updates

content/Math401/Extending_thesis/Math401_R2.md

@@ -252,19 +252,25 @@ $$
 
 Not very edible for undergraduates.
 
-## Crash course on Riemannian Geometry
+## Riemannian manifolds and geometry
 
 > This section is designed for stupids like me skipping too much essential materials in the book.
 
+> This part might be extended to a separate note, let's check how far we can go from this part.
+>
+> References:
+>
+> - [Riemannian Geometry by John M. Lee](https://www.amazon.com/Introduction-Riemannian-Manifolds-Graduate-Mathematics/dp/3319917544?dib=eyJ2IjoiMSJ9.88u0uIXulwPpi3IjFn9EdOviJvyuse9V5K5wZxQEd6Rto5sCIowzEJSstE0JtQDW.QeajvjQEbsDmnEMfPzaKrfVR9F5BtWE8wFscYjCAR24&dib_tag=se&keywords=riemannian+manifold+by+john+m+lee&qid=1753238983&sr=8-1)
+
 ### Manifold
 
-Unexpectedly, a good definition of the manifold is defined in the topology I.
-
-Check section 36. This topic extends to a wonderful chapter 8 in the book where you can hardly understand chapter 2.
+> Unexpectedly, a good definition of the manifold is defined in the topology I.
+>
+> Check section 36. This topic extends to a wonderful chapter 8 in the book where you can hardly understand chapter 2.
 
 #### Definition of m-manifold
 
-An $m$-manifold is a Hausdorff space $X$ with a countable basis such that each point of $x$ of $X$ has a neighborhood <text style="color: red;"> homeomorphic</text> to an open subset of $\mathbb{R}^m$.
+An $m$-manifold is a [Hausdorff space](../../Math4201/Math4201_L9#hausdorff-space) $X$ with a countable basis such that each point of $x$ of $X$ has a neighborhood [homeomorphic](../../Math4201/Math4201_L10#definition-of-homeomorphism) to an open subset of $\mathbb{R}^m$.
 
 Example is trivial that 1-manifold is a curve and 2-manifold is a surface.
 
@@ -274,17 +280,9 @@ If $X$ is a compact $m$-manifold, then $X$ can be imbedded in $\mathbb{R}^n$ for
 
 This theorem might save you from imagining abstract structures back to real dimension. Good news, at least you stay in some real numbers.
 
-### Riemannian manifold
+### Smooth manifold
 
-
-
-## Crash course on Riemannian manifolds
-
-> This part might be extended to a separate note, let's check how far we can go from this part.
->
-> References:
->
-> - [Riemannian Geometry by John M. Lee](https://www.amazon.com/Introduction-Riemannian-Manifolds-Graduate-Mathematics/dp/3319917544?dib=eyJ2IjoiMSJ9.88u0uIXulwPpi3IjFn9EdOviJvyuse9V5K5wZxQEd6Rto5sCIowzEJSstE0JtQDW.QeajvjQEbsDmnEMfPzaKrfVR9F5BtWE8wFscYjCAR24&dib_tag=se&keywords=riemannian+manifold+by+john+m+lee&qid=1753238983&sr=8-1)
+> This section is waiting for the completion of book Introduction to Smooth Manifolds by John M. Lee.
 
 ### Riemannian manifolds
 
@@ -296,7 +294,7 @@ An example of Riemannian manifold is the sphere $\mathbb{C}P^n$.
 
 A Riemannian metric is a smooth assignment of an inner product to each tangent space $T_pM$ of the manifold.
 
-An example of Riemannian metric is the Euclidean metric on $\mathbb{R}^n$.
+An example of Riemannian metric is the Euclidean metric, the bilinear form of $d(p,q)=\|p-q\|_2$ on $\mathbb{R}^n$.
 
 ### Notion of Connection
 
@@ -308,9 +306,12 @@ $$
 D_VX=\lim_{h\to 0}\frac{X(p+h)-X(p)}{h}
 $$
 
-### Nabla notation and Levi-Civita connection
+### Notion of Curvatures
 
+> [!NOTE]
+> 
+> Geometrically, the curvature of the manifold is radius of the tangent sphere of the manifold.
 
