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Trance-0
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If $Z_0=0$, then count $\infty$ as root.
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Using stereographic projection of each root we can get a unordered collection of $S^2$. Example: $\mathbb{C}P=S^2$, $\mathbb{C}p^2=S^2\times S^2\setminus S_2$ where $S_2$ is symmetric group.
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Using stereographic projection of each root we can get a unordered collection of $S^2$. Example: $\mathbb{C}P=S^2$, $\mathbb{C}p^2=S^2\times S^2\setminus S_2$ where $S_2$ is symmetric group.
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> [!NOTE]
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>
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> TODO: Check more definition from different area of mathematics (algebraic geometry, complex analysis, etc.) of the Majorana representation of quantum states.
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>
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> Read Chapter 5 and 6 of [Geometry of Quantum states](https://www.cambridge.org/core/books/geometry-of-quantum-states/46B62FE3F9DA6E0B4EDDAE653F61ED8C) for more details.
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content/Math4201/Math4201_L7.md
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content/Math4201/Math4201_L7.md
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# Math 4201 Lecture 7
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## Review from last lecture
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Not every open set in subspace topology is open in the original space
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Let $X=\mathbb{R}$ with standard topology and $Y=[0,1]\cup [2,3]$. equipped with subspace topology generated by the standard basis for $\mathbb{R}$.
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so $[0,1]=(-1,\frac{3}{2})\cap Y$ In particular, $[0,1]$ is open set in $Y$, but not an open set in $\mathbb{R}$.
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## New materials
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### Closed sets in topological space
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#### Proposition of open set in subspace topology
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If $X$ is a topological space, then $Y\subseteq X$ is open and is with the subspace topology. If $Z\subset Y$ is open subspace of $Y$, then $Z$ is also an open subspace of $X$.
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<details>
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<summary>Proof</summary>
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Since $Z\subset Y$ is open in the subspace topology, there is an open $U\subset Y$ such that $Z=U\cap Y$.
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SInce $Z$ is the intersection of open sets in $X$, then $Z$ is open in $X$.
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</details>
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#### Definition of closed set
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For any topology $\mathcal{T}$ on a set $X$, a subset $Z\subseteq X$ is said to be **closed** if its complement $Z^c$ is an open set (in $X$).
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> Note the complement is defined $Z=X\setminus Z$.
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<details>
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<summary>Example of closed set in standard topology of real numbers</summary>
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For example, $[a,b]$ is a closed set in the standard topology of real numbers. since $\mathbb{R}-[a,b]=(-\infty,a)\cup (b,\infty)$ is an open set.
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</details>
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<details>
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<summary>Example of closed set in trivial topology</summary>
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For any set $X$, the trivial topology is $\mathcal{T}_0=\{\emptyset, X\}$.
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Since $X^c=\emptyset$ is an open set, $X$ is a closed set.
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Since $\emptyset^c=X$ is an open set, $\emptyset$ is a closed set.
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</details>
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<details>
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<summary>Example of closed set in finite complement topology</summary>
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For any set $X$, the finite complement topology is $\mathcal{T}_1=\{U\subseteq X\mid X\setminus U\text{ is finite}\}$.
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Then the set of all finite subsets of $X$ is a closed set.
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</details>
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For general, if $\mathcal{T}$ is a topology on $X$, then:
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1. $\emptyset, X$ are closed sets.
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2. $\mathcal{T}$ is closed with respect to arbitrary unions.
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Let $\{Z_\alpha\}_{\alpha \in I}$ be an arbitrary collection of closed sets in $X$. Then $X-Z_\alpha$ is open for each $\alpha \in I$. So $\forall \alpha \in I$. $\bigcup_{\alpha \in I} (X-Z_\alpha)=X-\bigcap_{\alpha \in I} Z_\alpha$ is open. So $\bigcap_{\alpha \in I} Z_\alpha$ is closed.
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So the corollary is: an arbitrary intersection of closed sets is closed.
