\input{decls.tex}
\title{Sets}
\begin{document}
\maketitle
Set comprehensions can be written $\{ x | x \in \N \}$ or $\{ x : x \in \N \}$ - '$:$' or '$|$' \\
Sets are defined entirely by the values of $x$ for which $x \in A$
\begin{description}
	\item[Axiom of Extensionality / Set Equality] $A = B \iff \forall x. (x \in A \iff x \in B)$
	\item[$A \subseteq B$] \quad $\forall x \in A. x \in B$ \\
		Is transitive, reflexive, antisymmetric
	\item[$A \subset B$] \quad $(\forall x \in A.~x \in B) \land (\exists x \in B.~x \not\in A)$ \\
		Is transitive, antisymmetric
	\item[$\varnothing$] $\{\}$
	\item[$\cup$] Union
	\item[$\cap$] Intersection
	\item[$A \setminus B$] \quad $\{ x \in A : x \not\in B \}$
		\item[$A^\complement$]\quad $U \setminus A$
	\item[$[a, b)$] \quad $\{ x \in \R : a \leq x < b \}$
\end{description}
\begin{align*}
	C \setminus (A \cup B) \equiv           & ~ (C \setminus A) \cap (C \setminus B) \\
	C \setminus (A \cap B) \equiv           & ~ (C \setminus A) \cup (C \setminus B) \\
	(A^\complement)^\complement \equiv      & ~ A                                    \\
	A^\complement \cup B^\complement \equiv & ~ (A \cap B)^\complement               \\
	A^\complement \cap B^\complement \equiv & ~ (A \cup B)^\complement
\end{align*}
\section*{Families of Sets}
A family of sets indexed by a set $I$ (the indexing set): $A_i ~~\forall~i\in I (\equiv (A_i)_{i\in I})$ \\
$A_i$ is a set for every element $i \in I$ \\
A family of sets indexed by $\N$ is called a sequence of sets. Also written $(B_i)^{\inf}_{i=0}$ or $(B_i)_{i \geq 0}$
\begin{align*}
	\bigcup_{i \in I}~A_i \equiv \{x | \exists i \in I. x \in A_i \}                                      \\
	\bigcap_{i \in I}~A_i \equiv \{x | \forall i \in I. x \in A_i \} & \text{ Exists iff } \exists~i\in I
\end{align*}
\begin{align*}
	 & \forall i \in I. A_i \subseteq \cup_{j \in I}A_j                        \\
	 & \forall i \in I. A_i \subseteq B \implies \cup_{j \in I}A_j \subseteq B \\
	 & \forall i \in I.\cap_{j \in I}A_j \subseteq A_i                         \\
	 & \forall i \in I. B \subseteq A_i \implies B \subseteq \cap_{j \in I}A_j \\ \\
	 & B \cup \cap_{i \in I}A_i = \cup_{i \in I}(B \cap A_i)                   \\
	 & B \cap \cup_{i \in I}A_i = \cap_{i \in I}(B \cup A_i)                   \\
	 & B \setminus \cup_{i \in I}A_i = \cap_{i\in I}(B\setminus A_i)           \\
	 & B \setminus \cap_{i \in I}A_i = \cup_{i\in I}(B \setminus A_i)
\end{align*}
\section*{Cartesian Products}
Ordered pairs can be represented as $(x, y) \equiv \{x, \{x, y\}\}$ \\
$X \times Y = \{ (x, y) |~ \forall x, y.~x \in X \land y \in Y\}$ for sets $X$, $Y$ \\
$X^n$ is $X\times X^{n-1}$ for set $X$ and natural $n$ \\
$\card{X\times Y} = \card{X} \times \card{Y}$ (for finite $X$, $Y$)
\section*{Functions}
For sets $X$, $Y$:
\begin{description}
	\item[A function $F: X \to Y$] $\subseteq X\times Y$ \text { where } \\
		$\forall x \in X.~\exists \text{ a unique } F(x) \in Y.~ (x, F(x)) \in F$ \\
		There exist $\card{Y}^{\card{X}}$ functions $F: X \to Y$
	\item[$\operatorname{dom}(F)$] The domain of $F$, i.e. $X$
	\item[$\operatorname{incl}^X_A : A \to X$] $= a \quad \forall A,X. \text{ where } A \subseteq X$
	\item[Constant function] $\exists y_0 \in Y.~\forall x \in X.~ f(x) = y_0$
	\item[Characteristic Function of a set $A \subseteq X$: $\chi_A: X \to \{0, 1\}$]
		\[
			\chi_A: X \to \{0, 1\} = \left\{\begin{array}{lr}
				0 & \text{ if } x \not\in A \\
				1 & \text{ if } x \in A
			\end{array} \right.
