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content/Math4201/Math4201_L5.md

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+# Math4201 Lecture 5 Bonus

content/Math4201/Math4201_L6.md

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+# Math4201 Lecture 6
+
+## Product topology
+
+### Define topological spaces on cartesian product of two topological spaces
+
+Let $(X,\mathcal{T}_X)$ and $(Y,\mathcal{T}_Y)$ be two topological spaces.
+
+$$
+X\times Y=\{(x,y)|x\in X,y\in Y\}
+$$
+
+Goal: Define a topology on $X\times Y$.
+
+One way is to take $\mathcal{B}_{X\times Y}=\{(U\times V)|U\in \mathcal{T}_X,V\in \mathcal{T}_Y\}$ is a basis for the topology on $X\times Y$.
+
+> There are two ways to define the topology on $\mathbb{R}^2$: By rectangles ($\mathcal{B}_{rect}=\{(a,b)\times (c,d)|a,b,c,d\in \mathbb{R},a<b,c<d\}$) or by open disks ($\mathcal{B}_{disk}=\{(x,y)\in \mathbb{R}^2|d((x,y),(a,b))<r\}$). (check bonus video)
+
+<details>
+<summary>Proof</summary>
+
+Take $U=X,V=Y$, then $(x,y)\in (U\times V)=X\times Y$.
+
+So the first property of basis is satisfied.
+
+Check the second property of basis:
+
+Let $B_1=U_1\times V_1,B_2=U_2\times V_2$ be two basis elements, and $(x,y)\in B_1\cap B_2$.
+
+$$
+B_1\cap B_2=(U_1\times V_1)\cap (U_2\times V_2)=(U_1\cap U_2)\times (V_1\cap V_2)
+$$
+
+Since $U_1\cap U_2\in \mathcal{T}_X$ and $V_1\cap V_2\in \mathcal{T}_Y$, we have $(U_1\cap U_2)\times (V_1\cap V_2)\in \mathcal{B}_{X\times Y}$.
+
+Take $B_3=(U_1\cap U_2)\times (V_1\cap V_2)$, then $(x,y)\in B_3=B_1\cap B_2$.
+
+</details>
+
+> [!CAUTION]
+>
+> $\mathcal{B}_{X\times Y}$ is not a topology on $X\times Y$.
+>
+> $\mathcal{B}_{X\times Y}$ is not closed with respect to arbitrary unions.
+
+#### Definition of product topology
+
+Let $(X,\mathcal{T}_X)$ and $(Y,\mathcal{T}_Y)$ be two topological spaces.
+
+$$
+\mathcal{B}_{X\times Y}=\{(U\times V)|U\in \mathcal{T}_X,V\in \mathcal{T}_Y\}
+$$
+
+The product topology or $X\times Y$ is the topology generated by $\mathcal{B}_{X\times Y}$.
+
+Let $\mathcal{B}_X$ be a basis for $\mathcal{T}_X$ and $\mathcal{B}_Y$ be a basis for $\mathcal{T}_Y$.
+
+If we define
+
+$$
+\mathcal{B}'_{X\times Y}=\{(U\times V)|U\in \mathcal{B}_X,V\in \mathcal{B}_Y\}
+$$
+
+#### Proposition
+
+$\mathcal{B}'_{X\times Y}$ is a basis for the product topology on $X\times Y$.
+
+> [!NOTE]
+>
+> This basis is smaller than $\mathcal{B}_{X\times Y}$.
+>
+> Consider $X=Y=\mathbb{R}$ and $U=(a,b)\cap (c,d)$ and $V=(e,f)\cap (g,h)$. (Assume $a<b,c<d,e<f,g<h$) The union of $U$ and $V$ is four rectangles, which is not in $\mathcal{B}'_{X\times Y}$. but it is in $\mathcal{B}_{X\times Y}$.
+
+<details>
+<summary>Proof</summary>
+
+Using the [lemma](https://notenextra.trance-0.com/Math4201/Math4201_L4#theorem-of-basis-of-topology) from Friday. it suffices to show that:
+
