.

MA1006/decls.tex

@@ -9,7 +9,7 @@
 \author{}
 
 \newcommand{\paren}[1]{\left(#1\right)}
-\newcommand{\powerset}[1]{\mathbb{P}\paren{#1}}
+\newcommand{\powerset}[1]{\mathcal{P}\paren{#1}}
 \renewcommand{\Re}[1]{\operatorname{\mathbb{R}e}\paren{#1}}
 \renewcommand{\Im}[1]{\operatorname{\mathbb{{I}}m}\paren{#1}}
 \newcommand{\C}{\mathbb{C}}

MA1511/decls.tex

@@ -9,7 +9,7 @@
 \author{}
 
 \newcommand{\paren}[1]{\left(#1\right)}
-\newcommand{\powerset}[1]{\mathbb{P}\paren{#1}}
+\newcommand{\powerset}[1]{\mathcal{P}\paren{#1}}
 \renewcommand{\Re}[1]{\operatorname{\mathbb{R}e}\paren{#1}}
 \renewcommand{\Im}[1]{\operatorname{\mathbb{{I}}m}\paren{#1}}
 \newcommand{\C}{\mathbb{C}}

MA1511/sets.tex

@@ -239,15 +239,124 @@ $f: A \to X/\sim$ defined by $f(a) = [a]$
 \\\\\\
 A function $f: X \to Y$ is \emph{compatible} iff $x \sim y \implies f(x) = f(y)~\forall x, y \in X$ \\
 For a \emph{compatible} function, $\Bar{f}: X/\sim~\to Y$ exists and is defined by $\Bar{f}([x]) = f(x)$
-\section*{Integers modulo $k$}
+\subsection*{Integers modulo $k$}
 Fix some $k \in \N^+$ \\
 Define $\sim$ on $\Z$ by $n \sim m \iff n - m$ is a multiple of $k$ \\
 $\sim$ is an equivalence relation \\
 $[0, k)\cap\N$ is a complete set of representatives for $\sim$ \\
 $Z/k$ is the set of integers modulo $k$, $n \equiv m~(\operatorname{mod} k) \iff n \sim m$ \\
+\([m] = [m]_k\) \\
 $+$ and $\times$ on $\Z/k$:
 \begin{align*}
 	[n] + [m] = [n + m] \\
 	[n][m] = [nm]
 \end{align*}
+\section*{Countable Sets}
+A set $X$ is finite if $\exists n \geq 0. $ a bijection $\{1, ...n\} \to X$ \\
+Pigeonhole Principle: for finite $X$, any injective $f: X \to X$ is also surjective. \\
+$\N$ is infinite. Proof: $f: \N \to \N$ is trivially injective, and $\neg\exists x.~f(x) = 0$, and so not surjective. \\
+By the inverse of the Pigeonhole Principle, $\N$ is infinite.\\
+A set $X$ is \emph{countably infinite} iff there exists a bijection $\N \to X$. \\
+A set is \emph{countable} iff it is finite or countably infinite. \\
+Any subset of $\N$ is countable. Proof: Let $X \in \N$.\\
+If $X$ is finite, it's trivially countable.
+Otherwise, $X$ is infinite and it must be shown that $X$ is countably infinite. \\
+For $k \in \N$, $X_{>k} = \{ n \in X | n > k \}$. Then $X_{>k} \not= \varnothing$, as $X$ would be a subset of $\{1..k\}$ and would be finite. \\
+Then $min(X_{>k})$ exists, and $h: X \to X$ can be defined by $h(x) = min(X_{>x})$, and $f: \N \to X$ by recursion on $h$ with $f(0) = min(X)$. \\
+\\
+If an injection $f: A \to X$ exists, $A$ is countable if $X$ is. Proof: \\
+If $A$ is finite, it's countable. Otherwise, \\
+Since $X$ is countable, there exists a bijection $g: X \to \N$, and $g \circ f: A \to \N$
+exists and is a composite of 2 injective functions, and therefore is itsself injective. \\
+\\
+Any subset of a countable set is countable, by above with the inclusion function.
+
+$N^2$ is countable. Proof: \\
+Take $f: \N^2 \to \N$ defined by $f(a, b) = 2^a3^b$. \\
+$f$ is injective. Proof is simple -- prime factor decompositions are unique. \\
+
