updates

content/Math4201/Math4201_L4.md

@@ -115,7 +115,7 @@ An open set with respect to $\mathcal{T}_{\mathcal{S}}$ is a subset of $X$ such
 
 <details>
 
-<summary>Example (standard topology on $\mathbb{R}$)</summary>
+<summary>Example (standard topology on real numbers)</summary>
 
 Let $X=\mathbb{R}$. Take $\mathcal{S}=\{(-\infty, a)|a\in \mathbb{R}\}\cup \{(a,+\infty)|a\in \mathbb{R}\}$.
 

content/Math4201/Math4201_L5.md

@@ -1 +1,106 @@
 # Math4201 Lecture 5 Bonus
+
+## Comparison of two types of topologies
+
+Let $X=\mathbb{R}^2$ and the two types of topologies are:
+
+The "circular topology":
+
+$$
+\mathcal{T}_c=\{B_r(p)\mid p\in \mathbb{R}^2,r>0\}
+$$
+
+The "rectangle topology":
+
+$$
+\mathcal{T}_r=\{(a,b)\times (c,d)\mid a,b,c,d\in \mathbb{R},a<b,c<d\}
+$$
+
+> Are these two topologies the same?
+
+### Comparison of two topologies
+
+#### Definition of finer and coarser
+
+Let $\mathcal{T}$ and $\mathcal{T}'$ be two topologies on $X$. We say $\mathcal{T}$ is finer than $\mathcal{T}'$ if $\mathcal{T}'\subseteq \mathcal{T}$. We say $\mathcal{T}$ is coarser than $\mathcal{T}'$ if $\mathcal{T}\subseteq \mathcal{T}'$. We say $\mathcal{T}$ and $\mathcal{T}'$ are equivalent if $\mathcal{T}=\mathcal{T}'$.
+
+$\mathcal{T}$ is strictly finer than $\mathcal{T}'$ if $\mathcal{T}'\subsetneq \mathcal{T}$. (that is, $\mathcal{T}'$ is finer and not equivalent to $\mathcal{T}$)
+$\mathcal{T}$ is strictly coarser than $\mathcal{T}'$ if $\mathcal{T}\subsetneq \mathcal{T}'$. (that is, $\mathcal{T}$ is coarser and not equivalent to $\mathcal{T}'$)
+
+<details>
+<summary>Example (discrete topology is finer than the trivial topology)</summary>
+
+Let $X$ be an arbitrary set. The discrete topology is $\mathcal{T}_1 = \mathcal{P}(X)=\{U \subseteq X\}$
+
+The trivial topology is $\mathcal{T}_0 = \{\emptyset, X\}$
+
+Clearly, $\mathcal{T}_1 \subseteq \mathcal{T}_0$.
+
+So the discrete topology is finer than the trivial topology.
+
+</details>
+
+#### Lemma
+
+> [!TIP]
+>
+> Motivating condition:
+>
+> We want $U$ be an open set in $\mathcal{T}'$, then $U$ has to be open with respect to $\mathcal{T}$. In other words, $\forall x\in U, \exists$ some $B\in \mathcal{B}$ such that $x\in B\subseteq U$.
+
+Let $\mathcal{T}$ and $\mathcal{T}'$ be topologies on $X$ associated with bases $\mathcal{B}$ and $\mathcal{B}'$. Then
+
+$$
+\mathcal{T}\text{ is finer than } \mathcal{T}'\iff \text{ for any } B'\in \mathcal{B}', \exists B\in \mathcal{B} \text{ such that } B'\subseteq B
+$$
+
+<details>
+<summary>Proof</summary>
+
+$(\Rightarrow)$
+
+Let $B'\in \mathcal{B}'$. If $x\in B'$, then $B'\in \mathcal{T}'$ and $T$ is finer than $T'$, so $B'\in \mathcal{T}$.
+
+Take $T=\mathcal{T}_{\mathcal{B}}$. $\exists B\in \mathcal{B}$ such that $x\in B\subseteq B'$.
+
+$(\Leftarrow)$
+
+Let $U\in \mathcal{T}$. Then $U=\bigcup_{\alpha \in I} B_\alpha'$ for some $\{B_\alpha'\}_{\alpha \in I}\subseteq \mathcal{B}'$.
+
+For any $B_\alpha'$ and any $x\in \mathcal{B}_\alpha'$, $\exists B_\alpha\in \mathcal{B}$ such that $x\in B_\alpha\subseteq B_\alpha'$.
+
+Then $B_\alpha'$ is open set in $\mathcal{T}$.
+
+So $U$ is open in $\mathcal{T}$.
+
+$T$ is finer than $T'$.
+
+</details>
+
+Back to the example:
+
+For every point in open circle, we can find a rectangle that contains it.
+
+For every point in open rectangle, we can find a circle that contains it.
+
+So these two topologies are equivalent.
+
+#### Standard topology in $\mathbb{R}^2$
+
+The standard topology in $\mathbb{R}^2$ is the topology generated by the basis $\mathcal{B}_{st}=\{(a,b)\times (c,d)\mid a,b,c,d\in \mathbb{R},a<b,c<d\}$. This is equivalent to the topology generated by the basis $\mathcal{B}_{disk}=\{(x,y)\in \mathbb{R}^2|d((x,y),(a,b))<r\}$.
+
+<details>
+
+<summary>Example (lower limit topology is strictly finer than the standard topology)</summary>
+
+The lower limit topology is the topology generated by the basis $\mathcal{B}_{ll}=\{[a,b)\mid a,b\in \mathbb{R},a<b\}$.
+
+This is finer than the standard topology.
+
+Since $(a,b)\in \mathcal{B}_{st}$, we have $\forall x\in (a,b), \exists B=[x,b)\in \mathcal{B}_{ll}$ such that $x\in B\subsetneq (a,b)$.
+
+So the lower limit topology is strictly finer than the standard topology.
+
+$[0,1)$ is not open in the standard topology. but it is open in the lower limit topology.
+
+</details>

content/Math4201/_meta.js

@@ -7,6 +7,6 @@ export default {
     Math4201_L2: "Topology I (Lecture 2)",
     Math4201_L3: "Topology I (Lecture 3)",
     Math4201_L4: "Topology I (Lecture 4)",
-    Math4201_L5: "Topology I (Lecture 5) Bounus",
+    Math4201_L5: "Topology I (Lecture 5) Bonus",
     Math4201_L6: "Topology I (Lecture 6)"
 }

content/index.md

@@ -2,7 +2,9 @@
 
 > [!WARNING]
 >
-> 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.
+> 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.
+>
+> This site updates in a daily basis. But the sidebar is not. **If you find some notes are not shown on sidebar but the class already ends more than 24 hours**, please try to access the page directly via the URL. (for example, change the URL to `https://notenextra.trance-0.com/Math4201/Math4201_L{number}` to access the note of the lecture `Math4201_L{number}`)
  
 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.