From 0200ab7eed835ac0de932b2a862d43776385a97d Mon Sep 17 00:00:00 2001
From: Trance-0 <60459821+Trance-0@users.noreply.github.com>
Date: Wed, 10 Sep 2025 11:51:58 -0500
Subject: [PATCH] updates
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.../Math401/Extending_thesis/Math401_S2.md | 8 +-
content/Math4201/Math4201_L7.md | 163 ++++++++++++++++++
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@@ -22,4 +22,10 @@ $$
If $Z_0=0$, then count $\infty$ as root.
-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.
\ No newline at end of file
+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.
+
+> [!NOTE]
+>
+> TODO: Check more definition from different area of mathematics (algebraic geometry, complex analysis, etc.) of the Majorana representation of quantum states.
+>
+> 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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+# Math 4201 Lecture 7
+
+## Review from last lecture
+
+Not every open set in subspace topology is open in the original space
+
+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}$.
+
+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}$.
+
+## New materials
+
+### Closed sets in topological space
+
+#### Proposition of open set in subspace topology
+
+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$.
+
+
+Proof
+
+Since $Z\subset Y$ is open in the subspace topology, there is an open $U\subset Y$ such that $Z=U\cap Y$.
+
+SInce $Z$ is the intersection of open sets in $X$, then $Z$ is open in $X$.
+
+
+
+#### Definition of closed set
+
+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$).
+
+> Note the complement is defined $Z=X\setminus Z$.
+
+
+Example of closed set in standard topology of real numbers
+
+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.
+
+
+
+Example of closed set in trivial topology
+
+For any set $X$, the trivial topology is $\mathcal{T}_0=\{\emptyset, X\}$.
+Since $X^c=\emptyset$ is an open set, $X$ is a closed set.
+Since $\emptyset^c=X$ is an open set, $\emptyset$ is a closed set.
+
+
+
+
+Example of closed set in finite complement topology
+
+For any set $X$, the finite complement topology is $\mathcal{T}_1=\{U\subseteq X\mid X\setminus U\text{ is finite}\}$.
+
+Then the set of all finite subsets of $X$ is a closed set.
+
+
+For general, if $\mathcal{T}$ is a topology on $X$, then:
+
+1. $\emptyset, X$ are closed sets.
+2. $\mathcal{T}$ is closed with respect to arbitrary unions.
+
+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.
+
+So the corollary is: an arbitrary intersection of closed sets is closed.
+
+3. $\mathcal{T}$ is closed with respect to finite intersections. This also implies that a finite union of closed sets is closed.
+
+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.
+
+> [!NOTE]
+>
+> We can also define the topology using the closed sets instead of the open sets.
+>
+> 1. $\emptyset, X$ are closed sets.
+> 2. $\mathcal{T}$ is closed with respect to arbitrary intersections.
+> 3. $\mathcal{T}$ is closed with respect to finite unions.
+>
+> This yields the same topology.
+
+#### Theorem of closed set in subspace topology
+
+Let $X$ is a topological space and $Y\subseteq X$ equipped with the subspace topology.
+
+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$.
+
+
+Proof
+
+$\Rightarrow$
+
+If $Z$ is closed in $Y$, then $Y-Z$ is open in $Y$.
+
+Then, there is open set $U\subseteq X$ such that $Y-Z=U\cap Y$.
+
+So $Z=(X-U)\cap Y$, $X-U$ is closed in $X$ because $U$ is open in $X$.
+
+Take $W=X-U$.
+
+$\Leftarrow$
+
+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$.
+So $Z$ is closed in $Y$.
+
+
+
+#### Lemma of closed in closed subspace
+
+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$.
+
+> [!WARNING]
+>
+> Not any subset of a topological space $X$ is either open or closed.
+
+
+Example of open and closed subset
+
+Let $X=\mathbb{R}$ with standard topology.
+
+$(a,b)$ is open, but not closed.
+$[a,b]$ is closed, but not open.
+$[a,b)$ is neither open nor closed.
+$\emptyset,\mathbb{R}$ is both open and closed.
+
+
+
+
+Example of open and closed subset in other topologies
+
+Let $X=[0,1]\cup (2,3)$ induced by the standard topology of $\mathbb{R}$.
+
+$Z=[0,1]$ is an open subset of $X$.
+
+$Z=[0,1]$ is also closed subset of $X$ since $Z=[0,1]\cap X$ is open in $\mathbb{R}$.
+
+
+
+We can associate an open and a closed to any subset $A$ of a topological space $X$.
+
+#### Interior and closure of a set
+
+The interior of a set $A$ is defined as follows:
+
+$$
+\operatorname{Int}(A)=\bigcup_{U\subseteq A, U\text{ is open in }X} U
+$$
+
+Also denoted as $A^\circ$.
+
+The interior of a set $A$ is the largest open subset of $A$.
+
+That is $\forall U\subseteq A, U\text{ is open in }X$, then $U\subseteq \operatorname{Int}(A)$. (by definition that $U$ must be in collection of open sets that is a subset of $A$)
+
+#### Closure of a set
+
+The closure of a set $A$ is the smallest closed subset of $X$ that contains $A$.
+
+> Note that if we change the definition as the intersection of all closed subsets of $X$ that **contained in $A$**, we will get the empty set.
+
+$$
+\overline{A}=\bigcap_{A\subseteq C, C\text{ is closed in }X} C
+$$
+
+The closure of a set $A$ is the smallest closed subset of $X$ that contains $A$. (follows the same logic as the previous definition)