update notations and fix typos

pages/Math4111/Exam_reviews/Math4111_E2.md

@@ -22,7 +22,7 @@ Suppose $S$ and $T$ are compact, let $\{G_\alpha\}_{\alpha\in A}$ be an open cov
 
 ...
 
-EOP
+QED
 
 ## K-cells are compact
 
@@ -63,7 +63,7 @@ Let $n\in \mathbb{N}$ be such that $\frac{1}{2^n}<r$. Then by $(c)$, $I(n)\subse
 
 Then $\{G_{\alpha_0}\}$ is a cover of $I_n$ which contradicts with (b)
 
-EOP
+QED
 
 ## Redundant subcover question
 

pages/Math4111/Math4111_L1.md

@@ -36,7 +36,7 @@ So $m^2$ is divisible by 4, $2n^2$ is divisible by 4.
 
 So $n^2$ is even. but they are not both even.
 
-EOP
+QED
 
 ### Theorem (No closest rational for a irrational number)
 

pages/Math4111/Math4111_L10.md

@@ -47,7 +47,7 @@ $\impliedby$ Suppose $K$ is compact relative to $Y$. Let $\{G_\alpha\}_{\alpha\i
 
 Since $k$ is compact relative to $Y$, $\{G_\alpha\cap Y\}_\alpha$ has a finite subcover $\{G_{\alpha_i}\cap Y\}_{i=1}^n$. Then $\{G_{\alpha_i}\}_{i=1}^n$ is a finite subcover of $\{G_\alpha\}_{\alpha}$ of $K$.
 
-EOP
+QED
 
 #### Theorem 2.24
 

pages/Math4111/Math4111_L11.md

@@ -31,7 +31,7 @@ And $(\bigcup_{\alpha\in A} G_\alpha)\cup F^c=X\supset K$ is an open cover of $K
 
 Since $F\subset K$, So $\Phi$ is a cover of $F$ then $\Phi\backslash \{F^c\}$ is a finite subcover of $\{G_\alpha\}_{\alpha \in A}$ of $F$.
 
-EOP
+QED
 
 ##### Corollary 2.35
 
@@ -57,7 +57,7 @@ Let $l=\max\{n_1,...,n_m\}$, then $K_l=\bigcap^{m}_{i=i} K_{n_i}$. So $K_1\subse
 
 which contradicts with $K_l\subset K_1$
 
-EOP
+QED
 
 #### Theorem 2.37
 
@@ -75,7 +75,7 @@ For each $q\in K, q\notin E'$, so $\exists$ neighborhood $\forall q\in B_{r_q}(q
 
 Then $\{V_q\}_{q\in K}$ is an open cover of $K$ , so it has a finite subcover $\{V_{q_i}\}^n_{n=1}$. Then $E\subset K\subset \bigcup_{i=1}^n$, and $\forall i,V_{q_i}\cap E\subset \{q_i\}$, then $E\subset\{q_1,...,q_n\}$
 
-EOP
+QED
 
 #### Theorem 2.38
 
@@ -101,4 +101,4 @@ Since $x$ is an upper bound of $E$ and $a_n\in E$, then $a_n\leq x$.
 
 Since $x$ is the least upper bound of $E$, and $b_n$ is an upper bound of $E$, then $x\leq b_n$. $x\in E,E\neq \phi$
 
-EOP
+QED

pages/Math4111/Math4111_L12.md

@@ -20,7 +20,7 @@ Proof:
 
 Fix $p\in X$, then $\{B_n(p)\}_{n\in \mathbb{N}}$ (specific open cover) is an open cover of $S$ (Since $\bigcup_{n\in \mathbb{N}}=X$). Since $S$ is compact, then $\exists$ a finite subcover ${n\in \mathbb{N}}_{i=1}^k=S$, let $r=max(n_1,...n_k)$, Then $S\subset B_r(p)$
 
-EOP
+QED
 
 #### Definition k-cell
 
@@ -74,7 +74,7 @@ Let $n\in \mathbb{N}$ be such that $\frac{1}{2^n}<r$. Then by $(c)$, $I(n)\subse
 
 Then $\{G_{\alpha_0}\}$ is a cover of $I_n$ which contradicts with (b)
 
-EOP
+QED
 
 #### Theorem 2.41
 
@@ -131,4 +131,4 @@ $$
 
 So $B_r(y)\cap S$ is finite. By **Theorem 2.20**, $y\notin  S$, this proves the claim so $S'\cap E=\phi$
 
-EOP
+QED

pages/Math4111/Math4111_L13.md

@@ -90,7 +90,7 @@ Thus, $\exists r>0$ such that $(w-r,w+r)\cap B=\phi$.
 
