## Monday, June 16, 2014

### Some Set Theory

Set theory is a fundamental branch of mathematics that studies sets, that is, collections of objects that can be treated as wholes.
"A set is a Many that allows itself to be thought of as a One."
- Georg Cantor.

A set has various elements or members, which may either be individual objects or themselves sets. An extension to this is the inclusion of the empty set (the set containing no objects, symbolized $\emptyset$ or $\left \{\right\}$) as an allowable set. If set A has a, b and c as elements, we symbolize that as A={a,b,c}. Note that, in set theory, order and multiplicity don't matter, so A is also equal to {b,c,a,a,b}.

### Definitions and Notation

 $x \in A$ x is an element/member of A. $x \notin A$ x is not an element/member of A. $A \subseteq B$ Set A is a subset of B. That is, for all x, if x is an element of A, then x is an element of B. Note that "A is a subset of B" is equivalent to "B is a superset of A". $A = B$ A is equivalent to B. That is, for all x, x is an element of A if and only if x is an element of B. $A \neq B$ A is not equivalent to B. $A \subset B$ $A \subseteq B$ and $A \neq B$. If $A \subset B$, we say A is a proper subset of B. $A \cap B$ The intersection of sets A and B. That is, the set that contains all of the elements that are in both A and B. Sets A and B are said to be disjoint if $A \cap B = \emptyset$. $A \cup B$ The union of sets A and B. That is, the set that contains all of the elements that are in either A or B. $A \setminus B$ The set difference of sets A and B. That is, the set that contains all of the elements in A that are not in B. $\mathbb{Z} =\left \{ ...,-2,-1,0,1,2,... \right \}$ The set of all integers. That is, the set that includes 0 and every other number that can be achieved by successively adding or subtracting 1 from 0. $\mathbb{N} =\left \{1,2,3,... \right \}$ The set of all natural numbers. That is, the set that includes 1 and every other number that can be achieved by successively adding 1 to 1. $\mathbb{W} =\left \{0,1,2,3,... \right \}$ The set of all whole numbers. That is, the set that includes 0 and every other number that can be achieved by successively adding 1 to 0. $\left\{\begin{matrix} x \end{matrix}\right|\left.\begin{matrix} Px \end{matrix}\right\}$ The set of all elements of the form $x$ such that $Px$ is true of x. For instance $\mathbb{N}=\left\{\begin{matrix} x \end{matrix}\right|\left.\begin{matrix} x \in \mathbb{Z} \wedge x>0 \end{matrix}\right\}$. $\mathbb{Q}= \left\{\begin{matrix} \frac{x}{y} \end{matrix}\right|\left.\begin{matrix} x \in \mathbb{Z} \wedge y \in \mathbb{N} \end{matrix}\right\}$ The set of all rational numbers. That is, the set of all real numbers expressible as a ratio of integers with non-zero denominator. $\mathbb{N}_{n}=\left\{1,2,3,...,n\right\}$ The truncated set of the natural numbers up to $n$. Another way of expressing it would be: $\mathbb{N}_{n}=\left\{\begin{matrix} x \end{matrix}\right|\left.\begin{matrix} x \in \mathbb{N} \wedge x \leq n \end{matrix}\right\}$. $\mathbb{R}=\left (-\infty, \infty \right)$ The set of all real numbers. As a note, we often see the shorthand $(a,b)$ (not to be confused with an ordered pair!) to mean $\left\{\begin{matrix} x \end{matrix}\right|\left.\begin{matrix} x \in \mathbb{R} \wedge a < x \wedge x < b \end{matrix}\right\}$. We may also see the shorthand $[a,b]$ to mean $\left\{\begin{matrix} x \end{matrix}\right|\left.\begin{matrix} x \in \mathbb{R} \wedge a \leq x \wedge x \leq b \end{matrix}\right\}$. The combined notation $(a,b]$ should be clear enough. $\mathbb{W} =\left \{0,1,2,3,... \right \}$ The set of all whole numbers. That is, the set that includes 0 and every other number that can be achieved by successively adding 1 to 0. $\mathbb{P} =\left \{2,3,5,7,11,... \right \}$ The set of all prime numbers. That is, the subset of the natural numbers that have exactly two factors: 1 and themselves. By a well-known theorem, we know that there are infinitely many prime numbers. They are often represented as a sequence of integers in ascending order: $\mathbb{P}=\{p_{0},p_{1},p_{2},p_{3},...\}$. $\mathbb{A}$ The set of all algebraic numbers. That is, the set of numbers that solve polynomials with integer coefficients. We can also introduce the notation $\mathbb{A}_{n}$ which is the set of all algebraic numbers of order $n$, that is the set of numbers that solve polynomials of order $n$ with integer coefficients. In other words: $\mathbb{A}_n=\left\{\begin{matrix} x \\ \; \end{matrix}\right|\left.