+If a function $f: X \to Y$ exists where $X$ is countable, $f(X)$ is countable. Proof: \\
+For $y \in f(X)$, \emph{choose} an $x_y \in f^{-1}(\{y\})$ and define $g: f(X) \to X$ by $g(y) = x_y$. \\
+$g$ is injective, so $f(X)$ injects into the countable set $X$ and is itself countable. \\
+In particular, for any surjection $f: X \to Y$, $Y$ is countable if $X$ is \\
+\\
+The union over a countable set of countable sets is countable. Proof: For family $X_{i \in I}$, $X_i$ countable, $I$ countable \\
+There is an injection $h: I \to \N$, and $f_i: X_i \to \N \forall i \in I$, as these are countable sets. \\
+Define $Y = \bigcup_{i \in I}X_i$, and $g: Y \to \N$ by $g(y) = (h(i), f_i(y))$ where $i \in I$ is chosen so that $y \in X_i$.
+Then $g$ is injective because equality distributes over pairs, and $h$ and $f_i$ are injective.\\
+
+If $X$ and $Y$ are countable, so is $X\times Y$. Proof: \\
+Define for $x \in X$ a subset $Y_x = \{(x, y) | y \in Y\}$ of $X \times Y$, then $\{Y_x\}_{x \in X}$ is a countable family of countable sets.
+\\
+$\Z$ is countable, as a union of $\N$ and $(\times {-1})(\N^+)$, or $\N \times \{0, 1\}$ \\
+
+$\Q$ is countable. Proof:
+\begin{align*}
+	 & \text{Define } f: \Z\times (\Z \setminus \{ 0 \}) \text{ by } f(a, b) = \frac{a}{b}      \\
+	 & f \text{ is surjective, by definition of } \Q.                                           \\
+	 & \Z \text{ is countable, as above }                                                       \\
+	 & \Z \setminus \{ 0 \} \text{ is countable, as a subset of a countable set } \Z            \\
+	 & \Z\times (\Z \setminus \{ 0 \}) \text{ is countable, as a product of countable sets }    \\
+	 & \text{Since } f \text{ is surjective with a countable domain, } \Q \text{ is countable }
+\end{align*}
+
+For set family $X_{n \in \N^+}$, $\times_{n \in \N^+} X_n$ is countable if $\forall n \in \N.~X_n$ is countable. Proof: \\
+Base Case: $n = 1$: $X_1$ is countable, since $X_1$ is countable. \\
+Induction: Assume $\times_{n \in \N^+, \leq k} X_n$ is countable. \\
+Then the result for $k + 1$ is $\times_{n \in \N^+, \leq k + 1} X_n$, which is $(\times_{n \in \N^+, \leq k} X_n) \times X_{k + 1}$, which is the product of countable sets and is therefore countable. \\
+By induction, the result holds for $n \in N^+$ \\
+This generalizes to all countable indexing sets $I$, by constructing an injection $f: I \to \N$.
+\\ \\
+Let $X$ be countable. Define $\mathbb{P}_{<\infty}\paren{X}$ as the set of all finite subsets of $X$. $\mathbb{P}_{<\infty}\paren{X}$ is countable. Proof: \\
+For $n \in \N$, let $\mathbb{P}_{\leq n}(X)$ denote the set of all nonempty subsets of $X$ with $n$ elements or less. \\
+The function $p_n: X^n \to \mathbb{P}_{\leq n}(X)$ defined by $p_n(x_1, x_2, ... x_n) = \{ x_1, x_2, ... x_n\}$ is surjective. \\
+Then, since $X$ is countable, so is $\mathbb{P}_{\leq n}(X)$ (as $X^n$ is countable and there exists a surjection $X^n \to \mathbb{P}_{\leq n}(X)$) \\
+\[\mathbb{P}_{\leq\infty}(X) = \bigcup_{n \in \N^+}\mathbb{P}_{\leq n}(X) \cup \{\varnothing\} \]
+This is a countably-sized union of countable sets and so is itself countable, as required.