 Let $z=w+\frac{r}{2}$, then $x,y,z$ satisfy the desired properties.
 
-EOP
+QED
 
 ## Chapter 3: Numerical Sequences and Series
 
@@ -140,4 +140,4 @@ Let $n\geq N$ (arbitrary)
 
 Then $|s_n-q|=\frac{1}{n}\leq \frac{1}{N}\leq \epsilon$
 
-EOP
+QED

pages/Math4111/Math4111_L14.md

@@ -74,7 +74,7 @@ And $\forall \epsilon>0,d(p,p')<2\epsilon\implies d(p,p')=0$. So $p=p'$
 
 Let $\epsilon>0$. Choose $N\in \mathbb{N}$ such that $N>\frac{1}{\epsilon}$. Then if $n\geq N$, $d(p_n,p)<\frac{1}{n}\leq \frac{1}{N}\leq \epsilon$
 
-EOP
+QED
 
 #### Theorem 3.3
 

pages/Math4111/Math4111_L15.md

@@ -14,7 +14,7 @@ $$
     Proof:  
     Let $\epsilon>0$ be arbitrary, then $\exists N$ such that $\forall n\geq N,|a_n-a|<\epsilon$.
     Then if $n\geq N$, $|a_n-a|\leq \sqrt{|a_n-a|^2}\leq\sqrt{|a_n-a|^2+|b_n-b|^2}=|x_n-(a,b)|<\epsilon$.  
-    EOP  
+    QED  
     (b) If $x_n\to (a,b)$, then $b_n\to b$.
     This follows from the same argument from (a)
 2. Prove the $\implies$ direction.
@@ -28,7 +28,7 @@ $$
     Since $a_n\to a$, $\exists N_1$ such that $\forall n\geq N_1,|a_n-a|<\frac{\epsilon}{\sqrt{2}}$.  
     Since $b_n\to b$, $\exists N_2$ such that $\forall n\geq N_2,|b_n-b|<\frac{\epsilon}{\sqrt{2}}$.  
     Let $N=\max\{N_1,N_2\}$. Then if $n\geq N$, $|a_n-a|<\epsilon$ and $|b_n-b|<\sqrt{\frac{\epsilon^2}{2}+\frac{\epsilon^2}{2}}=\epsilon$.  
-    EOP  
+    QED  
 
 ## New Materials
 
@@ -82,7 +82,7 @@ $$
 \left|\frac{1}{s_n}-\frac{1}{s}\right|=\frac{|s-s_n|}{|s||s_n|}<\frac{\frac{\epsilon|s|^2}{2}}{|s|^2}=\epsilon
 $$
 
-EOP
+QED
 
 ### Subsequences
 
@@ -108,7 +108,7 @@ $\implies$:
 
 Thought process: show what if the sequence does not converge to $p$, then there exists a subsequence that does not converge to $p$.
 
-EOP
+QED
 
 #### Theorem 3.6
 

pages/Math4111/Math4111_L16.md

@@ -17,7 +17,7 @@ Let $n\geq N$. Since $(s_n)$ is monotonic increasing, $s_n\geq s_N>t-\epsilon$.
 
 So $(s_n)$ converges to $t$.
 
-EOP
+QED
 
 ## New materials
 
@@ -47,7 +47,7 @@ Step 3: By induction, we can get a sequence $n_1,n_2,\cdots$ such that $\forall
 
 Then $(p_{n_i})$ is a subsequence of $(p_n)$ and $p_{n_i}\to q$.
 
-EOP
+QED
 
 ### Cauchy Sequences
 
@@ -93,7 +93,7 @@ If $m,n\geq N$, then $d(p_m,p_n)\leq d(p_m,p)+d(p,p_n)<\epsilon+\epsilon=2\epsil
 
 *You can also use $\frac{\epsilon}{2}$ instead of $\epsilon$ in the above proof, just for fun.*
 
-EOP
+QED
 
 #### Lemma 3.11 (b)
 
@@ -107,7 +107,7 @@ Let $r=max\{d(p_i,p_j);1\leq i,j\leq N\}+1$.
 