		\]
	\item[Restriction of a function $f: X \to Y$] $\restr{f}{A}$ is $f$ specialized contravariantly to $A \subseteq X$
	\item[$f(A)$: Image of $A$ under $f$] $f$ mapped over $A$ \quad for function $f: X \to Y$, $A \subseteq X$
	\item[$\operatorname{ran}(f)$ / image of $f$ / range of $f$] $\{ f(x) | x \in X \}$, $f(X)$, i.e. all possible values of $f(x)$
	\item[Preimage of $B$ under $f$] $\{ x \in X ~|~ f(x) \in B \}$ \\
		written $f^{-1}(B)$, but is \emph{not} the inverse of f
\end{description}
For $f: X \to Y$, $A \subseteq A' \subseteq X$, $B \subseteq B' \subseteq Y$:
\begin{align*}
	f(A) \subseteq~          & f(A')      \\
	f^{-1}(B) \subseteq ~    & f^{-1}(B') \\
	f^{-1}(f(A)) \supseteq ~ & A          \\
	f(f^{-1}(B)) \subseteq ~ & B
\end{align*}
For set families $(A_i \subseteq X)_{i \in I}, (B_j \subseteq Y)_{j \in J}$:
\begin{align*}
	f(\cup_{i \in I} A_i) = ~         & \cup_{i \in I}f(A_i)       \\
	f(\cap_{i \in I} A_i) \subseteq ~ & \cap_{i \in I}f(A_i)       \\
	f^{-1}(\cup_{j \in J} B_j) = ~    & \cup_{j \in J} f^{-1}(B_j) \\
	f^{-1}(\cap_{j \in J} B_j) = ~    & \cap_{j \in J} f^{-1}(B_j)
\end{align*}
\section*{Function Composition}
For functions $f: X \to Y$, $g: Y \to Z$:
\begin{align*}
	 & (g \circ f): X \to Z                      \\
	 & (g\circ f)(x) = g(f(x)) & \forall x \in X \\ \\
\end{align*}
\section*{Surjection and Injection}
For $f: X \to Y$, $f$ is
\begin{description}
	\item[surjective] iff $f(X) = Y$, i.e. $\forall y \in Y.~ \exists x \in X.~ f(x) = y$ \\
		Range is codomain, 'onto'
	\item[injective] iff $\forall x, x' \in X. ~ f(x) = f(x') \implies x = x'$
	\item[bijective] iff $f$ is injective and $f$ is surjective \\
		one-to-one
\end{description}
Given $f: X \to Y$, $g: Y \to Z$: \\
\quad If f and g are injective, so is $g\circ f$ \\
\quad If f and g are surjective, so is $g \circ f$ \\
\quad If $g \circ f$ is injective, so is $f$ \\
\quad If $g \circ f$ is surjective, so is $g$
\section*{Inverse Functions}
The inverse function $f^{-1}$ of $f: X \to Y$ exists iff $f$ is bijective, and is defined by \[f^{-1}(y) = x \text{ where } \exists! x \in X. f(x) = y \qquad \forall y \in Y\]
\begin{align*}
	(f \circ f^{-1}) = \operatorname{Id} \\
	(f^{-1} \circ f) = \operatorname{Id}
\end{align*}
\section*{Power Sets}
Powerset of S: $\powerset{S}$ has $2^{\card{S}}$ elements is the set of all subsets of $S$ \\
$Fun(X, \{0, 1\})$ is the set of functions $X \to \{0, 1\}$ \\
$\Phi: Fun(X, \{0, 1\}) \to \powerset{X}$ \\
$\Phi(f) = \{ x \in X | f(x) = 1 \}$ \\
$\Phi$ is bijective.
