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# Math4201 Lecture 2
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## Topology
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Distance in $\mathbb{R}$, or more generally in $\mathbb{R}^n$. ([metric space](https://notenextra.trance-0.com/Math4111/Math4111_L9#metric-space))
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Intervals in $\mathbb{R}$, or more generally open balls in $\mathbb{R}^n$. (topological space)
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### Topological Spaces
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#### Definition of Topological Space
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A topological space is a pair of set $X$ and a collection of subsets of $X$, denoted by $\mathcal{T}$ (imitates the set of "open sets" in $X$), satisfying the following axioms:
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1. $\emptyset \in \mathcal{T}$ and $X \in \mathcal{T}$
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2. $\mathcal{T}$ is closed with respect to arbitrary unions. This means, for any collection of open sets $\{U_\alpha\}_{\alpha \in I}$, we have $\bigcup_{\alpha \in I} U_\alpha \in \mathcal{T}$
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3. $\mathcal{T}$ is closed with respect to finite intersections. This means, for any finite collection of open sets $\{U_1, U_2, \ldots, U_n\}$, we have $\bigcap_{i=1}^n U_i \in \mathcal{T}$
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The elements of $\mathcal{T}$ are called open sets.
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The topological space is denoted by $(X, \mathcal{T})$.
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#### Examples
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Trivial topology: Let $X$ be arbitrary. The trivial topology is $\mathcal{T}_0 = \{\emptyset, X\}$
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Discrete topology: Let $X$ be arbitrary. The discrete topology is $\mathcal{T}_1 = \mathcal{P}(X)=\{U \subseteq X\}$
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Understanding all possible topologies on a set.
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Let's say $X=\{a,b\}$
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The trivial topology is $\mathcal{T}_0 = \{\emptyset, \{a,b\}\}$
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The discrete topology is $\mathcal{T}_1 = \{\emptyset, \{a\}, \{b\}, \{a,b\}\}$
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Other topologies:
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$\mathcal{T}_2 = \{\emptyset, \{a\}, \{a,b\}\}$
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$\mathcal{T}_3 = \{\emptyset, \{b\}, \{a,b\}\}$
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Non-examples:
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Let $X=\{a,b,c\}$
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The set $\mathcal{T}_1=\{\emptyset, \{a\}, \{b\}, \{a,b,c\}\}$ is not a topology because it is not closed under union $\{a\} \cup \{b\} = \{a,b\} \notin \mathcal{T}_1$
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#### Definition of Complement finite topology
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Let $X$ be arbitrary. The complement finite topology is $\mathcal{T}\coloneqq \{U\subseteq X|X\setminus U \text{ is finite}\}\cup \{\emptyset\}$
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The topology is valid because:
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<details>
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<summary>Proof</summary>
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1. $\emptyset \in \mathcal{T}$ because $X\setminus \emptyset = X$ is finite.
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2. Let $\{U_\alpha\}_{\alpha \in I}$ be an arbitrary collection such that $X\setminus U_\alpha$ is finite for each $\alpha \in I$.
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Without loss of generality, we can assume that $U_\alpha \neq \emptyset$ for each $\alpha \in I$, since the union of arbitrary set with $\emptyset$ is the set itself.
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If all of them are empty, then the union is empty, which complement is $X\subset \mathcal{T}$.
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Otherwise,
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$$
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X\setminus \bigcup_{\alpha \in I} U_\alpha = \bigcap_{\alpha \in I} (X\setminus U_\alpha)
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$$
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is finite because each $X\setminus U_\alpha$ is finite. Therefore, $\bigcup_{\alpha \in I} U_\alpha \in \mathcal{T}$.
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3. Let $\{U_1, U_2, \ldots, U_n\}$ be a finite collection such that $X\setminus U_i$ is finite for each $i=1,2,\ldots,n$.
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Without loss of generality, we can assume that $U_i \neq \emptyset$ for each $i=1,2,\ldots,n$, since the intersection of arbitrary set with $\emptyset$ is $\emptyset$.
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If all of them are empty, then the intersection is $X\subset \mathcal{T}$.
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Otherwise,
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$$
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X\setminus \bigcap_{i=1}^n U_i = \bigcup_{i=1}^n (X\setminus U_i)
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$$
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is finite because each $X\setminus U_i$ is finite. Therefore, $\bigcap_{i=1}^n U_i \in \mathcal{T}$.
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</details>
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Another non-example is $\mathcal{T} = \{U\subseteq X|U \text{ is finite}\}\cup \{X\}$
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The topology is invalid because an arbitrary union of points is not a finite set.
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Consider $X=\mathbb{Z}$ but take $U_1=\{1\}, U_2=\{2\}, U_3=\{3\}, \ldots$
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Then $\bigcup_{i=1}^\infty U_i = \mathbb{Z}^+$ is not a finite set and is not $\{X\}$.
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> [!NOTE]
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>
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> If $X$ is finite, then finite complement topology is the same as the discrete topology.
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#### Definition of Topology basis
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For a set $X$, a topology basis, denoted by $\mathcal{B}$, is a collection of subsets of $X$, such that the following properties are satisfied:
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1. For any $x \in X$, there exists a $B \in \mathcal{B}$ such that $x \in B$
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2. If $B_1, B_2 \in \mathcal{B}$ and $x \in B_1 \cap B_2$, then there exists a $B_3 \in \mathcal{B}$ such that $x \in B_3 \subseteq B_1 \cap B_2$
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@@ -4,4 +4,5 @@ export default {
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type: 'separator'
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},
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Math4201_L1: "Topology I (Lecture 1)",
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Math4201_L2: "Topology I (Lecture 2)",
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}
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@@ -1,6 +1,7 @@
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# Welcome to NoteNextra
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> [!WARNING]
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>
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> This site use [Algolia Search](https://www.algolia.com/) to search the content. However, due to some unknown reasons, when the index page is loaded, the search bar is calling default PageFind package from Nextra. If you find the search bar is not working, please try to redirect to another page and then back to the index page or search in another page.
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This was originated from another project [NoteChondria](https://github.com/Trance-0/Notechondria) that I've been working on for a long time but don't have a stable release yet.
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