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Let $A \subseteq X$ be dense such that
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$f\defon{A} = g\defon{A} $.
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Then $f = g$.
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\end{fact}
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\begin{proof}
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Consider $(f,g)^{-1}(\Delta) \supseteq A$.
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The LHS is a dense closed set, i.e.~the entire space.
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\end{proof}
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We can uniquely extend a continuous $f\colon X \to Y$
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to a continuous $\overline{f}\colon \beta X \to Y$
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by setting $\overline{f}(\cU) \coloneqq \lim_\cU f$.
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% Let $V$ be an open neighbourhood of $y \in \overline{f}\left( U \right)$.
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% Consider $f^{-1}(V)$.
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% Then
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% \[
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% \{\cF \in \beta\N : \cF \ni f^{-1}(V)\}
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% \]
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% is a basic open set.
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\todo{I missed the last 5 minutes}
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