diff --git a/content/Math4201/Exam_reviews/Math4201_E3.md b/content/Math4201/Exam_reviews/Math4201_E3.md new file mode 100644 index 0000000..84111ac --- /dev/null +++ b/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 $f(a)0$ 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. + +
+Space that every convergent sequence not converges to a point in the closure of the sequence. + +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 $x_n$. + +
+ +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.* + + diff --git a/content/Math4201/Math4201_L29.md b/content/Math4201/Math4201_L29.md index de9bd65..ffaf02c 100644 --- a/content/Math4201/Math4201_L29.md +++ b/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 $x$, $(U-\{x\})\cap A\neq \emptyset$._ diff --git a/content/Math4201/Math4201_L32.md b/content/Math4201/Math4201_L32.md index c277935..2d74db3 100644 --- a/content/Math4201/Math4201_L32.md +++ b/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 $X$. diff --git a/content/Math4201/Math4201_L33.md b/content/Math4201/Math4201_L33.md index 8bd420a..913591e 100644 --- a/content/Math4201/Math4201_L33.md +++ b/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 $A$. $$ \rho(\underline{x},\underline{x}')=\sup_{i\in \mathbb{N}}\overline{d}(\underline{x}_\alpha,\underline{x}'_\alpha)=1 diff --git a/content/Math4201/Math4201_L34.md b/content/Math4201/Math4201_L34.md index cb5e7f5..df4463d 100644 --- a/content/Math4201/Math4201_L34.md +++ b/content/Math4201/Math4201_L34.md @@ -98,7 +98,7 @@ A $T_0$ space is regular if for any $x\in X$ and any close set $A\subseteq X$ su A $T_0$ space is normal if for any disjoint closed sets, $A,B\subseteq X$, there are **disjoint open sets** $U,V$ such that $A\subseteq U$ and $B\subseteq V$.
- +Finer topology may not be normal Let $\mathbb{R}_K$ be the topology on $\mathbb{R}$ generated by the basis: @@ -106,7 +106,7 @@ $$ \mathcal{B}=\{(a,b)\mid a,b\in \mathbb{R},a > 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. + +
+Proof that Sorgenfrey plane is not normal + +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 $\mathbb{R}_\ell^2$ + +
+Proof +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 $\mathbb{R}_\ell^2$. +
+ +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. + +
+Proof +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. +
+ +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 $X$. + +
+Proof +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 $(0.1,0.9)$. +
+ +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 +Proof +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 + +5. Show that if $q$ is a rational number with $a +Proof +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 $B$. +
+
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. diff --git a/content/Math4201/Math4201_L38.md b/content/Math4201/Math4201_L38.md index 931688e..9d89208 100644 --- a/content/Math4201/Math4201_L38.md +++ b/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 $\overline{B_n}\subseteq B_m$. diff --git a/content/Math4201/Math4201_L39.md b/content/Math4201/Math4201_L39.md index 645569d..2d9bb5b 100644 --- a/content/Math4201/Math4201_L39.md +++ b/content/Math4201/Math4201_L39.md @@ -25,7 +25,7 @@ $T=\mathbb{R}^2/\mathbb{Z}^2$ is a $2$-dimensional manifold. -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.