Fix typo

Fix typos introduces more

pages/CSE347/CSE347_L5.md

@@ -6,7 +6,7 @@
 - In general, we can design an algorithm to map instances of a new problem to instances of known solvable problem (e.g., max-flow) to solve this new problem!
 - Mapping from one problem to another which preserves solutions is called reduction.
 
-## Reduction: Basic Idea
+## Reduction: Basic Ideas
 
 Convert solutions to the known problem to the solutions to the new problem
 

pages/CSE347/CSE347_L7.md

@@ -45,7 +45,7 @@ Assumption: No clause contains both a literal and its negation.
 
 Need to: construct $S$ of positive numbers and a target $t$
 
-Idea of construction:
+Ideas of construction:
 
 For 3-SAT instance $\Psi$:
 
@@ -276,7 +276,7 @@ Consider an instance of SSS: $\{ a_1,a_2,\cdots,a_n\}$ and sum $b$. We can creat
 
 Then we prove that the scheduling instance is a "yes" instance if and only if the SSS instance is a "yes" instance.
 
-Idea of proof:
+Ideas of proof:
 
 If there is a subset of $\{a_1,a_2,\cdots,a_n\}$ that sums to $b$, then we can schedule the jobs in that order on one machine.
 

pages/CSE347/CSE347_L9.md

@@ -38,7 +38,7 @@ Answer: The adversary can make the runtime of each operation $\Theta(n)$ by simp
 
 We don't want the adversary to know the hash function based on just looking at the code.
 
-Idea: Randomize the choice of the hash function.
+Ideas: Randomize the choice of the hash function.
 
 ### Randomized Algorithm
 
@@ -57,7 +57,7 @@ $$O(n)=E[T(n)]$$ or some other probabilistic quantity.
 
 #### Randomization can help
 
-Idea: Randomize the choice of hash function $h$ from a family of hash functions, $H$.
+Ideas: Randomize the choice of hash function $h$ from a family of hash functions, $H$.
 
 If we randomly pick a hash function from this family, then the probability that the hash function is bad on **any particular** set $S$ is small.
 

pages/CSE442T/CSE442T_L12.md

@@ -82,7 +82,7 @@ The NBT(Next bit test) is complete.
 
 If $\{X_n\}$ on $\{0,1\}^{l(n)}$ passes NBT, then it's pseudorandom.
 
-Idea of proof: full proof is on the text.
+Ideas of proof: full proof is on the text.
 
 Our idea is that we want to create $H^{l(n)}_n=\{X_n\}$ and $H^0_n=\{U_{l(n)}\}$
 
@@ -137,7 +137,7 @@ The other part of proof will be your homework, damn.
 
 If one-way function exists, then Pseudorandom Generator exists.
 
-Idea of proof:
+Ideas of proof:
 
 Let $f:\{0,1\}^n\to \{0,1\}^n$ be a strong one-way permutation (bijection).
 

pages/CSE442T/CSE442T_L13.md

@@ -16,7 +16,7 @@ $$
 Pr[x\gets \{0,1\}^n;y=f(x);A(1^n,y)=h(x)]\leq \frac{1}{2}+\epsilon(n)
 $$
 
-Idea: $f:\{0,1\}^n\to \{0,1\}^*$ is a one-way function.
+Ideas: $f:\{0,1\}^n\to \{0,1\}^*$ is a one-way function.
 
 Given $y=f(x)$, it is hard to recover $x$. A cannot produce all of $x$ but can know some bits of $x$.
 
@@ -46,7 +46,7 @@ $\langle x,1^n\rangle=x_1+x_2+\cdots +x_n\mod 2$
 
 $\langle x,0^{n-1}1\rangle=x_ n$
 
-Idea of proof:
+Ideas of proof:
 
 If A could reliably find $\langle x,1^n\rangle$, with $r$ being completely random, then it could find $x$ too often.
 

pages/CSE442T/CSE442T_L14.md

@@ -123,7 +123,7 @@ $Enc_F(m):$ let $r\gets U_n$; output $(r,F(r)\oplus m)$.
 
