fix typos

content/Math4201/Exam_reviews/Math4201_E3.md

@@ -0,0 +1,649 @@
+# Math 4201 Final Exam 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.
+
+## Topological space
+
+### Basic definitions
+
+#### Definition for topological space
+
+A topological space is a pair of set $X$ and a collection of subsets of $X$, denoted by $\mathcal{T}$ (imitates the set of "open sets" in $X$), satisfying the following axioms:
+
+1. $\emptyset \in \mathcal{T}$ and $X \in \mathcal{T}$
+2. $\mathcal{T}$ is closed with respect to arbitrary unions. This means, for any collection of open sets $\{U_\alpha\}_{\alpha \in I}$, we have $\bigcup_{\alpha \in I} U_\alpha \in \mathcal{T}$
+3. $\mathcal{T}$ is closed with respect to finite intersections. This means, for any finite collection of open sets $\{U_1, U_2, \ldots, U_n\}$, we have $\bigcap_{i=1}^n U_i \in \mathcal{T}$
+
+#### Definition of open set
+
+$U\subseteq X$ is an open set if $U\in \mathcal{T}$
+
+#### Definition of closed set
+
+$Z\subseteq X$ is a closed set if $X\setminus Z\in \mathcal{T}$
+
+> [!WARNING]
+>
+> A set is closed is not the same as its not open.
+>
+> In all topologies over non-empty sets, $X, \emptyset$ are both closed and open.
+
+### Basis
+
+#### Definition of topological basis
+
+For a set $X$, a topology basis, denoted by $\mathcal{B}$, is a collection of subsets of $X$, such that the following properties are satisfied:
+
+1. For any $x \in X$, there exists a $B \in \mathcal{B}$ such that $x \in B$ (basis covers the whole space)
+2. If $B_1, B_2 \in \mathcal{B}$ and $x \in B_1 \cap B_2$, then there exists a $B_3 \in \mathcal{B}$ such that $x \in B_3 \subseteq B_1 \cap B_2$ (every non-empty intersection of basis elements are also covered by a basis element)
+
+#### Definition of topology generated by basis
+
+Let $\mathcal{B}$ be a basis for a topology on a set $X$. Then the topology generated by $\mathcal{B}$ is defined by the set as follows:
+
+$$
+\mathcal{T}_{\mathcal{B}} \coloneqq \{ U \subseteq X \mid \forall x\in U, \exists B\in \mathcal{B} \text{ such that } x\in B\subseteq U \}
+$$
+
+> This is basically a closure of $\mathcal{B}$ under arbitrary unions and finite intersections
+
+#### Lemma of topology generated by basis
+
+$U\in \mathcal{T}_{\mathcal{B}}\iff \exists \{B_\alpha\}_{\alpha \in I}\subseteq \mathcal{B}$ such that $U=\bigcup_{\alpha \in I} B_\alpha$
+
+#### Definition of basis generated from a topology
+
+Let $(X, \mathcal{T})$ be a topological space. Then the basis generated from a topology is $\mathcal{C}\subseteq \mathcal{B}$ such that $\forall U\in \mathcal{T}$, $\forall x\in U$, $\exists B\in \mathcal{C}$ such that $x\in B\subseteq U$.
+
+#### Definition of subbasis of topology
+
+A subbasis of a topology is a collection $\mathcal{S}\subseteq \mathcal{T}$ such that $\bigcup_{U\in \mathcal{S}} U=X$.
+
+#### Definition of topology generated by subbasis
+
+Let $\mathcal{S}\subseteq \mathcal{T}$ be a subbasis of a topology on $X$, then the basis generated by such subbasis is the closure of finite intersection of $\mathcal{S}$
+
+$$
+\mathcal{B}_{\mathcal{S}} \coloneqq \{B\mid B\text{ is the intersection of a finite number of elements of }\mathcal{S}\}
+$$
+
+Then the topology generated by $\mathcal{B}_{\mathcal{S}}$ is the subbasis topology denoted by $\mathcal{T}_{\mathcal{S}}$.
+
+Note that all open set with respect to $\mathcal{T}_{\mathcal{S}}$ can be written as a union of finitely intersections of elements of $\mathcal{S}$
+
+### Comparing topologies
+
+#### Definition of finer and coarser topology
+
+Let $(X,\mathcal{T})$ and $(X,\mathcal{T}')$ be topological spaces. Then $\mathcal{T}$ is finer than $\mathcal{T}'$ if $\mathcal{T}'\subseteq \mathcal{T}$. $\mathcal{T}$ is coarser than $\mathcal{T}'$ if $\mathcal{T}\subseteq \mathcal{T}'$.
+
+#### Lemma of comparing basis
+
