updates?

content/Math4201/Exam_reviews/Math4201_E1.md

@@ -1,4 +1,4 @@
-# Math 4201 Exam 1 review
+# Math 4201 Exam 1 Review
 
 > [!NOTE]
 >
@@ -343,5 +343,3 @@ $\sim$ is a subset of $X\times X$ with the following properties:
 3. If $(x,y)\in \sim$ and $(y,z)\in \sim$, then $(x,z)\in \sim$.
 
 The equivalence classes of $x\in X$ is denoted by $[x]=\{y\in X|y\sim x\}$.
-
-

content/Math4201/Exam_reviews/Math4201_E2.md

@@ -0,0 +1,194 @@
+# Math 4201 Exam 2 Review
+
+> [!NOTE]
+>
+> This is a review for definitions we covered in the classes. It may serve as a cheat sheet for the exam if you are allowed to use it.
+
+## Connectedness and compactness of metric spaces
+
+### Connectedness and separation
+
+#### Definition of separation
+
+Let $X=(X,\mathcal{T})$ be a topological space. A separation of $X$ is a pair of open sets $U,V\in \mathcal{T}$ that:
+
+1. $U\neq \emptyset$ and $V\neq \emptyset$ (that also equivalent to $U\neq X$ and $V\neq X$)
+2. $U\cap V=\emptyset$
+3. $X=U\cup V$ ($\forall x\in X$, $x\in U$ or $x\in V$)
+
+Some interesting corollary:
+
+- Any non-trivial (not $\emptyset$ or $X$) clopen set can create a separation.
+  - Proof: Let $U$ be a non-trivial clopen set. Then $U$ and $U^c$ are disjoint open sets whose union is $X$.
+- For subspace $Y\subset X$, a separation of $Y$ is a pair of open sets $U,V\in \mathcal{T}_Y$ such that:
+  1. $U\neq \emptyset$ and $V\neq \emptyset$ (that also equivalent to $U\neq Y$ and $V\neq Y$)
+  2. $U\cap V=\emptyset$
+  3. $Y=U\cup V$ ($\forall y\in Y$, $y\in U$ or $y\in V$)
+   - If $\overline{A}$ is closure of $A$ in $X$, same for $\overline{B}$, then the closure of $A$ in $Y$ is $\overline{A}\cap Y$ and the closure of $B$ in $Y$ is $\overline{B}\cap Y$. Then for separation $U,V$ of $Y$, $\overline{A}\cap B=A\cap \overline{B}=\emptyset$.
+
+#### Definition of connectedness
+
+A topological space $X$ is connected if there is no separation of $X$.
+
+> [!TIP]
+>
+> Connectedness is a local property. (That is, even the big space is connected, the subspace may not be connected. Consider $\mathbb{R}$ with the usual metric. $\mathbb{R}$ is connected, but $\mathbb{R}\setminus\{0\}$ is not connected.)
+>
+> Connectedness is a topological property. (That is, if $X$ and $Y$ are homeomorphic, then $X$ is connected if and only if $Y$ is connected. Consider if not, then separation of $X$ gives a separation of $Y$.)
+
+#### Lemma of connected subspace
+
+If $A,B$ is a separation of a topological space $X$, and $Y\subseteq X$ is a **connected** subspace with subspace topology, then $Y$ is either contained in $A$ or $B$.
+
+_Easy to prove by contradiction. Try to construct a separation of $Y$._
+
+#### Theorem of connectedness of union of connected subsets
+
+Let $\{A_\alpha\}_{\alpha\in I}$ be a collection of connected subsets of a topological space $X$ such that $\bigcap_{\alpha\in I} A_\alpha$ is non-empty. Then $\bigcup_{\alpha\in I} A_\alpha$ is connected.
+
+_Easy to prove by lemma of connected subspace._
+
+#### Lemma of compressing connectedness
+
+Let $A\subseteq X$ be a connected subspace of a topological space $X$ and $A\subseteq B\subseteq \overline{A}$. Then $B$ is connected.
+