 $Dec_F(m):$ Given $(r,c)$, output $m=F(r)\oplus c$.
 
-Idea: Adversary sees $r$ but has no idea about $F(r)$. (we choose all outputs at random)
+Ideas: Adversary sees $r$ but has no Ideas about $F(r)$. (we choose all outputs at random)
 
 If we could do this, this is MMS (multi-message secure).
 

pages/CSE442T/CSE442T_L16.md

@@ -77,7 +77,7 @@ With $g^a,g^b$ no one can compute $g^{ab}$.
 
 ### Public key encryption scheme
 
-Idea: The recipient Bob distributes opened Bob-locks
+Ideas: The recipient Bob distributes opened Bob-locks
 
 - Once closed, only Bob can open it.
 

pages/CSE442T/CSE442T_L17.md

@@ -110,7 +110,7 @@ $\{p\gets \tilde{\Pi_n};y\gets Gen_q;a,b,\bold{z}\gets \mathbb{Z}_q:(p,y,y^a,y^b
 
 So DDH assumption implies discrete logarithm assumption.
 
-Idea:
+Ideas:
 
 If one can find $a,b$ from $y^a,y^b$, then one can find $ab$ from $y^{ab}$ and compare to $\bold{z}$ to check whether $y^\bold{z}$ is a valid DDH tuple.
 

pages/CSE442T/CSE442T_L19.md

@@ -28,7 +28,7 @@ This is not more than one-time secure since the adversary can ask oracle for $Si
 
 We will show it is one-time secure
 
-Idea of proof:
+Ideas of proof:
 
 Say their query is $Sign_{sk}(0^n)$ and reveals $pk_0$. 
 

pages/CSE442T/CSE442T_L20.md

@@ -104,7 +104,7 @@ One-time secure:
 
 Then ($Gen',Sign',Ver'$) is one-time secure.
 
-Idea of Proof:
+Ideas of Proof:
 
 If the digital signature scheme ($Gen',Sign',Ver'$) is not one-time secure, then there exists an adversary $\mathcal{A}$ which can ask oracle for one signature on $m_1$ and receive $\sigma_1=Sign'_{sk'}(m_1)=Sign_{sk}(h_i(m_1))$.
 

pages/Math4111/Exam_reviews/Math4111_E3.md

@@ -2,9 +2,9 @@
 
 ## Relations between series and topology (compactness, closure, etc.)
 
-Limit points $E'=\{x\in\mathbb{R}:\forall r>0, B_r(x)\backslash\{x\}\cap E\neq\emptyset\}$
+Limit points $E'=\{x\in\mathbb{R}:\forall r>0, B_r(x)\backslash\{x\}\cap E\neq\phi\}$
 
-Closure $\overline{E}=E\cup E'=\{x\in\mathbb{R}:\forall r>0, B_r(x)\cap E\neq\emptyset\}$
+Closure $\overline{E}=E\cup E'=\{x\in\mathbb{R}:\forall r>0, B_r(x)\cap E\neq\phi\}$
 
 $p_n\to p\implies \forall \epsilon>0, \exists N$ such that $\forall n\geq N, p_n\in B_\epsilon(p)$
 
@@ -24,7 +24,7 @@ Rudin Proof:
 
 Rudin's proof uses a fact from Chapter 2.
 
-If $E$ is compact, and $S\subseteq E$ is infinite, then $S$ has a limit point in $E$ ($S'\cap E\neq\emptyset$).
+If $E$ is compact, and $S\subseteq E$ is infinite, then $S$ has a limit point in $E$ ($S'\cap E\neq\phi$).
 
 ## Examples of Cauchy sequence that does not converge
 

pages/Math4111/Math4111_L12.md

@@ -30,7 +30,7 @@ A 2-cell is a set of the form $[a_1,b_1]\times[a_2,b_2]$
 
 Theorem 2.38 replace with "closed and bounded intervals" to "k-cells".
 
-Idea of Proof:
+Ideas of Proof:
 
 Apply the Theorem to each dimension separately.
 

pages/Math4111/Math4111_L16.md

@@ -146,7 +146,7 @@ This proves the claim.
 