+Let $(X,\mathcal{T})$ and $(X,\mathcal{T}')$ be topological spaces with basis $\mathcal{B}$ and $\mathcal{B}'$. Then $\mathcal{T}$ is finer than $\mathcal{T}'$ if and only if for any $x\in X$, $x\in B'$, $B'\in \mathcal{B}'$, there exists $B\in \mathcal{B}$, such that $x\in B$ and $x\in B\subseteq B'$.
+
+### Product space
+
+#### Definition of cartesian product
+
+Let $X,Y$ be sets. The cartesian product of $X$ and $Y$ is the set of all ordered pairs $(x,y)$ where $x\in X$ and $y\in Y$, denoted by $X\times Y$.
+
+#### Definition of product topology
+
+Let $(X,\mathcal{T}_X)$ and $(Y,\mathcal{T}_Y)$ be topological spaces. Then the product topology on $X\times Y$ is the topology generated by the basis
+
+$$
+\mathcal{B}_{X\times Y}=\{U\times V, U\in \mathcal{T}_X, V\in \mathcal{T}_Y\}
+$$
+
+or equivalently,
+
+$$
+\mathcal{B}_{X\times Y}'=\{U\times V, U\in \mathcal{B}_X, V\in \mathcal{B}_Y\}
+$$
+
+> Product topology generated from open sets of $X$ and $Y$ is the same as product topology generated from their corresponding basis
+
+### Subspace topology
+
+#### Definition of subspace topology
+
+Let $(X,\mathcal{T})$ be a topological space and $Y\subseteq X$. Then the subspace topology on $Y$ is the topology given by
+
+$$
+\mathcal{T}_Y=\{U\cap Y|U\in \mathcal{T}\}
+$$
+
+or equivalently, let $\mathcal{B}$ be the basis for $(X,\mathcal{T})$. Then the subspace topology on $Y$ is the topology generated by the basis
+
+$$
+\mathcal{B}_Y=\{U\cap Y| U\in \mathcal{B}\}
+$$
+
+#### Lemma of open sets in subspace topology
+
+Let $(X,\mathcal{T})$ be a topological space and $Y\subseteq X$. Then if $U\subseteq Y$, $U$ is open in $(Y,\mathcal{T}_Y)$, then $U$ is open in $(X,\mathcal{T})$.
+
+> This also holds for closed set in closed subspace topology
+
+### Interior and closure
+
+#### Definition of interior
+
+The interior of $A$ is the largest open subset of $A$.
+
+$$
+A^\circ=\bigcup_{U\subseteq A, U\text{ is open in }X} U
+$$
+
+#### Definition of closure
+
+The closure of $A$ is the smallest closed superset of $A$.
+
+$$
+\overline{A}=\bigcap_{U\supseteq A, U\text{ is closed in }X} U
+$$
+
+#### Definition of neighborhood
+
+A neighborhood of a point $x\in X$ is an open set $U\in \mathcal{T}$ such that $x\in U$.
+
+#### Definition of limit points
+
+A point $x\in X$ is a limit point of $A$ if every neighborhood of $x$ contains a point in $A-\{x\}$.
+
+We denote the set of all limits points of $A$ by $A'$.
+
+$\overline{A}=A\cup A'$
+
+### Sequences and continuous functions
+
+#### Definition of convergence
+
+Let $X$ be a topological space. A sequence $(x_n)_{n\in\mathbb{N}_+}$ in $X$ converges to $x\in X$ if for any neighborhood $U$ of $x$, there exists $N\in\mathbb{N}_+$ such that $\forall n\geq N, x_n\in U$.
+
+#### Definition of Hausdoorff space
+
+A topological space $(X,\mathcal{T})$ is Hausdorff if for any two distinct points $x,y\in X$, there exist open neighborhoods $U$ and $V$ of $x$ and $y$ respectively such that $U\cap V=\emptyset$.
+
+#### Uniqueness of convergence in Hausdorff spaces
+
+In a Hausdorff space, if a sequence $(x_n)_{n\in\mathbb{N}_+}$ converges to $x\in X$ and $y\in X$, then $x=y$.
+
+#### Closed singleton in Hausdorff spaces
+
+In a Hausdorff space, if $x\in X$, then $\{x\}$ is a closed set.
+
+#### Definition of continuous function
+
+Let $(X,\mathcal{T}_X)$ and $(Y,\mathcal{T}_Y)$ be topological spaces. A function $f:X\to Y$ is continuous if for any open set $U\subseteq Y$, $f^{-1}(U)$ is open in $X$.
+
+#### Definition of point-wise continuity
+
+Let $(X,\mathcal{T}_X)$ and $(Y,\mathcal{T}_Y)$ be topological spaces. A function $f:X\to Y$ is point-wise continuous at $x\in X$ if for every openset $V\subseteq Y$, $f(x)\in V$ then there exists an open set $U\subseteq X$ such that $x\in U$ and $f(U)\subseteq V$.
+
+#### Lemma of continuous functions
+
+If $f:X\to Y$ is point-wise continuous for all $x\in X$, then $f$ is continuous.
+
+#### Properties of continuous functions
+
+If $f:X\to Y$ is continuous, then
