From 91f8359a5a4d75d44bb6de4053c3ee2effd0bd36 Mon Sep 17 00:00:00 2001 From: Trance-0 <60459821+Trance-0@users.noreply.github.com> Date: Mon, 3 Nov 2025 11:53:32 -0600 Subject: [PATCH] updates --- content/Math4201/Math4201_L28.md | 144 +++++++++++++++++++++++++++++++ content/Math4201/_meta.js | 1 + 2 files changed, 145 insertions(+) create mode 100644 content/Math4201/Math4201_L28.md diff --git a/content/Math4201/Math4201_L28.md b/content/Math4201/Math4201_L28.md new file mode 100644 index 0000000..cebe218 --- /dev/null +++ b/content/Math4201/Math4201_L28.md @@ -0,0 +1,144 @@ +# Math4201 Topology I (Lecture 28) + +## Compact spaces + +### Extreme value theorem + +#### Definition of diameter + +Let $(X,d)$ be a metric space and $A\subseteq X$. The diameter of $A$ is defined as +$$ +\operatorname{diam}(A) = \sup\{d(x,y):x,y\in A\} +$$ + +#### Lebesgue number lemma + +Let $X$ be a compact metric space and $\{U_\alpha\}_{\alpha\in I}$ be an open cover of $X$. Then there is $\delta>0$ such that for every subset $A\subseteq X$ with diameter less than $\delta$, there is $\alpha\in I$ such that $A\subseteq U_\alpha$. + +
+Proof + +Consider $x\in X$, there is an element $U_\alpha$ in the open covering such that $x\in U_\alpha$. + +In particular, there is $r_x$ such that $B_{r_x}(x)\subseteq U_\alpha$. + +Then the collection $\{B_{\frac{r_x}{2}}(x)\}_{x\in X}$ is an open covering of $X$. (each $x\in X$ is contained in some $B_{\frac{r_x}{2}}(x)$) + +Since $X$ is compact, there is a finite subcover $\{B_{\frac{r_{x_i}}{2}}(x_i)\}_{i=1}^n$ of $X$. Such that $\bigcup_{i=1}^n B_{\frac{r_{x_i}}{2}}(x_i)=X$. + +Let $\delta = \min\{r_{\frac{r_{x_1}}{2}}, ..., r_{\frac{r_{x_n}}{2}}\}>0$. + +Let $A\subseteq X$ be a subset with diameter less than $\delta$. + +Take $y\in A$, then $A\subseteq B_\delta(y)$. + +Take $x_i$ such that $y\in B_{\frac{r_{x_i}}{2}}(x_i)$. (such cover exists by definition of the subcover) + +And then $\alpha$ such that $B_{\frac{r_{x_i}}{2}}(x_i)\subseteq U_\alpha$. + +We claim that $B_\delta(y)\subseteq U_\alpha$, which would imply that $A\subseteq U_\alpha$. + +$y\in B_{\frac{r_{x_i}}{2}}(x_i)$, and we know that $B_{r_{x_i}}(x_i)\subseteq U_\alpha$. + +Since $\delta < \frac{r_{x_i}}{2}$, it suffices to show that $B_{\frac{r_{x_i}}{2}}(y)\subseteq U_\alpha$. + +For any $z\in B_{\frac{r_{x_i}}{2}}(y)$, we have $d(z,y)<\frac{r_{x_i}}{2}$, and $d(y,x_i)<\frac{r_{x_i}}{2}$. + +So $d(z,x_i)\leq d(z,y)+d(y,x_i)<\frac{r_{x_i}}{2}+\frac{r_{x_i}}{2}=r_{x_i}$, so $z\in B_{r_{x_i}}(x_i)\subseteq U_\alpha$. + +So $B_{\frac{r_{x_i}}{2}}(y)\subseteq U_\alpha$. + +
+ +#### Definition of finite intersection property + +A collection $\{C_\alpha\}_{\alpha\in I}$ of subsets of a set $X$ has finite intersection property if for every finite subcollection $\{C_{\alpha_1}, ..., C_{\alpha_n}\}$ of $\{C_\alpha\}_{\alpha\in I}$, we have $\bigcap_{i=1}^n C_{\alpha_i}\neq \emptyset$. + +#### Theorem + +A space $X$ is compact if and only if every collection $\{Z_\alpha\}_{\alpha\in I}$ of closed subsets of $X$ satisfies the **finite** intersection property has a non-empty intersection. + +$$ +\bigcap_{\alpha\in I} Z_\alpha \neq \emptyset +$$ + +