 By definition of supremum, the claim implies that $\forall \epsilon>0$, $diam(\overline{E})\leq 2\epsilon+diam E$. So $diam(\overline{E})\leq diam E$.
 
-(b) By **Theorem 2.36**, $\bigcap_{n=1}^{\infty}K_n\neq \emptyset$. 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.
+(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
 

pages/Math4111/Math4111_L20.md

@@ -78,7 +78,7 @@ So if $\limsup_{n\to\infty} t_n \leq \limsup_{n\to\infty} s_n$, then $\lim_{n\to
 
 Now we will show $\limsup_{n\to\infty} t_n \geq e$.
 
-Idea: (special case of the argument)
+Ideas: (special case of the argument)
 
 If $n\geq 2$, then
 

pages/Math4111/Math4111_L21.md

@@ -230,7 +230,7 @@ $$
 \sum_{n=0}^\infty a_n=\sum_{n=0}^\infty a_{f(n)}
 $$
 
-Idea of proof:
+Ideas of proof:
 
 Let $f:\mathbb{N}\to \mathbb{N}$ be a bijection.
 

pages/Math4111/Math4111_L24.md

@@ -1 +1,188 @@
 # Lecture 24
+
+## Reviews
+
+Let $f: X\to Y$. Consider the following statement:
+
+"$f$ is continuous $\iff$ for every open set $V\in Y$, $f^{-1}(V)$ is open in $X$."
+
+1. To give a direct proof of the $\implies$ direction, what must be the first few steps be?
+
+2. To give a direct proof of the $\impliedby$ direction, what must be the first few steps be?
+3. Try to complete the proofs of both directions.
+
+> A function $f:X\to Y$ is continuous if $\forall p\in X$, $\forall \epsilon > 0$, $\exists \delta > 0$ such that $f(B_\delta(p))\subset B_\epsilon(f(p))$. (_For every point in a ball of $B_\delta(p)$, there is a ball of $B_\epsilon(f(p))$ that contains the image of the point._)
+> 
+> A set $V\subset Y$ is open if $\forall q\in V$, $\exists r>0$ such that $B_r(q)\subset V$.
+
+## New materials
+
+### Continuity and open sets
+
+#### Theorem 4.8
+
+A function $f:X\to Y$ is continuous if and only if for every open set $V\subset Y$, $f^{-1}(V)$ is open in $X$.
+
+Proof:
+
+$\implies$: Suppose $f$ is continuous. Let $V\subset Y$ be open. Let $p\in f^{-1}(V)$. Since $f(p)\in V$, $\exists \epsilon > 0$ such that $B_\epsilon(f(p))\subset V$.
+
+Since $f$ is continuous, $\exists \delta > 0$ such that $f(B_\delta(p))\subset B_\epsilon(f(p))\subset V$. Therefore, $B_\delta(p)\subset f^{-1}(V)$. This shows that $f^{-1}(V)$ is open.
+
+$\impliedby$: Suppose for every open set $V\subset Y$, $f^{-1}(V)$ is open in $X$. Let $p\in X$ and $\epsilon > 0$. Let $B_\epsilon(f(p))\in V$. Then $f^{-1}(B_\epsilon(f(p)))$ 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
+
+#### Corollary 4.8
+
+$f$ is continuous if and only if for every closed set $C\subset Y$, $f^{-1}(C)$ is closed in $X$.
+
+Ideas of proof:
+
+- $C$ closed in $Y\iff Y\backslash C$ open in $Y$
+- $f^{-1}(C)$ closed in $X\iff f^{-1}(Y\backslash C)$ open in $X$
+- $f^{-1}(Y\backslash C) = X\backslash f^{-1}(C)$
+
+Continue this proof by yourself.
+
+#### Theorem 4.7
+
+Composition of continuous functions is continuous.
+
+Suppose $X,Y,Z$ are metric spaces, $E\subset X$, $f:E\to Y$ is continuous, and $g:Y\to Z$ is continuous. Then $g\circ f:E\to Z$ is continuous.
+
+Ideas of proof:
+
+- Let $B_\epsilon(g(f(p)))\subset Z$
+- $g(f(B_\delta(p)))\subset B_\epsilon(g(f(p)))$
+- $f(B_\delta(p))$ is open in $Y$
+- $g^{-1}(B_\epsilon(g(f(p)))$ is open in $Y$
+- $(g\circ f)^{-1}(B_\epsilon(g(f(p)))) = f^{-1}(g^{-1}(B_\epsilon(g(f(p)))))$
+- $f^{-1}(g^{-1}(B_\epsilon(g(f(p))))) = (g\circ f)^{-1}(B_\epsilon(g(f(p))))$
+
+Apply Theorem 4.8 to complete the proof.
+
+#### Theorem 4.9
+
+For $f:X\to \mathbb{C},g:X\to \mathbb{C}$ are continuous, then, $f+g,f/g$ are continuous.
+
+Ideas of proof:
+
+We can reduce this theorem to Theorem about limits and apply what you learned in chapter 3.
+
+#### Examples of continuous functions 4.11
+
+> $\forall p\in \mathbb{R}$, $\forall \epsilon > 0$, $\exists \delta > 0$ such that $\forall x\in \mathbb{R}$, $|x-p|<\delta\implies |f(x)-f(p)|<\epsilon$.