+_Easy to prove by lemma of connected subspace. Suppose $C,D$ is a separation of $B$, then $A$ lies completely in either $C$ or $D$. Without loss of generality, assume $A\subseteq C$. Then $\overline{A}\subseteq\overline{C}$ and $\overline{A}\cap D=\emptyset$ (from $\overline{C}\cap D=\emptyset$ by closure of $A$). (contradiction that $D$ is nonempty) So $D$ is disjoint from $\overline{A}$, and hence from $B$. Therefore, $B$ is connected._
+
+#### Theorem of connected product space
+
+Any finite cartesian product of connected spaces is connected.
+
+_Prove using the union of connected subsets theorem. Using fiber bundle like structure union with non-empty intersection._
+
+### Application of connectedness in real numbers
+
+Real numbers are connected.
+
+Using the least upper bound and greatest lower bound property, we can prove that any interval in real numbers is connected.
+
+#### Intermediate Value Theorem
+
+Let $f:[a,b]\to \mathbb{R}$ be continuous. If $c\in\mathbb{R}$ is such that $f(a)<c<f(b)$, then there exists $x\in [a,b]$ such that $f(x)=c$.
+
+_If false, then we can use the disjoint interval with projective map to create a separation of $[a,b]$._
+
+#### 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 $x'$.
+
+- Every connected space is path-connected.
+  - The converse may not be true, consider the topologists' sine curve.
+
+### Compactness
+
+#### Definition of compactness via open cover and finite subcover
+
+Let $X=(X,\mathcal{T})$ be a topological space. An open cover of $X$ is $\mathcal{A}\subset \mathcal{T}$ such that $X=\bigcup_{A\in \mathcal{A}} A$. A finite subcover of $\mathcal{A}$ is a finite subset of $\mathcal{A}$ that covers $X$.
+
+$X$ is compact if every open cover of $X$ has a finite subcover (i.e. $X=\bigcup_{A\in \mathcal{A}} A\implies \exists \mathcal{A}'\subset \mathcal{A}$ finite such that $X=\bigcup_{A\in \mathcal{A}'} A$).
+
+#### Definition of compactness via 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 $\bigcap_{i=1}^n C_{\alpha_i}\neq \emptyset$.
+
+Let $X=(X,\mathcal{T})$ be a topological space. $X$ is compact if every collection $\{Z_\alpha\}_{\alpha\in I}$ of closed subsets of $X$ satisfies the finite intersection property has a non-empty intersection (i.e. $\forall \{Z_{\alpha_1}, ..., Z_{\alpha_n}\}\subset \{Z_\alpha\}_{\alpha\in I}, \bigcap_{i=1}^n Z_{\alpha_i} \neq \emptyset\implies \bigcap_{\alpha\in I} Z_\alpha \neq \emptyset$).
+
+#### Compactness is a local property
+
+Let $X$ be a topological space. A subset $Y\subseteq X$ is compact if and only if every open covering of $Y$ (set open in $X$) has a finite subcovering of $Y$.
+
+- A space $X$ is compact but the subspace may not be compact.
+  - Consider $X=[0,1]$ and $Y=[0,1/2)$. $Y$ is not compact because the open cover $\{(0,1/n):n\in \mathbb{N}\}$ does not have a finite subcover.
+- A compact subspace may live in a space that is not compact.
+  - Consider $X=\mathbb{R}$ and $Y=[0,1]$. $Y$ is compact but $X$ is not compact.
+
+#### Closed subspaces of compact spaces
+
+A closed subspace of a compact space is compact.
+
+A compact subspace of Hausdorff space is closed.
+
+_Each point not in the closed set have disjoint open neighborhoods with the closed set in Hausdorff space._
+
+#### Theorem of compact subspaces with Hausdorff property
+
