From b50b42c01b92311069a6b28e3fc2f42ed38c44c4 Mon Sep 17 00:00:00 2001 From: bluepython508 Date: Thu, 28 Mar 2024 11:36:00 +0000 Subject: [PATCH] . --- MA1006/decls.tex | 2 +- MA1511/decls.tex | 2 +- MA1511/sets.tex | 141 +++++++++++++++++++++++++++++++++++++++++------ 3 files changed, 127 insertions(+), 18 deletions(-) diff --git a/MA1006/decls.tex b/MA1006/decls.tex index de46f2f..8c4ea6a 100644 --- a/MA1006/decls.tex +++ b/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}} diff --git a/MA1511/decls.tex b/MA1511/decls.tex index de46f2f..8c4ea6a 100644 --- a/MA1511/decls.tex +++ b/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}} diff --git a/MA1511/sets.tex b/MA1511/sets.tex index 54112e8..3d6e903 100644 --- a/MA1511/sets.tex +++ b/MA1511/sets.tex @@ -204,25 +204,25 @@ Define $\mathcal{Z}$ as the set of all Zeckendorff sets, and $\sigma: \mathcal{Z The theorem claims $\sigma$ is bijective. \\ \\ For nonempty $I \in \mathcal{Z}$ with largest element $k$: \begin{align*} - &\text{Let } J = I \setminus \{k\} \\ - &\sigma(I) = F_k + \sigma(J) \geq F_k \\ - &\text{If } J = \varnothing\text{, } \sigma(I) = F_k \leq F_{k + 1} \\ - &\text{Otherwise, we must show } \sigma(I) < F_{k + 1} \\ - &\equiv F_k + \sigma(J) < F_{k + 1} \\ - &\equiv \sigma(J) < F_{k + 1} - F_k = F_{k-1} \\ - &\text{But if $k' = \operatorname{max}(J)$, } \\ - &\sigma(J) < F_{k' + 1} \land k' < k - 2 \text{ (since $I$ is \emph{Zeckendorff})} \\ - & \sigma(J) < F_{k - 1} \text{ as required.} + & \text{Let } J = I \setminus \{k\} \\ + & \sigma(I) = F_k + \sigma(J) \geq F_k \\ + & \text{If } J = \varnothing\text{, } \sigma(I) = F_k \leq F_{k + 1} \\ + & \text{Otherwise, we must show } \sigma(I) < F_{k + 1} \\ + & \equiv F_k + \sigma(J) < F_{k + 1} \\ + & \equiv \sigma(J) < F_{k + 1} - F_k = F_{k-1} \\ + & \text{But if $k' = \operatorname{max}(J)$, } \\ + & \sigma(J) < F_{k' + 1} \land k' < k - 2 \text{ (since $I$ is \emph{Zeckendorff})} \\ + & \sigma(J) < F_{k - 1} \text{ as required.} \end{align*} Proof of theorem: $\forall n \in N.~ \sigma$ is bijective. \begin{align*} - n = 0: \quad& I = \varnothing \\ - n > 0: \quad& \text{Let } F_k \leq n < F_{k + 1},~ m = n - F_k \\ - & m = \sigma(J) \text{ for some } J \\ - & \text{If } J = \varnothing, I = \{k\}. \\ - & \text{Otherwise, } k' = \operatorname{max}(J). \\ - & \text{If } k' \leq k - 2, I = J \cup \{k\} \\ + n = 0: \quad & I = \varnothing \\ + n > 0: \quad & \text{Let } F_k \leq n < F_{k + 1},~ m = n - F_k \\ + & m = \sigma(J) \text{ for some } J \\ + & \text{If } J = \varnothing, I = \{k\}. \\ + & \text{Otherwise, } k' = \operatorname{max}(J). \\ + & \text{If } k' \leq k - 2, I = J \cup \{k\} \\ \end{align*} \section*{Equivalence Relations} Reflexive, Symmetric, Transitive \\ @@ -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}