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content/Math4201/Math4201_L23.md

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+# Math4201 Topology I (Lecture 23)
+
+## Connectedness of topological spaces
+
+### Connected space
+
+#### Definition of connected space
+
+Let $X$ be a topological space. $X$ is separated if there exist two disjoint nonempty open subsets $U,V\subset X$ such that $X=U\cup V$.
+
+If $X$ is not separated, then $X$ is connected.
+
+#### Any interval in $\mathbb{R}$ with standard topology is connected
+
+Let $I=[a,b]$ be an interval in $\mathbb{R}$ with standard topology. Then $I$ is connected.
+
+<details>
+<summary>Proof</summary>
+
+By contradiction, we assume that $U,V$ give a separation of $\mathbb{R}$. In particular, $\exists a\in U, b\in V$.
+
+Let $a_0\coloneqq \sup\{x\in U\cap [a,b]\}$. Note that $a\in U\cap [a,b]$, so $a_0\geq a$. Since any element of $U\cap [a,b]$ is less than or equal to $b$, $a_0\leq b$.
+
+**Case 1**: $a_0=a$
+
+Since $U$ is open, there is $\epsilon>0$ such that $[a,a+\epsilon)\subset U\cap [a,b]$. So $a_0\geq a+\epsilon>a$, which contradicts the definition of $a_0$.
+
+**Case 2**: $a_0=b$.
+
+Since $V$ is open, there is $\epsilon>0$ such that $(b-\epsilon,b]\subseteq V\cap [a,b]$. This implies that $b-\epsilon$ is also an upper bound of $U\cap [a,b]$, so $a_0\leq b-\epsilon<b$, which contradicts the definition of $a_0$.
+
+**Case 3**: $a<a_0<b$.
+
+**Subcase I**: $a_0\in U$.
+
+There is an $\epsilon>0$ such that $(a_0-\epsilon,a_0+\epsilon)\subset U\cap [a,b]$ because $U$ is open. 
+
+In particular, $a_0$ is greater than any element of $(a_0-\epsilon,a_0+\epsilon)$, which contradicts the definition of $a_0$.
+
+**Subcase II**: $a_0\in V$.
+
+There is an $\epsilon>0$ such that $(a_0-\epsilon,a_0+\epsilon)\subset V\cap [a,b]$ because $V$ is open. 
+
+In particular, $a_0$ is less than any element of $(a_0-\epsilon,a_0+\epsilon)$. Since $a_0$ is an upper bound of $U\cap [a,b]$, any point $>a_0$ is not in $U\cap [a,b]$. 
+
+So, $U\cap [a,b]\subseteq [a,a_0-\epsilon)$. This shows that $a_0-\epsilon$ is an upper bound of $U\cap [a,b]$, which contradicts the definition of $a_0$.
+
+</details>
+
+_Intuitively, since both sets in $\mathbb{R}$ are open, you cannot set a clear boundary between the two sets by least upper bound argument._
+
+#### Corollary as Intermediate Value Theorem
+
+If $f:[a,b]\to \mathbb{R}$ is continuous, and $c\in\mathbb{R}$ is such that $f(a)<c<f(b)$, then there exists $x\in [a,b]$ such that $f(x)=c$.
+
+<details>
+<summary>Proof</summary>
+
+Since $[a,b]$ is connected, since $f$ is continuous, $f([a,b])$ is connected.
+
+By contradiction, if $c\notin f([a,b])$, then $f([a,b])$ has two points $f(a),f(b)$ and $c$ is a point between that isn't in $f([a,b])$. This contradicts the connectedness of $f([a,b])$.
+
+So $f(a)<c<f(b)$ or $f(b)<c<f(a)$ must hold.
+
+</details>
+
+#### Definition of path-connected space
+
+A topological space $X$ is path-connected if for any two points $x,x'\in X$, there is a continuous map $\gamma:[0,1]\to X$ such that $\gamma(0)=x$ and $\gamma(1)=x'$. Any such continuous map is called a path from $x$ to $x'$.