\section*{Binary Operators}
A binary operator is $X^2 \to X$ \\
Union is $\powerset{X}^2 \to \powerset{X}$ \\
$\square$ is the unknown or indeterminate binop sigil \\
Unital: $\forall x \in X. \exists u \in X. u \square x = x \square u = x$
\section*{Construction of the Natural Numbers}
\begin{align*}
	x^{+} & = x \cup \{x\} & \text{successor of $x$}
\end{align*}
A set $X$ is \emph{inductive} if $\varnothing \in X \land (a \in X \implies a^+ \in X)$ \\
Axiom: There exists an inductive set. \\
Definition: A natural number is a set that is an element of all inductive sets. \\
Theorem: There exists a set whose elements are the natural numbers. \\
\begin{align*}
	 & \text{Given an inductive set $A$}                                                       \\
	 & \omega = \{ x \in A | x \text{ is a natural number}\}                                   \\
	 & \text{any natural number is in $A$ (since $A$ is inductive), and therefore in $\omega$} \\
	 & \omega \text{ is the natural numbers}
\end{align*}
\\
Define:
\begin{align*}
	0 & = \varnothing \\
	1 & = 0^+         \\
	2 & = 1^+         \\
	\ldots
\end{align*}
$\in$ is an ordering over $\omega$, as is $\subseteq$ \\
$\omega$ is an inductive set. Proof: \\
$\varnothing$ is a natural number.
$x \in \omega$ implies $x^+ \in \omega$, as $x$ is natural and is therefore in
every inductive set, and so $x^+$ is in every inductive set and is therefore a
natural number.
\subsection*{Principle of Induction}
If $A \subseteq \omega$ and $A$ is inductive, $A = \omega$.
Since $A$ is inductive, it contains every natural number, so $\omega \subseteq A$. \\
Since $A \subseteq \omega \land \omega \subseteq A$, $A = \omega$
\section*{Recursion on $\omega$}
\subsection*{Principle of Recursion}
For a set $X$, $x_0 \in X$, $h: X \to X$, there exists a unique function $f: \omega \to X$ where $f(0) = x_0$, $f(n^+) = h(f(n))$.
\section*{Relations}
Total order: $(X, \prec)$, requires $\forall x, y, z \in X$: \\
\begin{align*}
	 & x \prec y \land y \prec z \implies x \prec z                                         \\
	 & x = y \oplus x \prec y \oplus y \prec x      & \text{ (where $\oplus$ denotes XOR) }
\end{align*}
Lexicographic order $x <_L y$ on $\N\times\N = x_0 < y_0 \lor (x_0 = y_0 \land x_1 < y_1)$ \\
For a non-empty subset $A$ of $X$ given a total order $(X, \prec)$, a minimum/least element $a_0 \in A$
exists where $\forall a \in A.~a_0 \preceq a$.