+
+\subsection*{Cantor's Theorem}
+Let $X$ be a set. Then there exists no surjective function $f: X \to \powerset{X}$. Proof: \\
+Let $f: X \to \powerset{X}$ be a function. We will prove that $f$ is not surjective. \\
+Define $D \subseteq X$ by $\{ x \in X | x \not\in f(x) \}$ \\
+Then $D \in \powerset{X}$, and we will show that there is no element of $X$ for which $f(x) = D$. \\
+Suppose there was such an $x$. Either: \\
+
+$x \in D$. Then $x \in f(x)$, but by definition of $D$, $x \not\in f(x)$, which is a contradiction.
+
+$x \not\in D$. Then $x \not\in f(x)$, so $x \in D$ by definition of $D$, which is a contradiction. \\\\
+Since both cases give a contradiction, there exists no such $x$, and $f$ is not surjective. \\
+This is the less famous diagonal argument. \\
+\\
+$\R$ is uncountable. Proof:
+
+Define $f: \powerset{\N} \to \R$ by $f(A) = \sum_{n \in A} 2(3^{-n}) \quad \forall A \subseteq \N$.
+Then $f$ is injective: \\
+
+Take $\alpha, \beta \in \powerset{\N}$, where $\alpha \not = \beta$, and we will show $f(\alpha) \not = f(\beta)$. \\
+Then, take $k \in \N$ as the smallest natural number in exactly one of $\alpha$, $\beta$, and assume it's in $\beta$ WLOG. \\
+Then
+\begin{align*}
+	f(\alpha) & = \sum_{n \in \alpha}\frac{2}{3^n}                                                              \\
+	          & = \sum_{n \in \alpha, < k} \frac{2}{3^n} + \sum_{n \in \alpha, > k} \frac{2}{3^n}               \\
+	          & \leq \sum_{n \in \alpha, < k} \frac{2}{3^n} + \sum_{n > k} \frac{2}{3^n}                        \\
+	          & = \sum_{n \in \alpha, < k} \frac{2}{3^n} + \frac{1}{3^k}                                        \\
+	          & = \sum_{n \in \beta, < k} \frac{2}{3^n} + \frac{2}{3^k}                                         \\
+	          & < \sum_{n \in \beta, < k} \frac{2}{3^n} + \frac{2}{3^k} + \sum_{n \in \beta, > k} \frac{2}{3^n} \\
+	          & = f(\beta) \\ \\
+	f(\alpha) < f(\beta) & \implies f(\alpha) \neq f(\beta)
+\end{align*}
+Then there is no surjection $\N \to \powerset{\N}$ and so $\powerset{\N}$ is uncountable.
+Since there's an injection $\powerset{\N} \to \R$, $\R$ is uncountable, as if $\R$ was countable, $\powerset{\N}$ would be countable
+\\ \\
+The set of all polynomials with rational coefficients is countable. Proof:
+
+	Let $P$ be the set of all polynomials with rational coefficients, $P_n$ be the set of all polynomials with rational coefficients and degree $\leq$ n.
+
+	Then $P = \bigcup_{n \in \N} P_n$, $\N$ is countable, and $P_n$ is countable as there exists a surjection $\Q^n \to P_n$ by assigning each element of the tuple to a coefficient. \\
+
+\subsection*{Algebraic Numbers}
+The algebraic numbers ($\mathcal{A}$) are the real numbers which are roots of polynomials with rational coefficients.
+
+$\mathcal{A}$ is countable, as \[ \mathcal{A} = \bigcup_{p \in P} \{ x \in \R |~p(x) = 0 \} \] is a countable union of finite sets. \\
+Then $\R\setminus\mathcal{A}$ is uncountable, as if it were, $\R = (\R \setminus \mathcal{A}) \cup \mathcal{A}$, a union of countable sets.
 \end{document}