+If $Y$ is compact subspace of a **Hausdorff space** $X$, $x_0\in X-Y$, then there are disjoint open neighborhoods $U,V\subseteq X$ such that $x_0\in U$ and $Y\subseteq V$.
+
+#### Image of compact space under continuous map is compact
+
+Let $f:X\to Y$ be a continuous map and $X$ is compact. Then $f(X)$ is compact.
+
+#### Tube lemma
+
+Let $X,Y$ be topological spaces and $Y$ is compact. Let $N\subseteq X\times Y$ be an open set contains $X\times \{y_0\}$ for $y_0\in Y$. Then there exists an open set $W\subseteq Y$ is open containing $y_0$ such that $N$ contains $X\times W$.
+
+_Apply the finite intersection property of open sets in $X\times Y$. Projection map is continuous._
+
+#### Product of compact spaces is compact
+
+Let $X,Y$ be compact spaces, then $X\times Y$ is compact.
+
+Any finite product of compact spaces is compact.
+
+### Compact subspaces of real numbers
+
+#### Every closed and bounded subset of real numbers is compact
+
+$[a,b]$ is compact in $\mathbb{R}$ with standard topology.
+
+#### Good news for real numbers
+
+Any of the three properties is equivalent for subsets of real numbers (product of real numbers):
+
+1. $A\subseteq \mathbb{R}^n$ is closed and bounded (with respect to the standard metric or spherical metric on $\mathbb{R}^n$).
+2. $A\subseteq \mathbb{R}^n$ is compact.
+
+#### Extreme value theorem
+
+If $f:X\to \mathbb{R}$ is continuous map with $X$ being compact. Then $f$ attains its minimum and maximum. (there exists $x_m,x_M\in X$ such that $f(x_m)\leq f(x)\leq f(x_M)$ for all $x\in X$)
+
+#### Lebesgue number lemma
+
+For a compact metric space $(X,d)$ and an open covering $\{U_\alpha\}_{\alpha\in I}$ 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 $A\subseteq U_\alpha$.
+
+_Apply the extreme value theorem over the mapping of the averaging function for distance of points to the $X-U_\alpha$. Find minimum radius of balls that have some $U_\alpha$ containing the ball._
+
+#### Definition for uniform continuous function
+
+$f$ is uniformly continuous if for any $\epsilon > 0$, there exists $\delta > 0$ such that for any $x_1,x_2\in X$, if $d(x_1,x_2)<\delta$, then $d(f(x_1),f(x_2))<\epsilon$.
+
+#### Theorem of uniform continuous function
+
+Let $f:X\to Y$ be a continuous map between two metric spaces. If $X$ is compact, then $f$ is uniformly continuous.
+
+#### Definition of isolated point
+
+A point $x\in X$ is an isolated point if $\{x\}$ is an open subset of $X$.
+
+#### Theorem of isolated point in compact spaces
+
+Let $X$ be a nonempty compact Hausdorff space. If $X$ has no isolated points, then $X$ is uncountable.
+
+_Proof using infinite nested closed intervals should be nonempty._
+
+### Variation of compactness
+
+#### Limit point compactness
+
+A topological space $X$ is limit point compact if every infinite subset of $X$ has a limit point in $X$.
+
+- Every compact space is limit point compact.
+
+#### Sequentially compact
+
+A topological space $X$ is sequentially compact if every sequence in $X$ has a convergent subsequence.
+
+- Every compact space is sequentially compact.
+
+#### Equivalence of three in metrizable spaces
+
+If $X$ is a metrizable space, then the following are equivalent:
+
+1. $X$ is compact.
+2. $X$ is limit point compact.
+3. $X$ is sequentially compact.

