fix typos

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.