+
+1. $\forall A\subseteq Y$, $f^{-1}(A^c)=X\setminus f^{-1}(A)$ (complements maps to complements)
+2. $\forall A_\alpha\subseteq Y, \alpha\in I$, $f^{-1}(\bigcup_{\alpha\in I} A_\alpha)=\bigcup_{\alpha\in I} f^{-1}(A_\alpha)$
+3. $\forall A_\alpha\subseteq Y, \alpha\in I$, $f^{-1}(\bigcap_{\alpha\in I} A_\alpha)=\bigcap_{\alpha\in I} f^{-1}(A_\alpha)$
+4. $f^{-1}(U)$ is open in $X$ for any open set $U\subseteq Y$.
+5. $f$ is continuous at $x\in X$.
+6. $f^{-1}(V)$ is closed in $X$ for any closed set $V\subseteq Y$.
+7. Assume $\mathcal{B}$ is a basis for $Y$, then $f^{-1}(\mathcal{B})$ is open in $X$ for any $B\in \mathcal{B}$.
+8. $\forall A\subseteq X$, $\overline{f(A)}=f(\overline{A})$
+
+#### Definition of homeomorphism
+
+Let $(X,\mathcal{T}_X)$ and $(Y,\mathcal{T}_Y)$ be topological spaces. A function $f:X\to Y$ is a homeomorphism if $f$ is continuous, bijective and $f^{-1}:Y\to X$ is continuous.
+
+#### Ways to construct continuous functions
+
+1. If $f:X\to Y$ is constant function, $f(x)=y_0$ for all $x\in X$, then $f$ is continuous. (constant functions are continuous)
+2. If $A$ is a subspace of $X$, $f:A\to X$ is the inclusion map $f(x)=x$ for all $x\in A$, then $f$ is continuous. (inclusion maps are continuous)
+3. If $f:X\to Y$ is continuous, $g:Y\to Z$ is continuous, then $g\circ f:X\to Z$ is continuous. (composition of continuous functions is continuous)
+4. If $f:X\to Y$ is continuous, $A$ is a subspace of $X$, then $f|_A:X\to Y$ is continuous. (domain restriction is continuous)
+5. If $f:X\to Y$ is continuous, $Z$ is a subspace of $Y$, then $f:X\to Z$, $g(x)=f(x)\cap Z$ is continuous. If $Y$ is a subspace of $Z$, then $h:X\to Z$, $h(x)=f(x)$ is continuous (composition of $f$ and inclusion map).
+6. If $f:X\to Y$ is continuous, $X$ can be written as a union of open sets $\{U_\alpha\}_{\alpha\in I}$, then $f|_{U_\alpha}:X\to Y$ is continuous.
+7. If $X=Z_1\cup Z_2$, and $Z_1,Z_2$ are closed equipped with subspace topology, let $g_1:Z_1\to Y$ and $g_2:Z_2\to Y$ be continuous, and for all $x\in Z_1\cap Z_2$, $g_1(x)=g_2(x)$, then $f:X\to Y$ by $f(x)\begin{cases}g_1(x), & x\in Z_1 \\ g_2(x), & x\in Z_2\end{cases}$ is continuous. (pasting lemma)
+8. $f:X\to Y$ is continuous, $g:X\to Z$ is continuous if and only if $H:X\to Y\times Z$, where $Y\times Z$ is equipped with the product topology, $H(x)=(f(x),g(x))$ is continuous. (proved in homework)
+
+### Metric spaces
+
+#### Definition of metric
+
+A metric on $X$ is a function $d:X\times X\to \mathbb{R}$ such that $\forall x,y\in X$,
+
+1. $d(x,x)=0$
+2. $d(x,y)\geq 0$
+3. $d(x,y)=d(y,x)$
+4. $d(x,y)+d(y,z)\geq d(x,z)$
+
+#### Definition of metric ball
+
+The metric ball $B_r^{d}(x)$ is the set of all points $y\in X$ such that $d(x,y)\leq r$.
+
+#### Definition of metric topology
+
+Let $X$ be a metric space with metric $d$. Then $X$ is equipped with the metric topology generated by the metric balls $B_r^{d}(x)$ for $r>0$.
+
+#### Definition of metrizable
+
+A topological space $(X,\mathcal{T})$ is metrizable if it is the metric topology for some metric $d$ on $X$.
+
+#### Hausdorff axiom for metric spaces
+
+Every metric space is Hausdorff (take metric balls $B_r(x)$ and $B_r(y)$, $r=\frac{d(x,y)}{2}$).
+
+If a topology isn't Hausdorff, then it isn't metrizable.
+
+Prove by triangle inequality and contradiction.
+
+#### Common metrics in $\mathbb{R}^n$
+
+Euclidean metric
+
+$$
+d(x,y)=\sqrt{\sum_{i=1}^n (x_i-y_i)^2}
+$$
+
+Square metric
+
+$$
+\rho(x,y)=\max_{i=1}^n |x_i-y_i|
+$$
+
+Manhattan metric
+
+$$
+m(x,y)=\sum_{i=1}^n |x_i-y_i|
+$$
+
+These metrics are equivalent.
+
+#### Product topology and metric
+
+If $(X,d),(Y,d')$ are metric spaces, then $X\times Y$ is metric space with metric $d(x,y)=\max\{d(x_1,y_1),d(x_2,y_2)\}$.
+
+#### Uniform metric
+
+Let $\mathbb{R}^\omega$ be the set of all infinite sequences of real numbers. Then $\overline{d(x,y)}=\sup_{i=1}^\omega \min\{1,|x_i-y_i|\}$, the uniform metric on $\mathbb{R}^\omega$ is a metric.