content/Math4201/Math4201_L29.md

@@ -51,7 +51,9 @@ Therefore, $X$ is uncountable.
 
 #### Definition of limit point compact
 
-A space $X$ is limit point compact if any infinite subset of $X$ has a limit point.
+A space $X$ is limit point compact if any infinite subset of $X$ has a [limit point](./Math4201_L8#limit-points) in $X$.
+
+_That is, $\forall A\subseteq X$ and $A$ is infinite, there exists a point $x\in X$ such that $x\in U$, $\forall U\in \mathcal{T}$ containing $x$, $(U-\{x\})\cap A\neq \emptyset$._
 
 #### Definition of sequentially compact
 
@@ -59,7 +61,7 @@ A space $X$ is sequentially compact if any sequence has a convergent subsequence
 
 #### Theorem of limit point compact spaces
 
-If $(X,d)$ is a metric space, then the following are equivalent:
+If $(X,d)$ is a **metric space**, then the following are equivalent:
 
 1. $X$ is compact.
 2. $X$ is limit point compact.
@@ -85,6 +87,8 @@ $X$ is not sequentially compact because the sequence $\{(n,a)\}_{n\in\mathbb{N}}
 
 First, we show that 1. implies 2.
 
+We proceed by contradiction.
+
 Let $X$ be compact and $A\subseteq X$ be an infinite subset of $X$ that doesn't have any limit points.
 
 Then $X-A$ is open because any $x\in X-A$ isn't in the closure of $A$ otherwise it would be a limit point for $A$, and hence $x$ has an open neighborhood contained in the complement of $A$.
@@ -114,5 +118,22 @@ Continue with the proof that 2. implies 3. next time.
 
 Proof of 1. follows from the theorem of limit point compact spaces.
 
-
 </details>
+
+That means, sequentially compact is a stronger property than limit point compact, and compact is the stronger property than limit point compact.
+
+> [!WARNING]
+>
+> Hope you will not use it soon for your exams but here are some interesting examples.
+>
+> **There exists spaces that are sequentially compact but not compact.**
+>
+> Consider the interval $[0,1)$ with the standard topology over $\mathbb{R}$. This space is sequentially compact but not compact.
+>
+> [S000035](https://topology.pi-base.org/spaces/S000035)
+>
+> **There exists spaces that are compact but not sequentially compact.**
+>
+> Consider the space of functions $f:[0,1]\to [0,1]$ with the topology of pointwise convergence. This space is compact $I^I$ but not sequentially compact (You can always find a sequence of functions that does not converge to any function in the space, when there is uncountable many functions in the space).
+>
+> [S000103](https://topology.pi-base.org/spaces/S000103)