\section*{Ordering on $\omega$}
$(\omega, \in)$ and $(\omega, \subset)$ are both total orderings on $\omega$, such that $\varnothing$ is the minimum and $\forall x \in \omega.~x < x^+$
\section*{Strong Induction}
Every non-empty subset of $\N$ has a minimum. \\ \\
Let $\phi(x)$ be a predicate over $\N$ where $\forall n\in \N.~ (\forall m \in \N.~ m < n \implies \phi(m)) \implies \phi(n)$. \\
Then $\phi(0)$ holds, as $\neg\exists n \in \N.~n<0$, $\phi(1)$ holds as $\phi(0)$ holds, etc.
\begin{align*}
	 & \neg\exists x \in \N.~\neg\phi(x)                                                                                                   \\
	 & \text{Let } A \text{ be a subset of }N\text{ such that }\phi(n)\text{ is false }\forall n \in A                                     \\
	 & \text{If not, $\exists a_0 \in A$}                                                                                                  \\
	 & \forall n \in N.~n < a_0 \implies n \not\in A \implies \phi(n)                                                                      \\
	 & \text{Then $\phi(n)$ holds $\forall n < a_0$, so $\phi(a_0)$, then $a_0 \not\in A$, which is a contradiction, so $A = \varnothing$}
\end{align*}
\section*{Fibonacci}
Fibonacci Sequence: $F_n$ \\
Roots of $x^2 - x - 1$ are $\phi = \frac{1 + \sqrt{5}}{2}$ and $\psi = \frac{1 - \sqrt{5}}{2}$ \\
$F_n = \frac{\phi^{n+1} - \psi^{n+1}}{\sqrt{5}}$ \\
For $n=0$:

$F_0 = 1 = \frac{\frac{1 + \sqrt{5}}{2} - \frac{1-\sqrt{5}}{2}} = \frac{2\sqrt{5}}{2} = 1$ \\

For $n=1$: \\

$F_1 = 1 = \frac{\paren{\frac{1 + \sqrt{5}}{2}}^2 - \paren{\frac{1 - \sqrt{5}}{2}}^2}{\sqrt{5}} = \frac{\phi + 1 - \psi - 1}{\sqrt{5}}$ \\

For $n\geq 2$:
\begin{align*}
	 & F_n = F_{n-1} + F_{n-2} = \frac{\phi^n - \psi^n}{\sqrt{5}} + \frac{\phi^{n-1} - \psi^{n-1}}{\sqrt{5}}                      \\
	 & = \frac{\phi^{n-1}\paren{\phi + 1} - \psi^{n-1}\paren{\psi + 1}}{\sqrt{5}}                                                 \\
	 & = \frac{\phi^{n-1}\phi^2 - \psi^{n-1}\psi^2}{\sqrt{5}}                                                                     \\
	 & = \frac{\phi^{n+1} - \psi^{n+1}}{\sqrt{5}}                                                            & \text{as required}
\end{align*}

\subsection*{Zeckendorff's Theorem}
Every natural number can be written as a sum of non-adjacent Fibonacci numbers in a unique way (excluding $F_0$). \\
A finite subset $I$ of $\N^+$ is \emph{Zeckendorff} if it contains no adjacent elements ($\forall x \in I.~x^+ \not\in I$) \\
Define $\mathcal{Z}$ as the set of all Zeckendorff sets, and $\sigma: \mathcal{Z} \to \N^+$ by $\sigma(I) = \sum_{i \in I}F_i$.
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.}
\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\}       \\
\end{align*}
\section*{Equivalence Relations}
Reflexive, Symmetric, Transitive \\
Equivalence relations are usually called $\sim$ \\
In a set $X$,
\[ [x] = \{ y \in X | x \sim y \} \]
\[ [x] = [y] \equiv x \sim y \]
Any two equivalence classes of $X$ are either disjoint or equal. \\
\subsection*{Quotient Sets}
$X/\sim~ = \{[x] | x \in X \} \subset \powerset{X}$ \\
A complete set of representatives is a subset $A$ of $X$ where $\forall x \in X. \exists! a \in A. a \in [x].$ \\
I.E. a complete set of representatives contains exactly one element from each element of $X/\sim$ \\
$f: A \to X/\sim$ 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 $\Bar{f}([x]) = f(x)$
\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 $ defined by $ f(x) = x + 1$ 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 $X \subseteq \N$.\\
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.
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