generated from smyalygames/quartz
80 lines
3.2 KiB
Markdown
80 lines
3.2 KiB
Markdown
---
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lecture: 12
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date: 2025-02-24
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---
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# Weakest Topology
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Definition of [[Weakest Topology]] defined in the lecture.
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# Initial Topology
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Definition of [[Initial Topology]], defined in the lecture.
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# Proposition
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Let $X$ be a set with [[Initial Topology|initial topology]] induced by a family, $F$, of functions.
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Then a [[Nets|net]] in $\{ x_{i} \}$ in $X$ converges to $x \iff f(x_{i}) \to f(x) \; \forall f \in F$.
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## Proof
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### $\Rightarrow$:
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Is obvious.
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### $\Leftarrow$:
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Let $A$ be a neighbourhood of $x$.
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Hence there are finitely many [[Open Sets|open sets]] $A_{n} \subset Y_{f_{n}}$ such that $x \in \cap f^{-1}_{n}(A_{n}) \subset A$. Then $A_{n}$ is a neighbourhood of $f_{n}(x), \; \forall n$.
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By assumption $f_{n}(x_{i}) \to f_{n}(x)$, so $f_{n}(x_{i}) \in A_{n} \; \forall i \geq i(n)$ for some $i(n)$. Pick $j$ such that $j \geq i(n)$ for all the finitely many $n$s.
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Then $x_{i} \in \cap f^{-1}_{n}(A_{n}) \subset A$ for all $i \geq j$, so $x_i \to x$.
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QED.
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# Corollary
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$X$ has [[Initial Topology|initial topology]] induced by $F$. Say $Z$ is a [[Topological Space|topological space]], then:
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$g : Z \to X$ is [[Continuous|continuous]] $f \circ g$ is [[Continuous|continuous]] $\forall f \in F$.
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## Proof
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### $\Rightarrow$:
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Clear.
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### $\Leftarrow$:
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Say we have a [[Nets|net]] $\{ z_{i} \}$ that $z_{i} \to z$ in $Z$.
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Then $\underbrace{(f \circ g)(z_{i})}_{= f(g(z_{i}))} \to \underbrace{(f \circ g)(z)}_{f(g(z))} \; \forall f \in F$.
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Hence $g(z_{i}) \to g(z)$ by the proposition, so $g$ is [[Continuous|continuous]].
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# Product Topology
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Definition of the [[Product Topology]], defined in the lecture.
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![[Drawing 2025-02-24 11.58.37.excalidraw.dark.svg]]
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%%[[Drawing 2025-02-24 11.58.37.excalidraw.md|🖋 Edit in Excalidraw]], and the [[Drawing 2025-02-24 11.58.37.excalidraw.light.svg|light exported image]]%%
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$(x_{i}, y_{i}) \to (x,y)$
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By the previous proposition a net $\{ x_{i} \}$ in $\Pi X_{\lambda}$ converges to $x$ with respect to the [[Product Topology|product topology]] $\iff \pi_{\lambda} \to \pi_{\lambda}(x), \; \forall \lambda$.
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### Note
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$\pi_{\lambda}$ are [[Continuous|continuous]] (obvious) and [[Open Sets|open sets]] $\pi_{\lambda}(A)$ open in $X_{\lambda}$ for $A$ open in $\Pi X_{\lambda}$
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## Tychonoff
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It is the [[Tychonoff Theorem]] (defined in lecture).
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# Separating Family
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Defines [[Separating Points]] from the lecture.
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# Proposition
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A set $X$ with [[Initial Topology|initial topology]] induced from a [[Separating Points|separating family]] of functions $f : X \to Y_{f}$ is [[Hausdorff]] when all $Y_{f}$ are [[Hausdorff]].
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## Proof
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Say $x \neq y$ in $X$. Then $\exists f$ such that $f(x) \neq f(y)$ in $Y_{f}$.
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Can separate $f(x)$ and $f(y)$ by neighbourhoods $U$ and $V$ such that $U \cap V = \emptyset$.
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Then $f^{-1}(U)$ and $f^{-1}(V)$ will be disjoint neighbourhoods of $x$ and $y$.
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> [!note] Reasoning for $f^{-1}(U)$ and $f^{-1}(V)$ being disjoint
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> $$z \in f^{-1}(U) \cap f^{-1}(V)$$
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> $$f(z) \in U \cap f(z) \in V$$
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> Which is not possible
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QED.
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# Corollary
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A product of [[Hausdorff]] spaces is [[Hausdorff]] in the [[Product Topology|product topology]].
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$\Pi X_{\lambda}$
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$\Pi_{\lambda}$
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$f: X \to Y_{f}$
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$f : X_{f} \to Y$
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$q : X \to X \setminus \sim$
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$q(x) = [x]$ |