content/Math4201/Math4201_L30.md

@@ -0,0 +1,128 @@
+# Math4201 Topology I (Lecture 30)
+
+## Compactness
+
+### Compactness in Metric Spaces
+
+#### Limit point compactness
+
+A topological space $X$ is limit point compact if every infinite subset of $X$ has a limit point in $X$.
+
+- Every compact space is limit point compact.
+
+#### Sequentially compact
+
+A topological space $X$ is sequentially compact if every sequence in $X$ has a convergent subsequence.
+
+#### Theorem of equivalence of compactness in metrizable spaces
+
+If $(X,d)$ is a metric space then the following are equivalent:
+
+1. $X$ is compact.
+2. $X$ is limit point compact.
+3. $X$ is sequentially compact.
+
+<details>
+<summary>Proof</summary>
+
+(1) $\implies$ (2):
+
+We proceed by contradiction,
+
+Let $X$ be compact and $A\subseteq X$ be an infinite subset of $X$ that doesn't have any limit points.
+
+Then $X-A$ is open because any $x\in X-A$ isn't in the closure of $A$ otherwise it would be a limit point for $A$, and hence $x$ has an open neighborhood contained in the complement of $A$.
+
+Next, let $x\in A$. Since $x$ isn't a limit point of $A$, there is an open neighborhood $U_x$ of $x$ in $X$ that $U_x\cap A=\{x\}$. Now consider the open covering of $X$ given as
+
+$$
+\{X-A\}\cup \{U_x:x\in A\}
+$$
+
+This is an open cover because either $x\in X-A$ or $x\in A$ and in the latter case, $x\in U_x$ since $X$ is compact, this should have a finite subcover. Any such subcover should contain $U_x$ for any $x$ because $U_x$ is the element in the subcover for $x$.
+
+This implies that our finite cover contains infinite open sets, which is a contradiction.
+
+---
+
+(2) $\implies$ (3):
+
+Let $\{x_n\}_{n\in\mathbb{N}}$ be an arbitrary sequence in $X$.
+
+Since $d(z,x_{n_k})\leq \frac{1}{k}$ the subsequence $(x_{n_k})$ converges to $z$.
+
+This completes the proof.
+
+except possibly $z$.
+
+Now we consider
+
+$$
+B_{r_k}(z)\text { with } r_k=\min \left(\frac{1}{k}, d_k\right)
+$$
+
+This ball has a point $x_{n_k}$ from $\{x_n\}$ which isn't equal to $z$.
+
+$r_k\leq d_k\implies n_k\geq n_{k-1}$.
+
+Since $z$ is a limit point of $\{x_n\}$, there exists $x_{n_k}$ such that $d(z,x_{n_k})<\frac{1}{k}$. So $x_{n_k}\in B_{r_k}(z)$.
+
+So, we have a convergent subsequence $(x_{n_k})$.
+
+---
+
+(3) $\implies$ (1):
+
+First wee prove the analogue of Lebesgue number lemma for a sequentially compact space $(X,d)$.
+
+Let $\{U_\alpha\}_{\alpha\in I}$ be an open covering of $X$. **By contradiction**, assume that for any $\delta>0$, there are two points $x,x'$ with $d(x,x')<\delta$ don't belong to the same open set in the covering.
+
+Take $\delta=\frac{1}{n}$, and let $x_n,x_n'$ be the points as above, then $d(x_n,x_n')<\frac{1}{n}$.
+
+$x_n,x_n'$ don't belong to the same open set in $\{U_\alpha\}_{\alpha\in I}$.
+
+By assumption $\{x_n\}$ is convergent after passing to a subsequence
+
+$$
+\{x_{n_k}\}_i
+$$
+
+Let $y$ be the limit of this subsequence and $U_\alpha$ be an element of the open covering containing $y$. There is $\epsilon>0$ such that $B_\epsilon(y)\subseteq U_\alpha$.
+
+If $k$ is large enough, then $x_{n_k}\in B_{\epsilon/2}(y)$ and $d(x_{n_k},x_{n_k}')<\epsilon/2$. (take $k$ such that $\frac{1}{n_k}<\epsilon/2$)
+
+Then $d(x_{n_k}',y)<\epsilon/2$ this implies that $x_{n_k}'\in U_\alpha$.
+
+Thus, $d(x_{n_k}',y)\leq d(x_{n_k}',x_{n_k})+d(x_{n_k},y)<\epsilon/2+\epsilon/2=\epsilon$.
+
+So, $x_{n_k}'\in B_\epsilon(y)\subseteq U_\alpha$.
+
+This is a contradiction.
+
+Next we show that for any $\epsilon$, there are
+
+$$
+y_1,y_2,\cdots,y_k
+$$
+
+such that $X=\bigcup_{i=1}^k B_{\epsilon}(y_i)$.
+
+Let's assume that it's not true and construct a sequence of points inductively in the following way:
+
+- Pick $y_1$ be arbitrary point in $X$.
+- In the $k$-th step, if $X\neq B_{\epsilon}(y_1)\cup \cdots \cup B_{\epsilon}(y_k)$, then pick $y_{k+1}\notin B_{\epsilon}(y_1)\cup \cdots \cup B_{\epsilon}(y_k)$.
+- In particular, $d(y_{k+1},y_j)\geq \epsilon$ for all $j<k$.
+
+By iteration, this process we obtain a sequence such such that the distance between any two elements is at most $\epsilon$.
+
+This sequence cannot have a converging subsequence which is a contradiction.
+
+To prove the compactness of $X$, take an open covering $\{U_\alpha\}_{\alpha\in I}$ of $X$ and let be $\delta>0$ such that any set with diameter at least $\delta$ is one of the $U_\alpha$'s. Let also $y_1,y_2,\cdots,y_k\in X$ be chosen such that $B_{\frac{\delta}{2}}(y_1)\cup \cdots \cup B_{\frac{\delta}{2}}(y_k)=X$.
+
+Since diameter of $B_{\frac{\delta}{2}}(y_i)$ is less than $\delta$, it belongs to $U_\alpha$ for some $\alpha\in I$.
+
+Then $\{U_{\alpha_i}\}_{i=1}^k$ is a finite subcover of $X$.
+
+This completes the proof.
+
+</details>

content/Math4201/_meta.js

@@ -33,4 +33,5 @@ export default {
     Math4201_L27: "Topology I (Lecture 27)",
     Math4201_L28: "Topology I (Lecture 28)",
     Math4201_L29: "Topology I (Lecture 29)",
+    Math4201_L30: "Topology I (Lecture 30)",
 }