tutorial 02
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inputs/tutorial_02.tex
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\tutorial{02}{2023-10-24}{}
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% Points: 15 / 16
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\subsubsection{Exercise 4}
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\begin{fact}
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Let $X $ be a compact Hausdorffspace.
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Then the following are equivalent:
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\begin{enumerate}[(i)]
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\item $X$ is Polish,
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\item $X$ is metrisable,
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\item $X$ is second countable.
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\end{enumerate}
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\end{fact}
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\begin{proof}
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(i) $\implies$ (ii) clear
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(i) $\implies$ (iii) clear
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(ii) $\implies$ (i) Consider the cover $\{B_{\epsilon}(x) | x \in X\}$
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for every $\epsilon \in \Q$
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and chose a finite subcover.
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Then the midpoints of the balls from the cover
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form a countable dense subset.
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The metric is complete as $X$ is compact.
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(For metric spaces: compact $\iff$ seq.~compact $\iff$ complete and totally bounded)
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(iii) $\implies$ (ii)
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Use Urysohn's metrisation theorem and the fact that compact
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Hausdorff spaces are normal
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\end{proof}
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Let $X$ be compact Polish (compact metrisable $\implies$ compact Polish)
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and $Y $ Polish.
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Let $\cC(X,Y)$ be the set of continuous functions $X \to Y$.
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Consider the metric $d_u(f,g) \coloneqq \sup_{x \in X} |d(f(x), g(x))|$.
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Clearly $d_u$ is a metric.
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\begin{claim}
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$d_u$ is complete.
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\end{claim}
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\begin{subproof}
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Let $(f_n)$ be a Cauchy sequence in $\cC(X,Y)$.
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A $Y$ is complete,
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there exists a pointwise limit $f$.
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$f_n$ converges uniformly to $f$:
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\[
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d(f_n(x), f(x)) \le \overbrace{d(f_n(x), f_m(x))}^{\mathclap{\text{$(f_n)$ is Cauchy}}}
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+ \underbrace{d(f_m(x), f(x))}_{\mathclap{\text{small for appropriate $m$}}}.
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\]
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$f$ is continuous by the uniform convergence theorem.
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\end{subproof}
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\begin{claim}
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There exists a countable dense subset.
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\end{claim}
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