+
+#### Metric space and converging sequences
+
+Let $X$ be a topological space, $A\subseteq X$, $x_n\to x$ such that $x_n\in A$. Then $x\in \overline{A}$.
+
+If $X$ is a **metric space**, $A\subseteq X$, $x\in \overline{A}$, then there exists converging sequence $x_n\to x$ such that $x_n\in A$.
+
+### Metric defined for functions
+
+#### Definition for bounded metric space
+
+A metric space $(Y,d)$ is bounded if there is $M\in \mathbb{R}^{\geq 0}$ such that for all $y_1,y_2\in Y$, $d(y_1,y_2)\leq M$.
+
+#### Definition for metric defined for functions
+
+Let $X$ be a topological space and $Y$ be a bounded metric space, then the set of all maps, denoted by $\operatorname{Map}(X,Y)$, $f:X\to Y\in \operatorname{Map}(X,Y)$ is a metric space with metric $\rho(f,g)=\sup_{x\in X} d(f(x),g(x))$.
+
+#### Space of continuous map is closed
+
+Let $(\operatorname{Map}(X,Y),\rho)$ be a metric space defined above, then every continuous map is a limit point of some sequence of continuous maps.
+
+$$
+Z=\{f\in \operatorname{Map}(X,Y)|f\text{ is continuous}\}
+$$
+
+$Z$ is closed in $(\operatorname{Map}(X,Y),\rho)$.
+
+### Quotient space
+
+#### Quotient map
+
+Let $X$ be a topological space and $X^*$ is a set. $q:X\to X^*$ is a surjective map. Then $q$ is a quotient map.
+
+#### Quotient topology
+
+Let $(X,\mathcal{T})$ be a topological space and $X^*$ be a set, $q:X\to X^*$ is a surjective map. Then
+
+$$
+\mathcal{T}^* \coloneqq \{U\subseteq X^*\mid q^{-1}(U)\in \mathcal{T}\}
+$$
+
+is a topology on $X^*$ called quotient topology.
+
+**That is equivalent to say that $U\subseteq X^*$ is open in $X^*$ if and only if $p^{-1}(U)\subseteq X$ is open in $X$.**
+
+This is also called "strong continuity" since compared with the continuous condition, it requires if $p^{-1}(U)$ is open in $X$, then $U$ is open in $X^*$.
+
+$(X^*,\mathcal{T}^*)$ is called the quotient space of $X$ by $q$.
+
+#### Closed map and open map
+
+$f:X\to Y$ is a open map if for each open set $U$ of $X$, $f(U)$ is open in $Y$; it is a closed map if for each closed set $U$ of $X$, $f(U)$ is closed in $Y$.
+
+> [!WARNING]
+>
+> Not all quotient map are closed or open:
+>
+> 1. Example of quotient map that is not open nor closed:
+>
+> Consider the projection map $f:[0,1]\to S^1$, this map maps open set $[0,0.5)$ in $[0,1]$ to non open map $[0,\pi)$
+>
+> 2. Example of open map that is not closed:
+>
+> Consider projection map $f:\mathbb{R}\times \mathbb{R}\to \mathbb{R}$ to first coordinate, this map is open but not closed, consider $C\coloneqq\{x\times y\mid xy=1\}$ This set is closed in $\mathbb{R}\times \mathbb{R}$ but $f(C)=\mathbb{R}-\{0\}$ is not closed in $\mathbb{R}$.
+>
+> 3. Example of closed map that is not open:
+> 
+> Consider $f:[0,1]\cup[2,3]\to [0,2]$ by taking -1 to elements in $[2,3]$, this map is closed map but not open, since $f([2,3])=[1,2]$ is not open in $[0,2]$ but $[2,3]$ is open in $[0,1]\cup[2,3]$
+
+#### Equivalent classes
+
+$\sim$ is a subset of $X\times X$ with the following properties:
+
+1. $x\sim x$ for all $x\in X$.
+2. If $(x,y)\in \sim$, then $(y,x)\in \sim$.
+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 $[x]=\{y\in X|y\sim x\}$.
+
+We can use equivalent classes to define quotient space.
+
+#### Theorem 22.2
+
+Let $p:X\to Y$ be a quotient map. Let $Z$ be a space and let $g:X\to Z$ be a map that is constant on each set $p^{-1}(\{y\})$, for $y\in Y$. Then $g$ induces a map $f:Y\to Z$ such that $f\circ p=g$. The induced map $f$ is continuous if and only if $g$ is continuous; $f$ is a quotient map if and only if $g$ is a quotient map.
+
+*Prove by setting $f(p(x))=g(x)$, then $g^{-1}(V)=p^{-1}(f^{-1}(V))$ for $V$ open in $Z$.*
+
+## 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 $d(f(x_1),f(x_2))<\epsilon$.
+
+#### Uniform continuity theorem
+