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3. $\mathcal{T}$ is closed with respect to finite intersections. This also implies that a finite union of closed sets is closed.
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If $\{Z_1, Z_2, \ldots, Z_n\}$ is a closed subset of $X$, then $X-Z_i$ is open for each $i=1,2,\ldots,n$. So $\forall i=1,2,\ldots,n$. $\bigcap_{i=1}^n (X-Z_i)=X-\bigcup_{i=1}^n Z_i$ is open. So $\bigcup_{i=1}^n Z_i$ is closed.
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> [!NOTE]
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>
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> We can also define the topology using the closed sets instead of the open sets.
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>
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> 1. $\emptyset, X$ are closed sets.
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> 2. $\mathcal{T}$ is closed with respect to arbitrary intersections.
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> 3. $\mathcal{T}$ is closed with respect to finite unions.
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>
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> This yields the same topology.
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#### Theorem of closed set in subspace topology
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Let $X$ is a topological space and $Y\subseteq X$ equipped with the subspace topology.
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A subset $Z\subseteq Y$ is closed in $Y$ if and only if $Z$ is the intersection of a closed $W\subseteq X$ and $Y$. That is $Z=W\cap Y$.
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<details>
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<summary>Proof</summary>
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$\Rightarrow$
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If $Z$ is closed in $Y$, then $Y-Z$ is open in $Y$.
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Then, there is open set $U\subseteq X$ such that $Y-Z=U\cap Y$.
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So $Z=(X-U)\cap Y$, $X-U$ is closed in $X$ because $U$ is open in $X$.
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Take $W=X-U$.
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$\Leftarrow$
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If $Z=W\cap Y$ for some closed $W\subseteq X$, then $Y-Z=Y-(W\cap Y)=(Y-W)\cap Y$ is open in $Y$.
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So $Z$ is closed in $Y$.
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</details>
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#### Lemma of closed in closed subspace
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Let $X$ is a topological space and $Y\subseteq X$ is closed and is equipped with the subspace topology. Then any closed subset of $Y$ is also closed in $X$.
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> [!WARNING]
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>
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> Not any subset of a topological space $X$ is either open or closed.
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<details>
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<summary>Example of open and closed subset</summary>
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Let $X=\mathbb{R}$ with standard topology.
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$(a,b)$ is open, but not closed.
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$[a,b]$ is closed, but not open.
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$[a,b)$ is neither open nor closed.
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$\emptyset,\mathbb{R}$ is both open and closed.
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</details>
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<details>
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<summary>Example of open and closed subset in other topologies</summary>
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Let $X=[0,1]\cup (2,3)$ induced by the standard topology of $\mathbb{R}$.
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$Z=[0,1]$ is an open subset of $X$.
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$Z=[0,1]$ is also closed subset of $X$ since $Z=[0,1]\cap X$ is open in $\mathbb{R}$.
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</details>
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We can associate an open and a closed to any subset $A$ of a topological space $X$.
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#### Interior and closure of a set
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The interior of a set $A$ is defined as follows:
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$$
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\operatorname{Int}(A)=\bigcup_{U\subseteq A, U\text{ is open in }X} U
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$$
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Also denoted as $A^\circ$.
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The interior of a set $A$ is the largest open subset of $A$.
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That is $\forall U\subseteq A, U\text{ is open in }X$, then $U\subseteq \operatorname{Int}(A)$. (by definition that $U$ must be in collection of open sets that is a subset of $A$)
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#### Closure of a set
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The closure of a set $A$ is the smallest closed subset of $X$ that contains $A$.
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> Note that if we change the definition as the intersection of all closed subsets of $X$ that **contained in $A$**, we will get the empty set.
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$$
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\overline{A}=\bigcap_{A\subseteq C, C\text{ is closed in }X} C
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$$
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The closure of a set $A$ is the smallest closed subset of $X$ that contains $A$. (follows the same logic as the previous definition)
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