 Then $\forall n\in \mathbb{N}$, $p_n\in B_r(p_N)$.
 
-EOP
+QED
 
 > Note: This proof is nearly identical to the proof of convergent sequences implies bounded.
 
@@ -148,5 +148,5 @@ By definition of supremum, the claim implies that $\forall \epsilon>0$, $diam(\o
 
 (b) By **Theorem 2.36**, $\bigcap_{n=1}^{\infty}K_n\neq \phi$. Suppose for contradiction that there are at least two distinct points $p,q\in \bigcap_{n=1}^{\infty}K_n$. Then for all $n\in \mathbb{N}$, $x,y\in K_n$ so $diam K_n\geq d(p,q)>0$. Then diameter of $K_n$ does not converge to 0.
 
-EOP
+QED
 

pages/Math4111/Math4111_L17.md

@@ -44,7 +44,7 @@ Note that **Theorem 2.41** only works for $\mathbb{R}^k$.
 
 So by (b), $(p_n)$ converges to some $p\in \overline{B(0,R)}$.
 
-EOP
+QED
 
 #### Definition 3.12
 
@@ -89,7 +89,7 @@ If $(s_n)$ is monotonic and bounded, then by previous result, $(s_n)$ converges.
 
 If $(s_n)$ is monotonic and converges, then by **Theorem 3.2(c)**, $(s_n)$ is bounded.
 
-EOP
+QED
 
 ### Upper and lower limits
 

pages/Math4111/Math4111_L18.md

@@ -79,7 +79,7 @@ Case 2: $(s_n)$ is bounded above.
 
 Let $M$ be an upper bound for $(s_n)$. Then $\{n\in\mathbb{N}:s_n\in[M,x]\}$ is infinite, by **Theorem 3.6 (b)** ($\exists$ subsequence $(s_{n_k})$ in $[x,M)$ and $\exists t\in[x,M]$ such that $s_{n_k}\to t$. This implies $t\in E$, so $x\leq t\leq S^*$).
 
-EOP
+QED
 
 #### Theorem 3.19 ("one-sided squeeze theorem")
 
@@ -104,7 +104,7 @@ By **Theorem 3.17**, $\{n\in\mathbb{N}:t_n\geq x\}$ is finite $\implies$ $\{n\in
 
 Thus $\limsup_{n\to\infty} s_n\leq \limsup_{n\to\infty} t_n$.
 
-EOP
+QED
 
 > Normal squeeze theorem: If $s_n\leq t_n\leq u_n$ for all $n\in\mathbb{N}$, and $\lim_{n\to\infty} s_n=\lim_{n\to\infty} u_n=L$, then $\lim_{n\to\infty} t_n=L$.
 >

pages/Math4111/Math4111_L19.md

@@ -15,14 +15,14 @@
     &\geq\binom{n}{4}
     \end{aligned}
     $$
-    EOP
+    QED
 2. Using part 1, show that $\lim_{n\to\infty}\frac{n^3}{2^n}=0$.  
    Proof:
    $$
    \frac{n^3}{2^n}\leq\frac{n^3}{\binom{n}{4}}
    $$
    The value of $\frac{n^3}{\binom{n}{4}}$ is decreasing when $n\geq 4$.
-   EOP
+   QED
 
 ## New materials
 
@@ -53,7 +53,7 @@ $$
 |s_m-s_n|=\left|\sum_{k=n}^{m}a_k\right|<\epsilon.
 $$
 
-EOP
+QED
 
 Special case of this theorem.
 
@@ -94,7 +94,7 @@ $$
 \left|\sum_{k=n}^{m}c_k\right|\leq \sum_{k=n}^{m}c_k<\epsilon.
 $$
 
-EOP
+QED
 
 #### Theorem 3.26 (Geometric series)
 
@@ -115,7 +115,7 @@ So $s_n=\frac{1-x^{n+1}}{1-x}$.
 
 Since $|x|<1$, $x^{n+1}$ converges to 0. So $\lim_{n\to\infty}s_n=\frac{1}{1-x}$.
 