+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.
+
+#### Local compactness
+
+A space $X$ is locally compact if every point $x\in X$, **there is a compact subspace $K$ of $X$ containing a neighborhood $U$ of $x$** $x\in U\subseteq K$ such that $K$ is compact.
+
+#### Theorem of one point compactification
+
+Let $X$ be a locally compact Hausdorff space if and only if there exists a topological space $Y$ satisfying the following properties:
+
+1. $X$ is a subspace of $Y$.
+2. $Y-X$ has one point, usually denoted by $\infty$.
+3. $Y$ is compact and Hausdorff.
+
+The $Y$ is defined as follows:
+
+$U\subseteq Y$ is open if and only if one of the following holds.
+
+1. $U\subseteq X$ and $U$ is open in $X$
+2. $\infty \in U$ and $Y-U\subseteq X$, and $Y-U$ is compact.
+
+## Countability and Separation Axioms
+
+### Countability Axioms
+
+#### First countability axiom
+
+A topological space $(X,\mathcal{T})$ satisfies the first countability axiom if any point $x\in X$, there is a sequence of open neighborhoods of $x$, $\{V_n\}_{n=1}^\infty$ such that any open neighborhood $U$ of $x$ contains one of $V_n$.
+
+Apply the theorem above, we have if $(X,\mathcal{T})$ satisfies the first countability axiom, then:
+
+1. Every convergent sequence converges to a point in the closure of the sequence.
+
+<details>
+<summary>Space that every convergent sequence not converges to a point in the closure of the sequence.</summary>
+
+Consider $\mathbb{R}^\omega$ with the box topology.
+
+And $A=(0,1)\times (0,1)\times \cdots$ and $x=(0,0,\cdots)$.
+
+$x\in \overline{A}$ but no sequence converges to $x$.
+
+Suppose there exists such sequence, $\{x_n=(x_1^n,x_2^n,\cdots)\}_{n=1}^\infty$.
+
+Take $B=(-x_1^1,x_1^1)\times(-x_2^2,x_2^2)\times \cdots$, this is basis containing $x$ but none of $x_n$.
+
+</details>
+
+2. If $f:X\to Y$ such that for any sequence $\{x_n\}_{n=1}^\infty$ in $X$, $f(x_n)\to f(x)$, then $f$ is continuous.
+
+#### Second countability axiom
+
+Let $(X,\mathcal{T})$ be a topological space, then $X$ satisfies the second countability axiom if $X$ has a countable basis.
+
+If $X$ is second countable, then:
+
+1. Any discrete subspace $Y$ of $X$ is countable
+2. There exists a countable subset of $X$ that is dense in $X$.
+3. Every open covering of $X$ has a **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*)
+
+### Separation Axioms
+
+#### Hausdorff spaces
+
+A topological space $(X,\mathcal{T})$ is Hausdorff if for any two distinct points $x,y\in X$, there are **disjoint open sets** $U,V$ such that $x\in U$ and $y\in V$. (note that $U\cup V$ may not be $X$, compared with definition of separation)
+
+Some corollaries:
+
+1. A subspace of Hausdorff space is Hausdorff, and a product of Hausdorff spaces is Hausdorff.
+
+#### Regular spaces
+
+A topological space $(X,\mathcal{T})$ is regular if for any $x\in X$ and any closed set $A\subseteq X$ such that $x\notin A$, there are **disjoint open sets** $U,V$ such that $x\in U$ and $A\subseteq V$.
+
+Some corollaries:
+
+1. $X$ is regular if and only if given a point $x$ and a open neighborhood $U$ of $x$, there is open neighborhood $V$ of $x$ such that $\overline{V}\subseteq U$.
+2. A subspace of regular space is regular, and a product of regular spaces is regular.
+
+#### Normal spaces
+
+A topological space $(X,\mathcal{T})$ is normal if for any disjoint closed sets $A,B\subseteq X$, there are **disjoint open sets $U,V$** such that $A\subseteq U$ and $B\subseteq V$.
+
+Some corollaries:
+
+1. $X$ is normal if and only if given a closed set $A\subseteq X$, there is open neighborhood $V$ of $A$ such that $\overline{V}\subseteq U$.
+
+> [!CAUTION]
+>
+> Product of normal spaces may not be normal (consider Sorgenfrey plane)
+
+#### Regular space with countable basis is normal
+
+Let $X$ be a regular space with countable basis, then $X$ is normal.
+
+*Prove by taking disjoint open neighborhoods by countable cover.*
+
+