+Let $W\in X\times Y$ and $(x,y)\in W$, we need to show that there are $B\in \mathcal{B}_X, B'\in \mathcal{B}_Y$ such that $(x,y)\in (B\times B')\subseteq W$.
+
+Since $W$ is open with respect to the topology generated by $\mathcal{B}_{X\times Y}$, in particular, there is $U\times V$ such that $(x,y)\in U\times V\subseteq W$. And $x\in U$ and $y\in V$.
+
+Since $U\in \mathcal{B}_X$ and $V\in \mathcal{B}_Y$, by property of basis $\mathcal{B}_X$ and $\mathcal{B}_Y$, $\forall x\in U$, $\exists B\in \mathcal{B}_X$ such that $x\in B\subseteq U$ and $\forall y\in V$, $\exists B'\in \mathcal{B}_Y$ such that $y\in B'\subseteq V$.
+
+So $(x,y)\in (B\times B')\subseteq U\times V\subseteq W$.
+
+</details>
+
+#### Definition of standard topology on $\mathbb{R}^n$
+
+The standard topology on $\mathbb{R}^n$ is defined as the product topology on $\mathbb{R}^{n-1}\times \mathbb{R}$.
+
+## Subspace topology
+
+Let $(X,\mathcal{T})$ be a topological space and $Y\subseteq X$.
+
+Want to define a topology on $Y$.
+
+$$
+\mathcal{T}_Y=\{U\cap Y|U\in \mathcal{T}\}
+$$
+
+We claim that $\mathcal{T}_Y$ is a topology on $Y$, called as the subspace topology on $Y$.
+
+<details>
+<summary>Proof</summary>
+
+First, $\emptyset \cap Y=\emptyset \in \mathcal{T}_Y$ and $Y=X\cap Y\in \mathcal{T}_Y$.
+
+Second, let $\{U_\alpha\cap Y\}_{\alpha \in I}$ be collection of open sets in $\mathcal{T}_Y$. Note that $U_\alpha\in \mathcal{T}$ for all $\alpha \in I$.
+
+So, $\bigcup_{\alpha \in I} U_\alpha\cap Y=\left(\bigcup_{\alpha \in I} U_\alpha\right)\cap Y\in \mathcal{T}_Y$ because $\bigcup_{\alpha \in I} U_\alpha\in \mathcal{T}$.
+
+Third, let $\{U_i\cap Y\}_{i=1}^n$ be a finite collection of open sets in $\mathcal{T}_Y$. Note that $U_i\in \mathcal{T}$ for all $i=1,2,\ldots,n$.
+So, $\bigcap_{i=1}^n U_i\cap Y=\left(\bigcap_{i=1}^n U_i\right)\cap Y\in \mathcal{T}_Y$ because $\bigcap_{i=1}^n U_i\in \mathcal{T}$.
+
+</details>
+
+### Generate basis for subspace topology
+
+Let $\mathcal{B}$ is a basis for $(X,\mathcal{T})$. We'd like to use $\mathcal{B}$ to define a basis for $(Y,\mathcal{T}_Y)$.
+
+$$
+\mathcal{B}_Y=\{B\cap Y|B\in \mathcal{B}\}
+$$
+
+#### Proposition for basis of subspace topology
+
+$\mathcal{B}_Y$ is a basis for a topology on $Y$ that generates the subspace topology on $Y$.
+
+Proof as exercise. (same as the proof for the basis of product topology)
+
+<details>
+<summary>Example: not every open set in subspace topology is open in the original space</summary>
+
+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 $\mathbb{R}$.
+
+</details>
+
+#### Lemma of open set in subspace topology
+
+Let $(X,\mathcal{T})$ be a topological space and $Y\subseteq X$ is open. $Z\subseteq Y$ is open subset with respect to the subspace topology on $Y$. Then $Z$ is open in $X$.

content/Math4201/_meta.js

@@ -7,4 +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_L6: "Topology I (Lecture 6)"
 }