+Non-example + +Consider $X=(0,1)$ is not compact with the standard topology. + +Consider $Z_n=(0,\frac{1}{n}]$, each interval is closed in $X$. This satisfies the finite intersection property because $\bigcap_{i=1}^k Z_{n_i}\neq \emptyset$ for any finite subcollection $\{Z_{n_1}, ..., Z_{n_k}\}$. _We can find a smaller for any finite subcollection to get a non-empty intersection._ + +But $\bigcap_{n=1}^\infty Z_n = \emptyset$. + +
+ +
+Proof + +$\implies$ + +Let $U_\alpha=X-Z_\alpha$ is open for each $\alpha\in I$. By contradiction, suppose that $\bigcap_{\alpha\in I} Z_\alpha = \emptyset$. + +So $X-\bigcap_{\alpha\in I} Z_\alpha = X=\bigcup_{\alpha\in I} U_\alpha = \bigcup_{\alpha\in I} (X-Z_\alpha)=X$. + +So $\{U_\alpha\}_{\alpha\in I}$ is an open cover of $X$. Since $X$ is compact, there is a finite subcover $\{U_{\alpha_1}, ..., U_{\alpha_n}\}$. + +So $\bigcap_{i=1}^n U_{\alpha_i} = X-\bigcup_{i=1}^n Z_{\alpha_i} = X$, $\bigcap_{i=1}^n Z_{\alpha_i} = \emptyset$. This contradicts the finite intersection property. + +$\impliedby$ + +Proof is similar. + +
+ +#### Definition of isolated point + +A point $x\in X$ is an isolated point if $\{x\}$ is an open subset of $X$. + +
+Example of isolated point + +$X=[0,1]\cup \{2\}$ with subspace topology from $\mathbb{R}$. + +Then $\{2\}$ is an isolated point $\{2\}=X\cap (2-\frac{1}{2}, 2+\frac{1}{2})$. + +
+ +#### Theorem of compact Hausdorff spaces without isolated points + +Any non-empty compact Hausdorff space without an isolated point is uncountable. + +
+Proof + +Proof by contradiction. + +Let $X=\{x_n\}_{n\in\mathbb{N}}$ be a countable set. + +Since $x_1$ is not an isolated point, so there exists $y_1\in X$ such that $y_1\neq x_1$. Apply the Hausdorff property, there exists disjoint open neighborhoods $U_1$ and $V_1$ such that $x_1\in U_1$ and $y_1\in V_1$. + +In particular $\overline{V_1}$ does not contain $x_1$, but it contains $y_1$. (Follows from disjoint open neighborhoods) + +Since $x_2$ is not an isolated point, so there exists $y_2\in X$ such that $y_2\neq x_2$. Apply the Hausdorff property, there exists disjoint open neighborhoods $U_2$ and $V_2$ such that $x_2\in U_2$ and $y_2\in V_2$. + +If $x_2\notin V_1$, then we define $V_2$ as $V_1$. + +If $x_2\in V_1$, then by the assumption, there is another point $y_2$ in $V_1$ which isn't the same as $x_2$. + +CONTINUE NEXT TIME. + +
+ +#### Theorem real numbers is uncountable + +$\mathbb{R}$ is uncountable, and any interval in $\mathbb{R}$ is uncountable. + +
+Proof + +It suffices to prove this for a closed interval $[a,b]$ with $a \ No newline at end of file diff --git a/content/Math4201/_meta.js b/content/Math4201/_meta.js index 6186b48..47a0510 100644 --- a/content/Math4201/_meta.js +++ b/content/Math4201/_meta.js @@ -31,4 +31,5 @@ export default { Math4201_L25: "Topology I (Lecture 25)", Math4201_L26: "Topology I (Lecture 26)", Math4201_L27: "Topology I (Lecture 27)", + Math4201_L28: "Topology I (Lecture 28)", }