content/Math4201/Math4201_L29.md

@@ -51,7 +51,7 @@ 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](./Math4201_L8#limit-points) in $X$.
+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 $x$, $(U-\{x\})\cap A\neq \emptyset$._
 

content/Math4201/Math4201_L32.md

@@ -30,7 +30,7 @@ First, we prove that $X$ is a subspace of $Y$. (That is, every open set $U\subse
 
 Case 1: $U\subseteq X$ is open in $X$, then $U\cap X=U$ is open in $Y$.
 
-Case 2: $\infty\in U$, then $Y-U$ is a compact subspace of $X$, since $X$ is Hausdorff. So $Y-U$ is a closed subspace of $X$.
+Case 2: $\infty\in U$, then $Y-U$ is a compact subspace of $X$. Since $X$ is Hausdorff, $Y-U$ is a closed subspace of $X$. [Compact subspace of a Hausdorff space is closed](../Math4201_L25#proposition-of-compact-subspaces-with-hausdorff-property)
 
 So $X\cap U=X-(Y-U)$ is open in $X$.
 

content/Math4201/Math4201_L33.md

@@ -131,7 +131,7 @@ $$
 A=\{\underline{x}=(x_1,x_2,x_3,\dots)|x_i\in \{0,1\}\}
 $$
 
-Let $\underline{x} and \underline{x}'$ be two distinct elements of $A$.
+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

content/Math4201/Math4201_L34.md

@@ -98,7 +98,7 @@ A $T_0$ space is regular if for any $x\in X$ and any close set $A\subseteq X$ su
 A $T_0$ space is normal if for any disjoint closed sets, $A,B\subseteq X$, there are **disjoint open sets** $U,V$ such that $A\subseteq U$ and $B\subseteq V$.
 
 <details>
-<summary></summary>
+<summary>Finer topology may not be normal</summary>
 
 Let $\mathbb{R}_K$ be the topology on $\mathbb{R}$ generated by the basis:
 
@@ -106,7 +106,7 @@ $$
 \mathcal{B}=\{(a,b)\mid a,b\in \mathbb{R},a<b\}\cup \{(a,b)-K\mid a,b\in \mathbb{R},a<b\}
 $$
 
-where $K=\coloneqq \{\frac{1}{n}\mid n\in \mathbb{N}\}$.
+where $K\coloneqq \{\frac{1}{n}\mid n\in \mathbb{N}\}$.
 
 **This is finer than the standard topology** on $\mathbb{R}$.
 

content/Math4201/Math4201_L35.md

@@ -108,7 +108,100 @@ Apply the assumption to find $A\subseteq V\subseteq X$ and $V$ is open in $X$ an
 >
 > The above does not hold for normal.
 
