update notations and fix typos

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