diff --git a/content/Math4201/Math4201_L32.md b/content/Math4201/Math4201_L32.md index 6836ed6..c277935 100644 --- a/content/Math4201/Math4201_L32.md +++ b/content/Math4201/Math4201_L32.md @@ -105,7 +105,7 @@ Since $V$ is open in $Y$, then $K$ is closed in $Y$. Since $Y$ is compact, then Let $X$ be a topological space, then $X$ satisfies the first countability axiom if -For any $x\in X$, there is a countable collection $\{B_n}_n$ of open neighborhoods of $x$ such that any open neighborhood $U$ of $x$ contains one of $B_n$. +For any $x\in X$, there is a countable collection $\{B_n\}_n$ of open neighborhoods of $x$ such that any open neighborhood $U$ of $x$ contains one of $B_n$.
Example for metric space satisfies the first countability axiom diff --git a/content/Math4201/Math4201_L33.md b/content/Math4201/Math4201_L33.md new file mode 100644 index 0000000..8bd420a --- /dev/null +++ b/content/Math4201/Math4201_L33.md @@ -0,0 +1,184 @@ +# Math4201 Topology I (Lecture 33) + +## Countability axioms + +### First countability axiom + +For any $x\in X$, there is a countable collection $\{B_n\}_n$ of open neighborhoods of $x$ such that any open neighborhood $U$ of $x$ contains one of $B_n$. + +Example: Metric spaces + +### Second countability axiom + +There exists countable basis for a topology on $X$. + +Example: $\mathbb{R}$ and more generally $\mathbb{R}^n$ + +Consider the following topology on $\mathbb{R}^\omega$: + +There exists different topology can be defined on $\mathbb{R}^\omega$ + +- Product topology +- Box topology +- Union topology + +#### Definition of product topology + +Let $\{X_\alpha\}_{\alpha\in I}$ be a family of topological spaces: + +$$ +\prod_{\alpha\in I}X_\alpha=\{f:I\to \bigcup_{\alpha\in I} X_\alpha | \forall \alpha\in I, f(\alpha)\in X_\alpha\} +$$ + +The product topology defined on the basis that: + +The set of following forms: + +$$ +\mathcal{B}_{prod}=\{\prod_{\alpha\in I}U_\alpha| U_\alpha\subseteq X_\alpha\text{ is open and }U_\alpha=X_\alpha\text{ for all except finitely many }\alpha\} +$$ + +So $\mathbb{R}^\omega$ with product topology is second countable. + +
+Proof that product topology defined above is second countable +A countable basis for $\mathbb{R}^\omega$ is given by the union of following sets: + +$$ +B_1=\{(a_1,b_1)\times \mathbb{R}\times \mathbb{R}\times \dots |a_1,b_1\in \mathbb{Q}\}\\ +B_2=\{(a_1,b_1)\times(a_2,b_2) \mathbb{R}\times \dots |a_2,b_2\in \mathbb{Q}\}\\ +B_n=\{(a_1,b_1)\times(a_2,b_2)\times \dots (a_n,b_n)\times \mathbb{R}\times \dots |a_n,b_n\in \mathbb{Q}\}\\ +$$ + +Each $B_i$ is countable and there exists a bijection from $B_n\cong\mathbb{Q}^{2n}$ + +And the union of countably many countable sets is also countable. + +So $B_1\cup B_2\cup \dots$ is countable. This is also a baiss for the product topology on $\mathbb{R}^\omega$ + +
+ +#### Lemma of second countable spaces + +If $X_1,X_2,X_3,\dots$ are second countable topological spaces, then the following spaces are also second countable: + +1. $X_1\times X_2\times X_3\times \dots\times X_n$ (in this case, product topology = box topology) +2. $X_1\times X_2\times X_3\times \dots$ with the product topology + +
+Ideas for proof + +For $X_1\times X_2\times X_3\times \dots\times X_n$: + +$B_1\times B_2\times B_3\times \dots\times B_n$ is also countable basis for $X_1\times X_2\times X_3\times \dots\times X_n$ + +Let $\tilde{B}\coloneqq B_1\times B_2\times B_3\times \dots\times B_i\times X_{i+1}\times X_{i+2}\times \dots$ then $\tilde{B}$ is a countable basis for $X_1\times X_2\times X_3\times \dots$ + +
+ +#### Lemma of first countable spaces + +If $X_1,X_2,X_3,\dots$ are first countable topological spaces, then the following spaces are also first countable: + +1. $X_1\times X_2\times X_3\times \dots\times X_n$ (in this case, product topology = box topology) +2. $X_1\times X_2\times X_3\times \dots$ with the product topology + +