-Recall that $\mathbb{R}_{\ell}$ with lower limit topology is normal. But $\mathbb{R}_{\ell}\times \mathbb{R}_{\ell}$ with product topology is not normal. (In problem set 11)
+Recall that $\mathbb{R}_{\ell}$ with lower limit topology is normal. But $\mathbb{R}_{\ell}\times \mathbb{R}_{\ell}$ with product topology is not normal.
+
+<details>
+<summary>Proof that Sorgenfrey plane is not normal</summary>
+
+The goal of this problem is to show that $\mathbb{R}^2_\ell$ (the Sorgenfrey plane) is not normal. Recall that $\mathbb{R}_\ell$ is the real line with the lower limit topology, and $\mathbb{R}_\ell^2$ is equipped with the product topology. Consider the subset
+
+$$
+L = \{\, (x,-x) \mid x\in \mathbb{R}_\ell \,\} \subset \mathbb{R}^2_\ell.
+$$
+
+Let $A\subset L$ be the set points of the form $(x,-x)$ such that $x$ is rational and $B\subset L$ be the set points of the form $(x,-x)$ such that $x$ is irrational.
+
+1. Show that the subspace topology on $L$ is the discrete topology.  Conclude that $A$ and $B$ are closed subspaces of $\mathbb{R}_\ell^2$
+
+<details>
+<summary>Proof</summary>
+First we show that $L$ is closed.
+
+Consider $x=(a,b)\in\mathbb{R}^2_\ell-L$, by definition $a\neq -b$.
+
+If $a>-b$, then there exists open neighborhood $U_x=[\frac{\min\{a,b\}}{2},a+1)\times[\frac{\min\{a,b\}}{2},b+1)$ that is disjoint from $L$ (no points of form $(x,-x)$ in our rectangle), therefore $x\in U_x$.
+
+If $a<-b$, then there exists open neighborhood $U_x=[a,a+\frac{\min\{a,b\}}{2})\times[b,b+\frac{\min\{a,b\}}{2})$ that is disjoint from $L$, therefore $x\in U_x$.
+
+Therefore, $\mathbb{R}^2_\ell-L=\bigcup_{x\in \mathbb{R}_\ell^2-L} U_x$ is open in $\mathbb{R}^2_\ell$.
+
+So $L$ is closed in $\mathbb{R}^2_\ell$.
+
+To show $L$ with subspace topology on $\mathbb{R}^2_\ell$ is discrete topology, we need to show that every singleton of $L$ is open in $L$.
+
+For each $\{(x,-x)\}\in L$, $[x,x+1)\times [-x,-x+1)$ is open in $\mathbb{R}_\ell^2$ and $\{(x,-x)\}=([x,x+1)\times [-x,-x+1))\cap L$, therefore $\{(x,-x)\}$ is open in $L$.
+
+Since $A,B$ are disjoint and $A\cup B=L$, therefore $A=L-B$ and $B=L-A$, by definition of discrete topology, $A,B$ are both open therefore the complement of $A,B$ are closed. So $A,B$ are closed in $L$.
+
+since $L$ is closed in $\mathbb{R}^2_\ell$, by \textbf{Lemma \ref{closed_set_close_subspace_close}}, $A,B$ is also closed in $\mathbb{R}_\ell^2$. Therefore $A,B$ are closed subspace of $\mathbb{R}_\ell^2$.
+</details>
+
+2. Let $V$ be an open set of $\mathbb{R}^2_\ell$ containing $B$. Let $K_n$ consist of all irrational numbers $x\in [0,1]$ such that $[x, x+1/n) \times [-x, -x+1/n)$ is contained in $V$. Show that $[0,1]$ is the union of the sets $K_n$ and countably many one-point sets.
+
+<details>
+<summary>Proof</summary>
+Since $B$ is open in $L$, for each $b\in B$, by definition of basis in $\mathbb{R}_\ell^2$, and $B$ is open, there exists $b\in ([b,b+\epsilon)\times [-b,-b+\delta))\cap L\subseteq V$ and $0<\epsilon,\delta$, so there exists $n_b$ such that $\frac{1}{n_b}<\min\{\epsilon,\delta\}$ such that $b\in ([b,b+\frac{1}{n_b})\times [-b,-b+\frac{1}{n_b}))\cap L\subseteq V$.
+
+Therefore $\bigcup_{n=1}^\infty K_n$ covers irrational points in $[0,1]$
+
+Note that $B=L-A$ where $A$ is rational points therefore countable.
+
+So $[0,1]$ is the union of the sets $K_n$ and countably many one-point sets.
+</details>
+
+3. Use Problem 5-3 to show that some set $\overline{K_n}$ contains an open interval $(a,b)$ of $\mathbb{R}$. (You don't need to prove Problem 5.3, if it is not your choice of \#5.)
+
+#### Lemma
+
+Let $X$ be a compact Hausdorff space; let $\{A_n\}$ be a countable collection of closed sets of $X$. If each sets $A_n$ has empty interior in $X$, then the union $\bigcup_{n=1}^\infty A_n$ has empty interior in $X$.
+
+<details>
+<summary>Proof</summary>
+We proceed by contradiction, note that $[0,1]$ is a compact Hausdorff space since it's closed and bounded.
+
+And $\{\overline{K_n}\}_{n=0}^\infty$ is a countable collection of closed sets of $[0,1]$.
+
+Suppose for the sake of contradiction, $\overline{K_n}$ has empty interior in $X$ for all $n\in \mathbb{N}$, by \textbf{Lemma \ref{countable_closed_sets_empty_interior}}, then $\bigcup_{n=1}^\infty \overline{K_n}$ has empty interior in $[0,1]$, where $\Q\cap[0,1]$ are countably union of singletons, therefore has empty interior in $[0,1]$.
+
+Therefore $\bigcup_{n=1}^\infty \overline{K_n}$ has empty interior in $[0,1]$, since $\bigcup_{n=1}^\infty K_n\subseteq \bigcup_{n=1}^\infty \overline{K_n}$, $\bigcup_{n=1}^\infty K_n$ also has empty interior in $[0,1]$ by definition of subspace of $[0,1]$, therefore $\bigcup_{n=1}^\infty K_n\cup (\Q\cap[0,1])$ has empty interior in $[0,1]$. This contradicts that $\bigcup_{n=1}^\infty K_n\cup (\Q\cap[0,1])$ covers $[0,1]$ and should at least have interior $(0.1,0.9)$.
+</details>
+
+4. Show that $V$ contains the open parallelogram consisting of all points of the form
+
+$$
+x\times (-x+\epsilon)\quad\text{ for which }\quad a<x<b\text{ and }0<\epsilon<\frac{1}{n}.
+$$
+
+<details>
+<summary>Proof</summary>
+Since $V$ is open, by previous problem we know that there exists $n$ such that $\overline{K_n}$ contains the open interval $(a,b)$.
+
+If $x\in K_n$, $\forall a<x<b$, by definition of $K_n$ $[x,x+\frac{1}{n})\times [-x,-x+\frac{1}{n})\subseteq V$.
+
+If $x$ is a limit point of $K_n$, since $V$ is open, there exists $0<\epsilon<\frac{1}{n}$ such that $[x,x+\epsilon)\times [-x,-x+\epsilon)\subseteq V$.
+
+This gives our desired open parallelogram.
+</details>
+  
+5. Show that if $q$ is a rational number with $a<q<b$, then the point $q\times (-q)$ of $\mathbb{R}_\ell^2$ is a limit point of $V$. Conclude that there are no disjoint open neighborhoods $U$ of $A$ and $V$ of $B$.
+
+<details>
+<summary>Proof</summary>
+Consider all the open neighborhood of $q\times (-q)$ in $\mathbb{R}_\ell^2$, for all $\delta>0$, $[q,q+\delta)\times (-q,-q+\delta)$ will intersect with some $x\times [-x,-x+\epsilon)\subseteq V$ such that $0<\epsilon<\frac{1}{n}<\delta$.
+
+Therefore, any open set containing $q\times (-q)\in A$ will intersect with $V$, it is impossible to build disjoint open neighborhoods $U$ of $A$ and $V$ of $B$.
+</details>
+</details>
 
