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
+# 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)"
 }