Josia Pietsch 2023-10-19 20:18:40 +02:00
commit b636d722d4
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@ -4,7 +4,7 @@ in the summer term 2023 at the University Münster.
\begin{warning} \begin{warning}
This is not an official script. This is not an official script.
The official lecture notes can be found on The official lecture notes can be found
\href{https://sites.google.com/site/akwiatkmath/teaching/logic-3-abstract-topological-dynamics-and-descriptive-set-theory}{here}. \href{https://sites.google.com/site/akwiatkmath/teaching/logic-3-abstract-topological-dynamics-and-descriptive-set-theory}{here}.
\end{warning} \end{warning}

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@ -42,14 +42,14 @@ Recall the following notions:
and $U_i \subsetneq X_i$ for only finitely many $i$. and $U_i \subsetneq X_i$ for only finitely many $i$.
\end{definition} \end{definition}
\begin{fact} \begin{fact}
Countable products of separable spaces are separable, Countable products of separable spaces are separable.
\end{fact} \end{fact}
\begin{definition} \begin{definition}
A topological space $X$ is \vocab{second countable}, A topological space $X$ is \vocab{second countable},
if it has a countable base. if it has a countable base.
\end{definition} \end{definition}
If $X$ is a topological space. Let $X$ be a topological space.
Then if $X$ is second countable, it is also separable. If $X$ is second countable, it is also separable.
However the converse of this does not hold. However the converse of this does not hold.
\begin{example} \begin{example}
@ -81,7 +81,8 @@ However the converse of this does not hold.
For a metric space, the following are equivalent: For a metric space, the following are equivalent:
\begin{itemize} \begin{itemize}
\item compact, \item compact,
\item \vocab{sequentially compact} (every sequence has a convergent subsequence), \item \vocab{sequentially compact}
(every sequence has a convergent subsequence),
\item complete and \vocab{totally bounded} \item complete and \vocab{totally bounded}
(for all $\epsilon > 0$ we can cover the space with finitely many $\epsilon$-balls). (for all $\epsilon > 0$ we can cover the space with finitely many $\epsilon$-balls).
\end{itemize} \end{itemize}

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\tutorial{01}{202-10-17}{}
% TODO MAIL
\begin{fact}
A countable product of separable spaces $(X_n)_{n \in \N}$ is separable.
\end{fact}
\begin{proof}
Choose a countable dense subset $D_n \subseteq X_n$
Fix some point $(a_1,a_2,\ldots) \in \prod_n X_n$
and consider $\bigcup_{i \in \N} \prod_{n \le i} D_n \times \prod_{n > i} \{a_n\}$.
\end{proof}
\begin{fact}
\begin{itemize}
\item Let $X$ be a topological space.
Then $X$ 2nd countable $\implies$ X separable.
\item If $X$ is a metric space and separable,
then $X$ is 2nd countable.
\end{itemize}
\end{fact}
\begin{proof}
For the first point, choose some point from every basic open set.
For the second point consider balls of rational radius
around the points of a countable dense subset.
\end{proof}
\begin{definition}
A topological space is \vocab{Lindelöf}
if every open cover has a countable subcover.
\end{definition}
\begin{fact}
Let $X$ be a metric space.
If $X$ is Lindelöf,
then it is 2nd countable.
\end{fact}
\begin{proof}
For all $q \in \Q$
Consider the cover $B_q(x), x \in X$
and choose a countable subcover.
The union of these subcovers is
a countable base.
\end{proof}
\begin{fact}
Let $X$ be a topological space.
If $X$ is 2nd countable,
then it is Lindelöff.
\end{fact}
\begin{proof}
Let $A_0, A_1,\ldots$
be a countable base.
Let $\{U_i\}_{i \in I}$
be a cover.
Consider $J \coloneqq \{j : \exists i \in I.~A_j \in U_i\}$.
For every $j \in J$ choose a $U_i$ such that
$A_j \subseteq U_j$.
Let $I' \subseteq I$ be the subset of chosen indices.
Then $\{U_i\}_{i \in I'}$ is a countable subcover.
\end{proof}
\begin{remark}
For metric spaces the notions
of being 2nd countable, separable
and Lindelöf coincide.
In arbitrary topological spaces,
Lindelöf is the strongest of these notions.
\end{remark}