 This shows that $\mathbb{R}_{\ell}$ is not metrizable. Otherwise $\mathbb{R}_{\ell}\times \mathbb{R}_{\ell}$ would be metrizable. Which could implies that $\mathbb{R}_{\ell}$ is normal.
 

content/Math4201/Math4201_L38.md

@@ -40,7 +40,7 @@ If $X$ is a normal (regular and second countable) topological space, then $X$ is
 
 We will show that there exists embedding $F:X\to \mathbb{R}^\omega$ such that $F$ is continuous, injective and if $Z=F(X)$, $F:X\to Z$ is a open map.
 
-Recall that [regular and second countable spaces are normal](./Math4201_L36.md/#theorem-for-constructing-normal-spaces)
+Recall that [regular and second countable spaces are normal](../Math4201_L36#theorem-for-constructing-normal-spaces)
 
 1. Since $X$ is regular, then 1 point sets in $X$ are closed.
 2. $X$ is regular if and only if $\forall x\in U\subseteq X$, $U$ is open in $X$. There exists $V$ open in $X$ such that $x\in V\subseteq\overline{V}\subseteq U$.
@@ -49,7 +49,7 @@ Let $\{B_n\}$ be a countable basis for $X$ (by second countability).
 
 Pass to $(n,m)$ such that $\overline{B_n}\subseteq B_m$.
 
-By [Urysohn lemma](./Math4201_L37.md/#urysohn-lemma), there exists continuous function $g_{m,n}: X\to [0,1]$ such that $g_{m,n}(\overline{B_n})=\{1\}$ and $g_{m,n}(B_m)=\{0\}$.
+By [Urysohn lemma](../Math4201_L37#urysohn-lemma), there exists continuous function $g_{m,n}: X\to [0,1]$ such that $g_{m,n}(\overline{B_n})=\{1\}$ and $g_{m,n}(B_m)=\{0\}$.
 
 Therefore, we have $\{g_{m,n}\}$ is a countable set of functions, where $\overline{B_n}\subseteq B_m$.
 

content/Math4201/Math4201_L39.md

@@ -25,7 +25,7 @@ $T=\mathbb{R}^2/\mathbb{Z}^2$ is a $2$-dimensional manifold.
 
 </details>
 
-Recall the [Urysohn metirzation theorem](./Math4201_L38.md/#urysohn-metirzation-theorem). Any normal and second countable space is metrizable.
+Recall the [Urysohn metirzation theorem](../Math4201_L38#urysohn-metirzation-theorem). Any normal and second countable space is metrizable.
 
 In the proof we saw that any such space can be embedded into $\mathbb{R}^\omega$ with the product topology.