Josia Pietsch
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78 lines
2.2 KiB
TeX
78 lines
2.2 KiB
TeX
\tutorial{01}{2023-10-17}{}
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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 $X$ \nth{2} countable $\implies$ X separable.
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\item If $X$ is a metric space and separable,
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then $X$ is \nth{2} 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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iff 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 \nth{2} 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 \nth{2} 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 \nth{2} countable, separable
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and Lindelöf coincide.
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In arbitrary topological spaces,
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Lindelöf is the weakest of these notions.
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\end{remark}
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\begin{definition}+
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A metric space $X$ is \vocab{totally bounded}
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iff for every $\epsilon > 0$ there exists
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a finite set of points $x_1,\ldots,x_n$
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such that $X = \bigcup_{i=1}^n B_{\epsilon}(x_i)$.
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\end{definition}
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