w23-logic-3/inputs/tutorial_15.tex

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\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
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 $\cN_x$,
all sets containing an open neighbourhood of $x$,
is contained in $\cF$.
\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.
\end{fact}
\begin{proof}
Neighbourhood filters are compatible
iff the corresponding points
can not be separated by open subsets.
\end{proof}
\begin{fact}
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\label{fact:compactiffufconv}
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$X$ 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.
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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 $A \in \cU$.
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 $(f,g)^{-1}(\Delta) \supseteq A$.
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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 $\overline{f}\colon \beta X \to Y$
by setting $\overline{f}(\cU) \coloneqq \lim_\cU f$.
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% Let $V$ be an open neighbourhood of $y \in \overline{f}\left( U \right)$.
% Consider $f^{-1}(V)$.
% Then
% \[
% \{\cF \in \beta\N : \cF \ni f^{-1}(V)\}
% \]
% is a basic open set.
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\todo{I missed the last 5 minutes}