-### Ricci curvature
-
+#### Nabla notation and Levi-Civita connection
 
+#### Ricci curvature

content/Math4201/Math4201_L10.md

@@ -0,0 +1,124 @@
+# Math4201 Lecture 10
+
+## Continuity
+
+### Continuous functions
+
+Let $X,Y$ be topological spaces and $f:X\to Y$. For any $x\in X$ and any open neighborhood $V$ of $f(x)$ in $Y$, $f^{-1}(V)$ contains an open neighborhood of $x$ in $X$.
+
+
+#### Lemma for continuous functions
+
+Let $f:X\to Y$ be a function, then:
+
+1. $A\subseteq Y$: $f^{-1}(A^c) = (f^{-1}(A))^c$.
+2. $\{A_\alpha\}_{\alpha\in I}\subseteq Y$: $f^{-1}(\bigcup_{\alpha\in I} A_\alpha) = \bigcup_{\alpha\in I} f^{-1}(A_\alpha)$.
+3. $\{A_\alpha\}_{\alpha\in I}\subseteq Y$: $f^{-1}(\bigcap_{\alpha\in I} A_\alpha) = \bigcap_{\alpha\in I} f^{-1}(A_\alpha)$.
+
+<details>
+<summary>Proof</summary>
+
+1. By definition of continuous functions, $\forall V$ open in $Y$, $f^{-1}(V)$ is open in $X$.
+
+2. It is sufficient to shoa that $x\in f^{-1}(\bigcup_{\alpha\in I} A_\alpha)$ if and only if $x\in \bigcup_{\alpha\in I} f^{-1}(A_\alpha)$.
+
+This condition holds if and only if $\exists \alpha\in I$ such that $f(x)\in A_\alpha$.
+
+Which is equivalent to $\exists \alpha\in I$ such that $x\in f^{-1}(A_\alpha)$.
+
+So $x\in f^{-1}(\bigcup_{\alpha\in I} A_\alpha)$
+
+In particular, $f^{-1}(\bigcup_{\alpha\in I} A_\alpha) = \bigcup_{\alpha\in I} f^{-1}(A_\alpha)$.
+
+3. Similar to 2 but use forall.
+
+</details>
+
+#### Properties of continuous functions
+
+A function $f:X\to Y$ is continuous if and only if:
+
+1. $f^{-1}(V)$ is open in $X$ for any open set $V\subset Y$.
+2. $f$ is continuous at any point $x\in X$.
+3. $f^{-1}(C)$ is closed in $X$ for any closed set $C\subset Y$.
+4. Assume $\mathcal{B}$ is a basis for $Y$, then $f^{-1}(\mathcal{B})$ is open in $X$ for any $B\in \mathcal{B}$.
+5. For any $A\subseteq X$, $f(\overline{A})\subseteq \overline{f(A)}$.
+
+<details>
+<summary>Proof</summary>
+
+**Showing $1\iff 3$**:
+
+> Use the lemma for continuous functions (1)
+
+**Showing $1\iff 4$**:
+
+$1 \implies 4$: 
+
+Because any $B\in \mathcal{B}$ is open in $Y$, so $f^{-1}(B)$ is open in $X$.
+
+$4 \implies 1$:
+
+Let $V\subset Y$ be an open set. Then there are basis elements $\{B_\alpha\}_{\alpha\in I}$ such that $V=\bigcup_{\alpha\in I} B_\alpha$.
+
+So $f^{-1}(V) = f^{-1}(\bigcup_{\alpha\in I} B_\alpha) = \bigcup_{\alpha\in I} f^{-1}(B_\alpha)$ (by lemma (2)) is a union of open sets, so $f^{-1}(V)$ is open in $X$.
+
+**Showing $1\implies 5$**:
+
+Take $A\subseteq X$ and $x\in \overline{A}$. It suffices to show $f(x)$ is an element of the closure of $f(A)$. This is equivalent to say that any open neighborhood $V$ of $f(x)$ intersects $f(A)$ has a non-trivial intersection with $f(A)$.
+
+For any such $V$, 1 implies that $f^{-1}(V)$ is open in $X$. Moreover, $x\in f^{-1}(V)$ because $f(x)\in V$.
+
+This means that $f^{-1}(V)$ is an open neighborhood of $x$. Since $x\in \overline{A}$, we have $f^{-1}(V)\cap A\neq \emptyset$ and contains a point $x'\in X$.
+
+So $x'\in f^{-1}(V)\cap A$, this implies that $f(x')\in V$ and $f(x')\in f(A)$, so $f(x')\in V\cap f(A)$.
+
+> [!NOTE]
+>
+> This verifies our claim. Proof of $5\implies 1$ is similar and left as an exercise.
+
+</details>
+
+<details>
+<summary>Example of property 5</summary>
+
+Let $X=(0,1)\cup (1,2)$ and $Y=\mathbb{R}$ equipped with the subspace topology induced by the standard topology on $\mathbb{R}$.
+
+Let $f:X\to Y$ be the inclusion map, $f(x)=x$ for all $x\in X$. This is continuous.
+
+Let $A=(0,1)\cup (1,2)$. Then $\overline{A}=A$. So $f(\overline{A})=f(A)=(0,1)\cup (1,2)$.
+
+However, $\overline{f(A)}=\overline{(0,1)\cup (1,2)}=[0,2]$.
+
+So $f(\overline{A})\subsetneq \overline{f(A)}$.
+
+</details>
+
+#### Definition of homeomorphism
+
+A **homeomorphism** $f:X\to Y$ is a continuous map of topological spaces that is a bijection and $f^{-1}:Y\to X$ is also continuous.
+
+<details>
+<summary>Example of homeomorphism</summary>
+
+Let $X=\mathbb{R}$ and $Y=\mathbb{R}+$ with standard topology.
+
+$f:\mathbb{R}\to \mathbb{R}^+$ be defined by $f(x)=e^x$ is continuous and bijective.
+
+$f^{-1}:\mathbb{R}^+\to \mathbb{R}$ be defined by $f^{-1}(y)=\ln(y)$ is continuous and homeomorphism.
+
+</details>
+
+### Epsilon delta definition of continuity
+
+Let $f:\mathbb{R}\to \mathbb{R}$ be a continuous function where we use the standard topology on $\mathbb{R}$.
+
+Then [property 4](#properties-of-continuous-functions) implies that for any open interval $(a,b)\in \mathbb{R}$, $f^{-1}((a,b))$ is open in $\mathbb{R}$.
+
+Now take an arbitrary $x\in \mathbb{R}$ and $\epsilon > 0$. In particular $f^{-1}((f(x)-\epsilon, f(x)+\epsilon))$ is an open set containing $x$.
+
+In particular, there is an open interval (by the standard topology on $\mathbb{R}$) $(c,d)$ such that $x\in (c,d)\subseteq f^{-1}((f(x)-\epsilon, f(x)+\epsilon))$.
+
+Let $\delta = \min\{x-c, d-x\}$. Then $(x-\delta, x+\delta)\subseteq (c,d)\subseteq f^{-1}((f(x)-\epsilon, f(x)+\epsilon))$.
+
+This says that if $|y-x| < \delta$, then $|f(y)-f(x)| < \epsilon$.