-EOP
+QED
 
 #### Lemma 3.28 
 
@@ -149,7 +149,7 @@ $$
 
 > Fun fact: $\sum_{n=1}^{\infty}\frac{1}{n^2}=\frac{\pi^2}{6}$.
 
-EOP
+QED
 
 #### Theorem 3.27 (Cauchy condensation test)
 
@@ -191,4 +191,4 @@ So $(s_n)_{n=1}^{\infty}$ is a bounded above.
 
 By **Theorem 3.14**, $(s_n)_{n=1}^{\infty}$ converges if and only if $(t_k)_{k=0}^{\infty}$ converges.
 
-EOP
+QED

pages/Math4111/Math4111_L2.md

@@ -59,7 +59,7 @@ Suppose for contradiction $\alpha$ and $\beta$ are both LUB of $E$, then $\alpha
 
 WLOG $\alpha>\beta$ and $\beta>\alpha$.
 
-EOP
+QED
 
 We write $\sup E$ to denote the LUB of $E$.
 

pages/Math4111/Math4111_L20.md

@@ -114,7 +114,7 @@ Therefore, $e\leq \liminf_{n\to\infty} t_n\leq \limsup_{n\to\infty} t_n\leq e$.
 
 So $\lim_{n\to\infty} t_n$ exists and $\lim_{n\to\infty} t_n = e$.
 
-EOP
+QED
 
 #### Theorem 3.32
 
@@ -156,7 +156,7 @@ $$
 
 Contradiction.
 
-EOP
+QED
 
 ### The root and ratio tests
 
@@ -190,7 +190,7 @@ Thus $a_n\not\to 0$, $\sum_{n=0}^{\infty} a_n$ diverges.
 
 (c) $\sum_{n=0}^{\infty} \frac{1}{n}$ and $\sum_{n=0}^{\infty} \frac{1}{n^2}$ both have $\alpha = 1$. but the first diverges and the second converges.
 
-EOP
+QED
 
 #### Theorem 3.34 (Ratio test)
 
@@ -232,7 +232,7 @@ i.e. $\forall n\geq N, |a_n| < \beta^{n-N}|a_N|=\beta^n(\beta^{-N}|a_N|)$.
 
 Since $\sum_{n=N}^{\infty} \beta^n$ converges, by comparison test, $\sum_{n=0}^{\infty} a_n$ converges.
 
-EOP
+QED
 
 We will skip **Theorem 3.37**. One implication is that if ratio test can be applied, then root test can be applied.
 
@@ -267,4 +267,4 @@ $$
 
 By root test, the series converges absolutely for all $z\in\mathbb{C}$ with $|z| < R$.
 
-EOP
+QED

pages/Math4111/Math4111_L21.md

@@ -56,7 +56,7 @@ $$
 \end{aligned}
 $$
 
-EOP
+QED
 
 #### Theorem 3.42 (Dirichlet's test)
 
@@ -101,7 +101,7 @@ $$
 
 So $\sum a_nb_n$ converges.
 
-EOP
+QED
 
 #### Theorem 3.43 (Alternating series test)
 
@@ -122,7 +122,7 @@ So $|A_n|\leq 1$ for all $n\in \mathbb{N}$.
 
 By Theorem 3.42, $\sum_{n=1}^\infty a_n b_n$ converges.
 
-EOP
+QED
 
 Example:
 
@@ -161,7 +161,7 @@ So $|A_n|\leq \frac{2}{|1-z|}$ for all $n\in \mathbb{N}$.
 
 By Dirichlet's test, $\sum_{n=0}^\infty b_nz^n$ 
 
-EOP
+QED
 
 ### Absolute convergence
 
@@ -183,7 +183,7 @@ $$
 \sum_{n=0}^\infty |a_n|\geq \sum_{n=0}^\infty a_n
 $$
 
-EOP
+QED
 
 Rearrangement of series:
 
@@ -257,4 +257,4 @@ For every $n\in \mathbb{N}$, there exists a $p$ such that $\{1,2,\cdots,n\}\subs
 
 Then $|s_n-t_n|\leq \sum_{k=N+1}^\infty |a_k|$.
 
-EOP
+QED

pages/Math4111/Math4111_L22.md

@@ -77,7 +77,7 @@ This proves that $\lim_{n\to\infty} |s_n - t_n| = 0$.
 
