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115 lines
6.1 KiB
Markdown
115 lines
6.1 KiB
Markdown
---
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lecture: 13
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date: 2025-02-27
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---
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# Sigma Algebra
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Defined in the lecture: [[Sigma-Algebra]]
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(See also [[Sigma-Algebra#Measurable]])
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## Measure
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(See [[Sigma-Algebra#Measure]])
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> [!example]
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> $M = \wp(X)$ is a $\sigma$-algebra with the **counting measure** $\mu$ given by
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> $\mu(A) = \begin{cases}|A|, & \text{when}\ A\ \text{is finite}\\ \infty, & \text{when}\ A\ \text{is infinite}\end{cases}$
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> $\mu'(A) = 0,\ \forall A$
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Given a collection $N$ of subsets of any set $X$, the intersection $M$ of all $\sigma$-algebras containing $N$ is a $\sigma$-algebra; the one **generated** by $N$.
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## Borel Sets
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Defined in the lecture: [[Borel Sets]]
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>[!example] Example of [[Borel Sets]]
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> $\cap^{\infty}_{n=1}\left[ -\frac{1}{n}, \frac{1}{n} \right] = \{ 0 \}$
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# Proposition
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Say $X$ has a $\sigma$-algebra. Then all [[Borel Measurable|measurable]] functions $X \to \mathbb{C}$ is an algebra under [[Pointwise|pointwise]] operations.
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## Proof
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Say $f$, $g$ are [[Borel Measurable|measurable]] and $a, b \in \mathbb{C}$.
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Note that $k : \mathbb{C} \to \mathbb{C}$ given by $b \mapsto ab$ is [[Continuous|continuous]] then $(af)(x) = (\underbrace{k}_{\text{continuous}} \circ \underbrace{f}_{\text{measurable}})(x)$ ($= k(\underbrace{f(x)}_{b}) = ab = a \times f(x) = (af)(x)$)
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Since $\mathrm{Re}, \mathrm{Im} : \mathbb{C} \to \mathbb{R}$ by $\mathrm{Re}(a + ib) = a$ and $\mathrm{Im}(a + ib) = b$ are [[Continuous|continuous]] the real and imaginary parts of $f$ are [[Borel Measurable|measurable]].
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So to show that $f+g$, $f \times g$ are [[Borel Measurable|measurable]], we may assume that $f$ and $g$ are real.
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> [!note]
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>$$f + g = \mathrm{Re}(f+g) + i \mathrm{Im}(f+g) = \mathrm{Re} f + \mathrm{Re} g + i (\mathrm{Im} f + \mathrm{Im} g)$$
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>$(a, b) \mapsto x + i b$ is [[Continuous|continuous]].
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Let $h : x \to \mathbb{R}^2$; $h(x) = (f(x), g(x))$.
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**Claim** that $h$ is [[Borel Measurable|measurable]].
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Then use that $f + g$, $f \times g$ are compositions of $h$ with the [[Continuous|continuous]] functions $\mathbb{R}^2 \to \mathbb{R}$ given by $(s, t) \xmapsto{p} s + t$ and $(s, t) \xmapsto{q} s \times t$, so $p \circ h = f + g$ and $q \circ h = f \times g$.
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(All that needs to be proven now is that $h$ is [[Borel Measurable|measurable]])
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Show the claim: note that [[Open Sets|open]] $V \subset \mathbb{R}^2$ is a [[Countable|countable]] union of rectangles $I \times J$ for segments $I, J \subset \mathbb{R}$.
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Also note that (taking the inverse image of the rectangle) $h^{-1}(I \times J) = f^{-1}(I) \cap g^{-1}(J)$ is [[Borel Measurable|measurable]].
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Then so is $h^{V} = h^{-1}(\cup_{n}(I_{n} \times J_{n})) = \cup_{n}h^{-1}(I_{n} \times J_{n})$.
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QED.
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---
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Exercise Part of the session...
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# Question 3 (Exercise 6)
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Show that a sequence $E_{1} \supset E_{2} \supset E_{3} \supset \dots$ of a [[Compact|compact]] non-empty subsets of a [[Hausdorff]] space has a non-empty intersection $\cap E_{i} \neq \emptyset$
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## Proof
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### Net argument
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$X = \mathbb{R}$
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$E_{n} = \left[ -\frac{1}{n}, \frac{1}{n} \right]$
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$[-1,1] \supset \left[ -\frac{1}{2}, \frac{1}{2} \right] \supset \left[ -\frac{1}{3}, \frac{1}{3} \right] \supset \dots$
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$\cap_{n} [-\frac{1}{n}, \frac{1}{n}] = \{ \emptyset \} \neq \emptyset$
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$I = \{ E_{i} | i=1,\dots,\infty \}$
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Upward filtered ordered set under reverse inclusion.