+
+(a). $f(x) = \mathbb{R}\to \mathbb{R},f(x) = x$ is continuous. boring.
+
+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
+
+Therefore, by **Theorem 4.9**, $f(x) = x^2$ is continuous. $f(x) = x^3$ is continuous... So all polynomials are continuous.
+
+(b). $f:\mathbb{R}^k\to \mathbb{R},f(x)=|x|$ is continuous.
+
+Ideas of proof:
+
+- $|f(x)-f(p)| = ||x|-|p||\leq |x-p|$ By reverse triangle inequality.
+- Let $\epsilon > 0$. Let $\delta = \epsilon$.
+
+### Continuity and compactness
+
+#### Theorem 4.13
+
+A mapping of $f$ of a set $E$ into a metric space $Y$ is said to be **bounded** if there is a real number $M$ such that $|f(x)|\leq M$ for all $x\in E$.
+
+#### Theorem 4.14
+
+$f:X\to Y$ is continuous. If $X$ is compact, then $f(X)$ is compact.
+
+Proof strategy:
+
+For every open cover $\{V_\alpha\}_{\alpha\in A}$ of $f(X)$, there exists a corresponding open cover $\{f^{-1}(V_\alpha)\}_{\alpha\in A}$ of $X$.
+
+Since $X$ is compact, there exists a finite subcover $\{f^{-1}(V_\alpha)\}_{\alpha\in A}$ of $X$. Let the finite subcover be $\{f^{-1}(V_\alpha)\}_{i=1}^n$.
+
+Then, $\{V_\alpha\}_{i=1}^n$ is a finite subcover of $\{V_\alpha\}_{\alpha\in A}$ of $f(X)$.
+
+See the detailed proof in the textbook.
+
+#### Theorem 4.16 (Extreme Value Theorem)
+
+Suppose $X$ is a compact metric space and $f:X\to \mathbb{R}$ is continuous. Then $f$ has a maximum and a minimum on $X$.
+
+i.e.
+
+$$
+\exists p_0,q_0\in X\text{ such that }f(p_0) = \sup f(X)\text{ and }f(q_0) = \inf f(X).
+$$
+
+Proof:
+
+By Theorem 4.14, $f(X)$ is compact.
+
+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
+
+### Continuity and connectedness
+
+> **Definition 2.45**: Let $X$ be a metric space. $A,B\subset X$ are **separated** if $\overline{A}\cap B = \phi$ and $\overline{B}\cap A = \phi$.
+> 
+> $E\subset X$ is **disconnected** if there exist two separated sets $A$ and $B$ such that $E = A\cup B$.
+> 
+> $E\subset X$ is **connected** if $E$ is not disconnected.
+
+#### Theorem 4.22
+
+$f:X\to Y$ is continuous, $E\subset X$. If $E$ is connected, then $f(E)$ is connected.
+
+Proof:
+
+We will prove the contrapositive statement: if $f(E)$ is disconnected, then $E$ is disconnected.
+
+Suppose $f(E)$ is disconnected. Then there exist two separated sets $A$ and $B\in Y$ such that $f(E) = A\cup B$.
+
+Let $G = f^{-1}(A)\cap E$ and $H = f^{-1}(B)\cap E$.
+
+We have:
+
+$f(E)=A\cup B\implies E = G\cup H$
+
+Since $A$ and $B$ are nonempty, $A,B\subset f(E)$, this implies that $G$ and $H$ are nonempty.
+
+To complete the proof, we need to show $\overline{G}\cap H = \phi$ and $\overline{H}\cap G = \phi$.
+
+We have $G\subset f^{-1}(A)\cap E\subset f^{-1}(A)\subset f^{-1}(\overline{A})$ Since $\overline{A}$ is closed, $f^{-1}(\overline{A})$ is closed. This implies that $\overline{G}\subset f^{-1}(\overline{A})$.
+
+So $\overline{G}\subset f^{-1}(\overline{A})$ and $\overline{H}\subset f^{-1}(\overline{B})$.
+
+Since $A$ and $B$ are separated, $\overline{A}\cap B = \phi$ and $\overline{B}\cap A = \phi$.
+
+Therefore, $\overline{G}\cap H = \phi$ and $\overline{H}\cap G = \phi$.
+
+EOP
+
+#### Theorem 4.23 (Intermediate Value Theorem)
+
+Let $f:[a,b]\to \mathbb{R}$ be continuous. If $c$ is a real number between $f(a)$ and $f(b)$, then there exists a point $x\in [a,b]$ such that $f(x) = c$.
+
+Ideas of proof:
+
+Use Theorem 2.47. A subset $E$ of $\mathbb{R}$ is connected if and only if it has the following property: if $x,y\in E$ and $x<z<y$, then $z\in E$.
+
+Since $[a,b]$ is connected, by **Theorem 4.22**, $f([a,b])$ is connected.
+
+$f(a)$ and $f(b)$ are real numbers in $f([a,b])$, and $c$ is a real number between $f(a)$ and $f(b)$.
+
+By **Theorem 2.47**, $c\in f([a,b])$.
+
+EOP