 Since $\lim_{n\to\infty} s_n$ exists, $\lim_{n\to\infty} s_n = \lim_{n\to\infty} t_n$.
 
-EOP
+QED
 
 #### Theorem 3.54
 
@@ -137,7 +137,7 @@ Then: $(p_n)$ is a sequence in $E\backslash\{p\}$ with $d_X(p_n,p) = \frac{1}{n}
 
 So $\lim_{n\to\infty} f(p_n) \neq q$.
 
-EOP
+QED
 
 With this theorem, we can use the properties of limit of sequences to study limits of functions.
 

pages/Math4111/Math4111_L24.md

@@ -33,7 +33,7 @@ $\impliedby$: Suppose for every open set $V\subset Y$, $f^{-1}(V)$ is open in $X
 
 Since $p\in f^{-1}(B_\epsilon(f(p)))$ and $f^{-1}(B_\epsilon(f(p)))$ is open, $\exists \delta > 0$ such that $B_\delta(p)\subset f^{-1}(B_\epsilon(f(p)))$. Therefore, $f(B_\delta(p))\subset B_\epsilon(f(p))$. This shows that $f$ is continuous.
 
-EOP
+QED
 
 #### Corollary 4.8
 
@@ -82,7 +82,7 @@ Proof:
 
 Let $p\in \mathbb{R}$ and $\epsilon > 0$. Let $\delta = \epsilon$. Then, $\forall x\in \mathbb{R}$, if $|x-p|<\delta$, then $|f(x)-f(p)| = |x-p| < \delta = \epsilon$.
 
-EOP
+QED
 
 Therefore, by **Theorem 4.9**, $f(x) = x^2$ is continuous. $f(x) = x^3$ is continuous... So all polynomials are continuous.
 
@@ -131,7 +131,7 @@ By Theorem 2.41, $f(X)$ is closed and bounded.
 
 By Theorem 2.28, $\sup f(X)$ and $\inf f(X)$ exist and are in $f(X)$. Let $p_0\in X$ such that $f(p_0) = \sup f(X)$. Let $q_0\in X$ such that $f(q_0) = \inf f(X)$.
 
-EOP
+QED
 
 ---
 
@@ -151,7 +151,7 @@ Proof:
 
 See the textbook.
 
-EOP
+QED
 
 ---
 
@@ -191,7 +191,7 @@ Since $A$ and $B$ are separated, $\overline{A}\cap B = \phi$ and $\overline{B}\c
 
 Therefore, $\overline{G}\cap H = \phi$ and $\overline{H}\cap G = \phi$.
 
-EOP
+QED
 
 #### Theorem 4.23 (Intermediate Value Theorem)
 
@@ -207,4 +207,4 @@ $f(a)$ and $f(b)$ are real numbers in $f([a,b])$, and $c$ is a real number betwe
 
 By **Theorem 2.47**, $c\in f([a,b])$.
 
-EOP
+QED

pages/Math4111/Math4111_L25.md

@@ -99,7 +99,7 @@ _The $\epsilon$ bound would not hold if we only had pointwise convergence._
 
 $|f_N(x) - f_N(p)| < 3\epsilon$.
 
-EOP
+QED
 
 > Recall: If $(s_n)$ is a sequence in $\mathbb{R}$, then $(s_n)$ converges to $s$ if and only if it is Cauchy.  
 > i.e. $\forall \epsilon > 0$, $\exists N\in \mathbb{N}$, $\forall n, m\geq N$, $|s_n - s_m| < \epsilon$.
@@ -147,7 +147,7 @@ $$
 \end{aligned}
 $$
 
-EOP
+QED
 
 Example:
 

pages/Math4111/Math4111_L3.md

@@ -22,10 +22,12 @@ Let $S=\mathbb{Z}$.
 
 ## Continue
 
-### LUBP
+### LUBP (The least upper bound property)
 
 Proof that $LUBP\implies GLBP$.
 
+Proof:
+
 Let $S$ be an ordered set with LUBP. Let $B<S$ be non-empty and bounded below.
 
