diff --git a/content/Math4201/Math4201_L3.md b/content/Math4201/Math4201_L3.md new file mode 100644 index 0000000..a042d8c --- /dev/null +++ b/content/Math4201/Math4201_L3.md @@ -0,0 +1,164 @@ +# Math4201 Lecture 3 + +## Recall form last lecture + +### Topological Spaces + +#### Basis for a topology + +Let $X$ be a set. A basis for a topology on $X$ is a collection $\mathcal{B}$ (elements of $\mathcal{B}$ are called basis elements) of subsets of $X$ such that: + +1. $\forall x\in X$, $\exists B\in \mathcal{B}$ such that $x\in B$ +2. $\forall B_1,B_2\in \mathcal{B}$, $\forall x\in B_1\cap B_2$, $\exists B_3\in \mathcal{B}$ such that $x\in B_3\subseteq B_1\cap B_2$ + +
+Example 1 + +Let $X=\mathbb{R}$ and $\mathcal{B}=\{(a,b)|a,b\in \mathbb{R},a + +
+Example 2 + +Let $X=\mathbb{R}$ and $\mathcal{B}_{LL}=\{[a,b)|a,b\in \mathbb{R},a + +Extend this to $\mathbb{R}^2$. + +#### Definition for cartesian product + +Let $X$ and $Y$ be sets. The cartesian product of $X$ and $Y$ is the set $X\times Y=\{(x,y)|x\in X,y\in Y\}$. + +
+Example 3 + +Let $X=\mathbb{R}^2$ and $\mathcal{B}$ be the collection of rectangle of the form $(a,b)\times (c,d)$ where $a,b,c,d\in \mathbb{R}$ and $a + +
+Example 4 + +Let $X=\mathbb{R}^2$ and $\mathcal{B}$ be the collection of open disks. + +Check properties 1: + +for any $x\in \mathbb{R}^2$, $\exists B_1(x)\in \mathcal{B}$ such that $x\in B_1(x)$. + +Check properties 2: + +let $B_{r_1}(x)$ and $B_{r_2}(y)$ be two basis elements, for every $z\in B_{r_1}(x)\cap B_{r_2}(y)$, $\exists B_{r_3}(z)\in \mathcal{B}$ such that $z\in B_{r_3}(z)\subseteq B_{r_1}(x)\cap B_{r_2}(y)$. + +(even $B_{r_1}(x)\cap B_{r_2}(y)\notin \mathcal{B}$) +
+ +#### Topology generated by a basis + +Let $\mathcal{B}$ be a basis for a topology on $X$. The topology generated by $\mathcal{B}$, denoted by $\mathcal{T}_{\mathcal{B}}$. + +$U\in \mathcal{T}_{\mathcal{B}}\iff \forall x\in U, \exists B\in \mathcal{B}$ such that $x\in B\subseteq U$ + +
+Proof + +$\mathcal{T}_{\mathcal{B}}$ is a topology on $X$ because: + +1. $\emptyset \in \mathcal{T}_{\mathcal{B}}$ because $\emptyset \in \mathcal{B}$. $X\in \mathcal{T}_{\mathcal{B}}$ because $\forall x\in X, \exists B\in \mathcal{B}$ such that $x\in B\subseteq X$ (by definition of basis (property 1))) + +2. $\mathcal{T}_{\mathcal{B}}$ is closed under arbitrary unions. + + Want to show $\{U_\alpha | U_\alpha\in \mathcal{T}_{\mathcal{B}}\}_{\alpha \in I}\implies \bigcup_{\alpha \in I} U_\alpha\in \mathcal{T}_{\mathcal{B}}$. + + Because $\forall x\in \bigcup_{\alpha \in I} U_\alpha$, $\exists \alpha_0$ such that $x\in U_{\alpha_0}$. Since $U_{\alpha_0}\in \mathcal{T}_{\mathcal{B}}$, $\exists B\in \mathcal{B}$ such that $x\in B\subseteq U_{\alpha_0}\subseteq \bigcup_{\alpha \in I} U_\alpha$. + +3. $\mathcal{T}_{\mathcal{B}}$ is closed under finite intersections. + + Want to show $U_1,U_2,\ldots,U_n\in \mathcal{T}_{\mathcal{B}}\implies \bigcap_{i=1}^n U_i\in \mathcal{T}_{\mathcal{B}}$. + + If $n=2$, since $\forall x\in U_1\cap U_2$, $x\in U_1$ and $x\in U_2$, $\exists B_1\in \mathcal{B}$ such that $x\in B_1\subseteq U_1$ and $\exists B_2\in \mathcal{B}$. + + Applying the second property of basis, $\exists B_3\in \mathcal{B}$ such that $x\in B_3\subseteq B_1\cap B_2\subseteq U_1\cap U_2$. + + By induction, we can show that $\bigcap_{i=1}^n U_i\in \mathcal{T}_{\mathcal{B}}$. + +
+ +
+Example of topology generated by a basis + +Let $X$ be arbitrary. + +Let $\mathcal{B}=\{x|x\in X\}$ (collection of all singleton subsets of $X$). + +Then $\mathcal{T}$ is the discrete topology. + +
+ +#### Properties of basis in generated topology + +**Observation 1**: Any $B\in \mathcal{B}$ is an open set in $\mathcal{T}_{\mathcal{B}}$. + +By the defining property of basis, $\forall x\in B$, $\exists x\in B\subseteq B$. + +**Observation 2**: For any collection $\{B_\alpha\}_{\alpha \in I}$, $\bigcup_{\alpha \in I} B_\alpha\in \mathcal{T}_{\mathcal{B}}$. + +By observation 1, each $B_\alpha\in \mathcal{T}_{\mathcal{B}}$. Since $\mathcal{T}_{\mathcal{B}}$ is a topology, $\bigcup_{\alpha \in I} B_\alpha\in \mathcal{T}_{\mathcal{B}}$. + +#### Lemma + +Let $\mathcal{B}$ and $\mathcal{T}_{\mathcal{B}}$ be a basis and the topology generated by $\mathcal{B}$ on $X$. Then, + +$U\in \mathcal{T}_{\mathcal{B}}\iff$ there are basis elements $\{B_\alpha\}_{\alpha \in I}$ such that $U=\bigcup_{\alpha \in I} B_\alpha$. + +
+Proof + +$(\Rightarrow)$ + +If $U\in \mathcal{T}_{\mathcal{B}}$, we want to show that $U$ is a union of basis elements. + +For any $x\in U$, by the definition of $\mathcal{T}_{\mathcal{B}}$, there is a basis element $B_x$ such that $x\in B_x\subseteq U$. + +So, $U\subseteq \bigcup_{x\in U} B_x$. + +Since $\forall B_x\in \{B_x\}_{\alpha \in I}$, $B_x\subseteq U$, we have $U\supseteq\bigcup_{x\in U} B_x$. + +So, $U=\bigcup_{x\in U} B_x$. + +$(\Leftarrow)$ + +Applies observation 2. + +
+ +> [!NOTE] +> +> A basis for a topology is like a basis for a vector space in the sense that any open set/vector can be represented in terms of basis elements. +> +> But unlike linear algebra, it's not true that any open set can be written as a union of basis element in a **unique** way. diff --git a/content/Math4201/_meta.js b/content/Math4201/_meta.js index f10a8d1..ea5889e 100644 --- a/content/Math4201/_meta.js +++ b/content/Math4201/_meta.js @@ -5,4 +5,5 @@ export default { }, Math4201_L1: "Topology I (Lecture 1)", Math4201_L2: "Topology I (Lecture 2)", + Math4201_L3: "Topology I (Lecture 3)", }