pages/Math4111/Math4111_L26.md

@@ -1 +0,0 @@
-# Lecture 26

pages/Math4111/Math4111_L27.md

@@ -1 +0,0 @@
-# Lecture 27

pages/Math4111/Math4111_L28.md

@@ -1 +0,0 @@
-# Lecture 28

pages/Math4111/_meta.js

@@ -26,22 +26,7 @@ export default {
     Math4111_L20: "Lecture 20",
     Math4111_L21: "Lecture 21",
     Math4111_L22: "Lecture 22",
-    Math4111_L23: {
-        display: 'hidden'
-    },
-    Math4111_L24: {
-        display: 'hidden'
-    },
-    Math4111_L25: {
-        display: 'hidden'
-    },
-    Math4111_L26: {
-        display: 'hidden'
-    },  
-    Math4111_L27: {
-        display: 'hidden'
-    },
-    Math4111_L28: {
-        display: 'hidden'
-    }
+    Math4111_L23: "Lecture 23",
+    Math4111_L24: "Lecture 24",
+    Math4111_L25: "Lecture 25"
 }

pages/Math429/Math429_L10.md

@@ -35,7 +35,7 @@ Suppose $V,W$ are finite dimensional with $dim(V)>dim(W)$, then there are no inj
 
 Suppose $V,W$ are finite dimensional with $dim(V)<dim(W)$, then there are no surjective maps from $V$ to $W$.
 
-ideas of Proof: relies on **Theorem 3.21** $dim(null(T))>0$
+Ideas of Proof: relies on **Theorem 3.21** $dim(null(T))>0$
 
 ### Linear Maps and Linear Systems 3EX-1