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Defined by axiom of choice $X_{E_{i}} \in E_{i}$, so we get a net $\{ x_{E_{i}} \}_{E_{i} \in I} \subset E_{i}$
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Then it has a convergent subnet $\{ x_{{E_{i}}_{j}} \}_{j}$, say $x_{{E_{i}}_{j}} \to x \in E_{1}$.
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Claim $x \in \cap E_{i}$:
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If $A$ is a neighbourhood of $x$ in $E_{1}$, then $x_{{E_{i}}_{j}} \in A\ \forall j \geq k$.
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So $A \cap {E_{i}}_{j} \neq \emptyset\ \forall j \geq k$.
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![[Drawing 2025-02-27 13.19.24.excalidraw.dark.svg]]
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%%[[Drawing 2025-02-27 13.19.24.excalidraw.md|🖋 Edit in Excalidraw]], and the [[Drawing 2025-02-27 13.19.24.excalidraw.light.svg|light exported image]]%%
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If $x \notin \cap E_{i}$, then $\exists$ neighbourhood
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$B$ of $x$ such that $B \cap (\cap E_{i}) = \emptyset$.
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Then $B \cap E_{i} = \emptyset\ \forall i \geq m$.
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Pick $B=A$ and get a contradiction.
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### Subcover argument
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$V_{n} = E_{n}^{\complement}$ [[Open Sets|open]] in [[Hausdorff]] space if $\cap E_{i} = \emptyset$, then $\{ V_{n} \}$ will be an [[Open Cover|open cover]] of $E_{1}$ that has no finite [[Subcover|subcover]], contradicting that $E_{1}$ is compact.
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$\cup V_{n} = \cup E_{n}^{C} = (\cap E_{n})^{\complement} = \emptyset^{\complement} = X$
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# Question 2 (Exercise 7)
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Show that [[Compact|compact]] [[Hausdorff]] spaces are rigid; if $X$ is [[Compact|compact]] [[Hausdorff]] it has no weaker or stronger [[Topology|topology]] that is [[Compact|compact]] [[Hausdorff]].
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## Proof
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$\iota : (X, \overbrace{\tau}^{\text{Compact Hausdorff}}) \to (X, \overbrace{\tau}^{\text{Weaker}})$ ($\tau' \subset \tau$)
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$\iota(x) = x$, so $\iota$ is [[Bijective|bijective]]
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This is a [[Continuous|continuous]] map since $\tau' \subset \tau$.
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$\iota$ is also an [[Open Map|open map]], because it takes closed sets to closed sets, because if $E$ is closed in $(X, \tau)$, then it is [[Compact|compact]], and $\iota(E)$ is [[Compact|compact]] in $(X, \tau')$ as $\iota$ is [[Continuous|continuous]], so $\iota(E)$ is closed in $(X, \tau')$ as it is [[Hausdorff]]. So in this case $\tau' = \tau$.
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# Question 1 (Exercise 7)
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A [[Topological Space|topological space]] is **second [[Countable|countable]]** if we have a sequence $\{ V_{n} \}$ of [[Open Sets|open sets]] such that any open set is a union of some of these $V_{n}$s.
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## Proof
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$\langle c, d \rangle = \cup V_{\frac{1}{n}}^{a}$
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$V_{n} = \{ n \}$
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$\mathbb{R} = \cup_{a \in \mathbb{R}}V_{a}$
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$V_{a} = \{ a \}$
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(which is not [[Countable|countable]])
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Instead:
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$I \times J$
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Any open $A \subset \mathbb{R}$ [[Countable|countable]] union of $I$.
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$X$ is separable if it has a dense sequence $\{ x_{n} \} = X$.
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Claim:
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$X$ selectable countable $\implies X$ separable
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Axiom of choice def $x_{n} \in V_{n}. Then $\overline{\{ x_{n} \}} = X$. If $\overline{\{ x_{n} \}} \neq X$.
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$x \in \overline{\{ x_{n} \}}^{\complement}$ $\exists$ [[Open Sets|open]] $A$ with $x \in A$ and $A \cap \overline{\{ x_{n} \}} = \emptyset$. But $\exists m$ such that $x \in V_{m}$....... |