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@ -4,7 +4,7 @@ in the summer term 2023 at the University Münster.
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\begin{warning}
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This is not an official script.
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The official lecture notes can be found on
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The official lecture notes can be found
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\href{https://sites.google.com/site/akwiatkmath/teaching/logic-3-abstract-topological-dynamics-and-descriptive-set-theory}{here}.
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\end{warning}
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@ -42,14 +42,14 @@ Recall the following notions:
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and $U_i \subsetneq X_i$ for only finitely many $i$.
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\end{definition}
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\begin{fact}
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Countable products of separable spaces are separable,
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Countable products of separable spaces are separable.
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\end{fact}
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\begin{definition}
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A topological space $X$ is \vocab{second countable},
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if it has a countable base.
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\end{definition}
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If $X$ is a topological space.
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Then if $X$ is second countable, it is also separable.
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Let $X$ be a topological space.
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If $X$ is second countable, it is also separable.
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However the converse of this does not hold.
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\begin{example}
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@ -81,7 +81,8 @@ However the converse of this does not hold.
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For a metric space, the following are equivalent:
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\begin{itemize}
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\item compact,
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\item \vocab{sequentially compact} (every sequence has a convergent subsequence),
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\item \vocab{sequentially compact}
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(every sequence has a convergent subsequence),
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\item complete and \vocab{totally bounded}
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(for all $\epsilon > 0$ we can cover the space with finitely many $\epsilon$-balls).
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\end{itemize}
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70
inputs/tutorial_01.tex
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70
inputs/tutorial_01.tex
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@ -0,0 +1,70 @@
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\tutorial{01}{202-10-17}{}
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% TODO MAIL
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\begin{fact}
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A countable product of separable spaces $(X_n)_{n \in \N}$ is separable.
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\end{fact}
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\begin{proof}
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Choose a countable dense subset $D_n \subseteq X_n$
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Fix some point $(a_1,a_2,\ldots) \in \prod_n X_n$
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and consider $\bigcup_{i \in \N} \prod_{n \le i} D_n \times \prod_{n > i} \{a_n\}$.
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\end{proof}
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\begin{fact}
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\begin{itemize}
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\item Let $X$ be a topological space.
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Then $X$ 2nd countable $\implies$ X separable.
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\item If $X$ is a metric space and separable,
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then $X$ is 2nd countable.
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\end{itemize}
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\end{fact}
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\begin{proof}
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For the first point, choose some point from every basic open set.
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For the second point consider balls of rational radius
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around the points of a countable dense subset.
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\end{proof}
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\begin{definition}
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A topological space is \vocab{Lindelöf}
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if every open cover has a countable subcover.
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\end{definition}
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\begin{fact}
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Let $X$ be a metric space.
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If $X$ is Lindelöf,
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then it is 2nd countable.
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\end{fact}
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\begin{proof}
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For all $q \in \Q$
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Consider the cover $B_q(x), x \in X$
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and choose a countable subcover.
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The union of these subcovers is
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a countable base.
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\end{proof}
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\begin{fact}
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Let $X$ be a topological space.
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If $X$ is 2nd countable,
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then it is Lindelöff.
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\end{fact}
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\begin{proof}
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Let $A_0, A_1,\ldots$
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be a countable base.
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Let $\{U_i\}_{i \in I}$
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be a cover.
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Consider $J \coloneqq \{j : \exists i \in I.~A_j \in U_i\}$.
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For every $j \in J$ choose a $U_i$ such that
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$A_j \subseteq U_j$.
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Let $I' \subseteq I$ be the subset of chosen indices.
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Then $\{U_i\}_{i \in I'}$ is a countable subcover.
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\end{proof}
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\begin{remark}
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For metric spaces the notions
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of being 2nd countable, separable
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and Lindelöf coincide.
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In arbitrary topological spaces,
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Lindelöf is the strongest of these notions.
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\end{remark}
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