 Let $L=y\in S:y$ is a lower bound of $B$. From the picture, we expect $\sup L=\inf B$ First we'll show $\sup L$ exists.
@@ -55,6 +57,8 @@ Let's say $\alpha=sup\ L$. We claim that $\alpha=inf\ B$. We need to show $2$ th
 
 Thus $\alpha=inf\ B$
 
+QED
+
 ### Field
 
 |                | addition                                                    | multiplication                                                 |

pages/Math4111/Math4111_L4.md

@@ -39,7 +39,7 @@ This implies $(m+1)x>\alpha$
 
 Since $(m+1)x\in \alpha$, this contradicts the fact that $\alpha$ is an upper bound of $A$.
 
-EOP
+QED
 
 ### $\mathbb{Q}$ is dense in $\mathbb{R}$
 
@@ -59,7 +59,7 @@ By Archimedean property, $\exist n\in \mathbb{N}$ such that $n(y-x)>1$, and $\ex
 
 So $-m_2<nx<m_1$. Thus $\exist m\in \mathbb{Z}$ such that $m-1\leq nx<m$ (Here we use a property of $\mathbb{Z}$) We have $ny>1+nx\geq 1+(m-1)=m$
 
-EOP
+QED
 
 ### $\sqrt{2}\in \mathbb{R}$, $(\sqrt[n]{x}\in\mathbb{R})$
 

pages/Math4111/Math4111_L5.md

@@ -94,6 +94,8 @@ So want $k\leq \frac{y^n-x}{ny^{n-1}}$
 
 [For actual proof, see the text.]
 
+QED
+
 ### Complex numbers
 
 1. $=\{a+bi:a,b\in \mathbb{R}\}$.
@@ -149,7 +151,9 @@ $$
 (\sum a_j b_j)^2=(\sum a_j^2)(\sum b_j^2)
 $$
 
-Proof for real numbers:
+Proof:
+
+For real numbers:
 
 Let $A=\sum a_j^2,B=\sum b_j^2, C=\sum a_j b_j$, want to show $C^2\leq AB$
 
@@ -165,6 +169,8 @@ let $t=C/B$ to get $0\leq A-2(C/B)C+(C/B)^2B=A-\frac{C^2}{B}$
 
 to generalize this to $\mathbb{C}$, $A=\sum |a_j|^2,B=\sum |b_j|^2,C=\sum |a_j \bar{b_j}|$.
 
+QED
+
 ### Euclidean spaces
 
 Nothing much to say. lol.

pages/Math4111/Math4111_L6.md

@@ -126,7 +126,9 @@ $A$ is countable, $n\in \mathbb{N}$,
 
 $\implies A^n=\{(a_{1},...,a_{n}):a_1\in A, a_n\in A\}$, is countable.
 
-Proof: Induct on $n$, 
+Proof:
+
+Induct on $n$,
 
 Base case $n=1$,
 
@@ -136,14 +138,20 @@ Induction step: suppose $A^{n-1}$ is countable. Note $A^n=\{(b,a):b\in A^{n-1},a
 
 Since $b$ is fixed, so this is in 1-1 correspondence with $A$, so it's countable by Theorem 2.12.
 
+QED
+
 #### Theorem 2.14
 
 Let $A$ be the set of all sequences for 0s and 1s. Then $A$ is uncountable.
 
-Proof: Let $E\subset A$ be a countable subset. We'll show $A\backslash E\neq \phi$ (i.e.$\exists t\in A$ such that $t\notin E$)
+Proof:
+
+Let $E\subset A$ be a countable subset. We'll show $A\backslash E\neq \phi$ (i.e.$\exists t\in A$ such that $t\notin E$)
 
 $E$ is countable so we can list it's elements $S_1,S_2,S_3,...$.
 
 Then we define a new sequence $t$ which differs from $S_1$'s first bit and $S_2$'s second bit,...
 
 This is called Cantor's diagonal argument.
+
+QED

pages/Math4111/Math4111_L7.md

@@ -80,12 +80,16 @@ Let $(X,d)$ be  a metric space, $\forall p\in X,\forall r>0$, $B_r(p)$ is an ope
 
 *every ball is an open set*
 
-Proof: Let $q\in B_r(p)$.
+Proof:
+
+Let $q\in B_r(p)$.
 
 Let $h=r-d(p,q)$.
 