+Ideas for proof + +Basically the same as before but now you have analyze the basis for each $x\in X_1\times X_2\times X_3\times \dots$ + +
+ +#### Definition of box topology + +Let $\{X_\alpha\}_{\alpha\in I}$ be a family of topological spaces: + +$$ +\prod_{\alpha\in I}X_\alpha=\{f:I\to \bigcup_{\alpha\in I} X_\alpha | \forall \alpha\in I, f(\alpha)\in X_\alpha\} +$$ + +The product topology defined on the basis that: + +The set of following forms: + +$$ +\mathcal{B}_{box}=\{\prod_{\alpha\in I}U_\alpha| U_\alpha\subseteq X_\alpha\text{ is open}\} +$$ + + +#### Definition of uniform topology + +The uniform topology on $X$ is the topology induced by the uniform metric on $X$. + +$$ +\rho(x,y)=\sup_{i\in \mathbb{N}}\overline{d}(x_\alpha,y_\alpha) +$$ + +To get a finite number for $\rho(x,y)$, we define the bounded metric $\overline{d}$ on $X$ by $\overline{d}(x,y)=\min\{d(x,y),1\}$ where $d$ is the usual metric on $X$. + +where $x=(x_1,x_2,x_3,\dots), y=(y_1,y_2,y_3,\dots)\in X$ + +In particular, the $\mathbb{R}^\omega$ with the uniform topology is first countable because it's a metric space. + +However, it's not second countable. + +> Recall that $Y\subseteq X$ is a discrete subspace if the subspace topology on $Y$ is the discrete topology, i.e. any point of $y$ is open in $Y$. + +Define $A\subseteq \mathbb{R}^\omega$ be defined as follows: + +$$ +A=\{\underline{x}=(x_1,x_2,x_3,\dots)|x_i\in \{0,1\}\} +$$ + +Let $\underline{x} and \underline{x}'$ be two distinct elements of $A$. + +$$ +\rho(\underline{x},\underline{x}')=\sup_{i\in \mathbb{N}}\overline{d}(\underline{x}_\alpha,\underline{x}'_\alpha)=1 +$$ + +(since there exists at least one entry different in $\underline{x}$ and $\underline{x}'$) + +In particular, $B_1^\rho(\underline{x})\cap A=\{\underline{x}\}$, so $A$ is a discrete subspace of $\mathbb{R}^\omega$. + +This subspace is also uncountable ($A$ can create a surjective map to $(0,1)$ using binary representation) which implies that $\mathbb{R}^\omega$ is not second countable. + +#### Proposition of second countable spaces + +Let $X$ be a second countable topological space. Then the following holds: + +1. Any discrete subspace $Y$ of $X$ is countable +2. There exists a countable subset of $X$ that is dense in $X$ (_also called separable spaces_) +3. Every open covering of $X$ has **countable** subcover (That is if $X=\bigcup_{\alpha\in I} U_\alpha$, then there exists a **countable** subcover $\{U_{\alpha_1}, ..., U_{\alpha_\infty}\}$ of $X$) (_also called Lindelof spaces_) + +
+Proof + +First we prove that any discrete subspace $Y$ of $X$ is countable. + +Let $Y$ be a discrete subspace of $X$. In particular, for any $y\in Y$ we can find an element $B_y$ of the countable basis $\mathcal{B}$ for $Y$ such that $B_y\cap Y=\{y\}$. + +In particular, if $y\neq y'$, then $B_y\neq B_{y'}$. Because $\{y\}=B_y\cap Y\neq B_{y'}\cap Y=\{y'\}$. + +This shows that $\{B_y\}_{y\in Y}\subseteq B$ has the same number of elements as $Y$. + +So $Y$ has to be countable. + +--- + +Next we prove that there exists a countable subset of $X$ that is dense in $X$. + +For each basis element $B\in \mathcal{B}$, we can pick an element $x\in B$ and let $A$ be the union of all such $x$. + +We claim that $A$ is dense. + +To show that $A$ is dense, let $U$ be a non-empty open subset of $X$. + +Take an element $x\in U$. Note that by definition of basis, there is some element $B\in \mathcal{B}$ such that $x\in B$. So $x\in B\cap U$. $U\cap B\neq \emptyset$, so $A\cap U\neq \emptyset$. + +Since $A\cap U\neq \emptyset$ this shows that $A$ is dense. + +--- + +Third part next lecture. +
diff --git a/content/Math4201/_meta.js b/content/Math4201/_meta.js index 5bcaabe..b915733 100644 --- a/content/Math4201/_meta.js +++ b/content/Math4201/_meta.js @@ -36,5 +36,5 @@ export default { Math4201_L30: "Topology I (Lecture 30)", Math4201_L31: "Topology I (Lecture 31)", Math4201_L32: "Topology I (Lecture 32)", - + Math4201_L33: "Topology I (Lecture 33)", }