+
+> [!NOTE]
+>
+> Path-connectedness is a stronger condition than connectedness.
+
+#### Theorem of path-connectedness and connectedness
+
+Any path-connected space is connected.
+
+<details>
+<summary>Proof</summary>
+
+By contradiction, let $U,V$ be a separation of $X$. In particular, $\exists x\in U, x'\in V$.
+
+Since $X$ is path-connected, $\exists \gamma:[0,1]\to X$ such that $\gamma(0)=x$ and $\gamma(1)=x'$.
+
+Then since $\gamma$ is continuous, $\gamma^{-1}(U)$ and $\gamma^{-1}(V)$ are open in $[0,1]$ and $[0,1]=\gamma^{-1}(U)\cup \gamma^{-1}(V)$. We want to show that this gives a separation of $[0,1]$.
+
+Since $U\cap V=\emptyset$, $\gamma^{-1}(U)$ and $\gamma^{-1}(V)$ are disjoint.
+
+$U\cup V=X$ so $\gamma^{-1}(U)\cup \gamma^{-1}(V)=[0,1]$.
+
+Each of $\gamma^{-1}(U)$ and $\gamma^{-1}(V)$ is non-empty because $x\in U\implies 0\in \gamma^{-1}(U)$ and $x'\in V\implies 1\in \gamma^{-1}(V)$.
+
+This contradicts the assumption that $[0,1]$ is connected.
+
+</details>
+
+<details>
+<summary>Example of path-connected space</summary>
+
+A subspace $X$ of $\mathbb{R}^n$ is convex if for any two points $x,x'\in X$, the line segment connecting $x$ and $x'$ is contained in $X$.
+
+In particular $B_R(x)$ is convex. So $X$ is path-connected.
+
+---
+
+Let $X=\mathbb{R}^n\setminus\{0\}$. with $n\geq 2$. Then $X$ is path-connected. (simply walk around the origin)
+
+</details>
+
+#### Theorem for invariant property of connectedness
+
+If $f:X\to Y$ is a continuous and surjective map, and $X$ is connected, then $Y$ is connected.
+
+<details>
+<summary>Proof</summary>
+
+Take $y,y'\in Y$, since $f$ is surjective, $\exists x,x'\in X$ such that $f(x)=y$ and $f(x')=y'$. Let $\gamma:[0,1]\to X$ be a path from $x$ to $x'$.
+
+Then $f\circ \gamma:[0,1]\to Y$ is a continuous map. and $f\circ \gamma(0)=y$ and $f\circ \gamma(1)=y'$.
+
+</details>
+
+<details>
+<summary>Example of connected but not path-connected space</summary>
+
+Let $A=\{(x,y)\in \mathbb{R}^2\mid y=\sin(1/x), x>0\}$. Then $A$ is connected, and also path-connected.
+
+However, take $\overline{A}=A\cup \{0\}\times [-1,1]$. Then $\overline{A}$ is not path-connected but connected.
+
+_Show next time_
+
+</details>

content/Math4201/_meta.js

@@ -26,4 +26,5 @@ export default {
     Math4201_L20: "Topology I (Lecture 20)",
     Math4201_L21: "Topology I (Lecture 21)",
     Math4201_L22: "Topology I (Lecture 22)",
+    Math4201_L23: "Topology I (Lecture 23)",
 }

package.json

@@ -1,7 +1,7 @@
 {
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-    "build": "cross-env VERCEL_FORCE_NO_BUILD_CACHE=1 NODE_OPTIONS='--max-old-space-size=16384' next build",
+    "build": "cross-env VERCEL_FORCE_NO_BUILD_CACHE=1 NODE_OPTIONS='--max-old-space-size=32768' next build",
     "build:test": "cross-env ANALYZE=true NODE_OPTIONS='--inspect --max-old-space-size=4096' next build",
     "build:analyze": "cross-env ANALYZE=true NODE_OPTIONS='--max-old-space-size=16384' next build",
     "postbuild": "next-sitemap && pagefind --site .next/server/app --output-path public/_pagefind && cp -r ./public/_pagefind ./out",