 Since $q\in B_r(p),h>0$. We claim that $B_h(q)$. Then $d(q,s)<h$, so $d(p,s)\leq d(p,q)+d(q,s)<d(p,q)+h=r$. (using triangle inequality) So $S\in B_r(p)$.
 
+QED
+
 ### Closed sets
 
 1. $E\subset X,p\in X$. We say $p$ is a limit point of $E$ if $\forall r>0, (B_r(p)\cap E)\backslash {p}\neq \phi$.
@@ -94,7 +98,9 @@ Since $q\in B_r(p),h>0$. We claim that $B_h(q)$. Then $d(q,s)<h$, so $d(p,s)\leq
 
 2. $E$ is closed if $E'\subset E$
 
-Example: $X=\mathbb{R}^2$, $E=[0,1)\times [0,1)$.
+Example:
+
+$X=\mathbb{R}^2$, $E=[0,1)\times [0,1)$.
 
 $(1,1)$ is a limit point.
 

pages/Math4111/Math4111_L8.md

@@ -41,7 +41,7 @@ let $B_s(p)\cap E)\backslash \{p\}={q_1,...,q_n}$
 
 Then $(B_s(p)\cap E)\backslash \{p\}=\phi$, so $p\notin E$
 
-EOP
+QED
 
 #### Theorem 2.22 De Morgan's law
 
@@ -95,7 +95,7 @@ $$
 
 So $(E^c)'\subset E^c$
 
-EOP
+QED
 
 #### Theorem 2.24
 
@@ -105,7 +105,7 @@ Proof:
 
 Suppose $\forall \alpha, G_\alpha$ is open. Let $x\in \bigcup _{\alpha} G_\alpha$. Then $\exists \alpha_0$ such that $x\in G_{\alpha_0}$. Since $G_{\alpha_0}$ is open, $\exists r>0$ such that $B_r(x)\subset G_{\alpha_0}$ Then $B_r(x)\subset G_{\alpha_0}\subset \bigcup_{\alpha} G_\alpha$
 
-EOP
+QED
 
 ##### A finite intersection of open set is open
 
@@ -117,7 +117,7 @@ Let $x\in \bigcap^n_{i=1}G_i$, then $\forall i\in \{1,..,n\}$ and $G_i$ is open,
 
 Let $r=min\{r_1,...,r_n\}$. Then $\forall i\in \{1,...,n\}$. $B_r(x)\subset B_{r_i}(x)\subset G_i$. So $B_r(x)\subset \bigcup_{i=1}^n G_i$
 
-EOP
+QED
 
 The other two can be proved by **Theorem 2.22,2.23**
 
@@ -147,4 +147,4 @@ This proves (b)
 
 So $\bar{E}^c$ is open
 
-EOP
+QED

pages/Math4111/Math4111_L9.md

@@ -37,7 +37,7 @@ Let $h>0$. Since $y-h$ is not an upper bound of $E$, $\exists x\in E$ such that
 
 Since $y$ is an upper bound of $E$, $x\leq y$. So $x\in B_n(y)\cap E$, so $B_h(y)\cap E\neq \phi$.
 
-EOP
+QED
 
 #### Remark 2.29
 
@@ -75,7 +75,7 @@ To show $G\cap Y\subset E$.
 
 $G\cap Y=\left(\bigcup_{p\in E}V_p\right)\cap Y=\bigcup_{p\in E}(V_p\cap Y)\subset E$
 
-EOP
+QED
 
 ### Compact sets
 
@@ -101,4 +101,4 @@ as we can build an infinite open cover $\bigcup_{i\in Z} (i,i+2)$ and it does no
 
 Suppose we consider the sub-collection $\{n_i,n_i+2:i=1,..,k\}$, Then $N+3$ is not in the union, where $N=max\{n_1,...,n_k\}$.
 
-EOP
+QED

pages/Math4111/index.md

@@ -10,7 +10,7 @@ Topics include:
 4. Convergence of Series and Sequences
 5. Limits and Continuity
 
-The course is taught by [Alan Chang](https://math.wustl.edu/people/alan-chang).
+The course is taught by [Alan Chang](https://math.wustl.edu/pQEDle/alan-chang).
 
 It is easy in my semester perhaps, it is the first course I got 3 perfect scores in exams. (Unfortunately, I did not get the extra credit for the third midterm exam.)