\begin{matrix} x \in \mathbb{R} \wedge \exists A = \{a_{0},a_{1},a_{2},...a_{n}\} \sim \mathbb{N}_{n+1} \\ \wedge c \in A\Rightarrow c \in \mathbb{Z} \wedge 0=\sum_{k=0}^{n} a_{k} x^{k} \end{matrix}\right\}$. We then take the union: $\mathbb{A}=\mathbb{A}_{1} \cup \mathbb{A}_{2} \cup \mathbb{A}_{3} \cup ...$. A number in the set $\mathbb{R} \setminus \mathbb{A}$ is called a transcendental number. $(a,b)$ An ordered pair, that is a pairing of mathematical objects in which the order matters. We can define them in the following way, for technical purposes: $(a,b) = \left \{ \left \{ a \right \}, \left \{ a,b \right \} \right \}$. $A \times B$ The cartesian product of sets A and B. That is, the set of all ordered pairs $(a,b)$ such that $a \in A$ and $b \in B$. $\mathcal{P}(A)$ The power set of set A. That is, the set of all subsets of /(A/). We can write this as $\mathcal{P}(A)=\left\{\begin{matrix} X \end{matrix}\right|\left.\begin{matrix} X \subseteq A \end{matrix}\right\}$. $f: A \rightarrow B$ f is a function from A to B. That is, f takes an element of A and returns an element of B. Strictly speaking, a function from A to B is a subset of $A \times B$ with the properties $\left\{\begin{matrix} a \end{matrix}\right|\left.\begin{matrix}(a,b)\in f \end{matrix}\right\} = A$ If $(a,b)\in f$ and $(a,c)\in f$, then $b=c$. A is called the domain of f and B is called the range or codomain of f. Often we will see the notation $y=f(x)$, which means that f associates y with x, or $(x,y)\in f$. In this case, y is called the image of x under f. More generally, if $X \subseteq A$, then $f(X)=Y= \left\{\begin{matrix} y \end{matrix}\right|\left.\begin{matrix} y=f(x) \wedge x \in X \end{matrix}\right\}$. Y is likewise called the image of X under f. Clearly $Y \subseteq B$. $f: A \rightarrowtail B$ f is an injective function from A to B. That is, the function is one-to-one, or, if $f(a)=f(b)$ then $a=b$. Every injective function f has an inverse function, $f^{-1}: f(A) \rightarrow A$. We can say $f^{-1}=\left\{\begin{matrix} (b,a) \end{matrix}\right|\left.\begin{matrix} (a,b) \in f \end{matrix}\right\}$. Clearly the inverse of an injective function is injective. We can also see that $f(f^{-1}(a))=a$, and $f^{-1}(f(a))=a$. $f: A \twoheadrightarrow B$ f is a surjective function from A to B. That is, the function is onto. In other words, $f(A)=B$, i.e. $\left\{\begin{matrix} b \end{matrix}\right|\left.\begin{matrix}(a,b)\in f \end{matrix}\right\} = B$. $f: A \leftrightarrow B$ f is a bijective function from A to B. That is, f is both injective and surjective. The inverse of a bijective function is also bijective. $A \preceq B$ There exists an injective function from $A$ to $B$. $A \sim B$ There exists a bijective function from $A$ to $B$. $A \nsim B$ There does not exist a bijective function from $A$ to $B$. $A \prec B$ There exists an injective function from $A$ to $B$, but no bijective function from $A$ to $B$. That is, $A \preceq B$ and $A \nsim B$. $\left | A \right |$ The cardinality of $A$. That is, a value associated with a set that can be used to compare sets with respect to some sense of magnitude. $\left | A \right | \leq \left | B \right |$ $A \preceq B$. We say this as "the cardinality of A is less than or equal to the cardinality of B", though this is only something of a manner of speaking. $\left | A \right | = \left | B \right |$ The cardinality of $A$ is the same as the cardinality of $B$, meaning there exists a bijective function from $A$ to $B$. In other words, $A \sim B$. When this is the case, it is often convenient to say "A and B are equinumerous", but this is a technical term, and should not be naively interpreted. $\aleph_{0}$ $\aleph_{0}=\left | \mathbb{N} \right |$. $\aleph_{0}$ is the "smallest" infinite cardinal. In other words, if $\left | A \right |< \aleph_{0}$, then A is a finite set. Also, a set A is said to be countable iff $\left | A \right | \leq \aleph_{0}$. If $A \sim \mathbb{N}$, then A is said to be countably infinite. $\mathbb{S}_{n}$ $\mathbb{S}_{n}=\left | \mathbb{N}_{n}\right |$. A is a finite set iff for some $n$, $\left | A \right |= \mathbb{S}_{n}$, where $n \in \mathbb{N}$. Typically, if $\left | A \right |= \mathbb{S}_{n}$, where $n \in \mathbb{N}$, we say "A has n elements", though we will here avoid all notions of the "number of elements". $\left | A \right |+\left |B \right |$ $\left | A \right |+\left |B \right |=\left |A \cup \tilde{B} \right |$, where $\tilde{B} \sim B$ and $A \cap \tilde{B}=\emptyset$. $\left | A \right |\times\left |B \right |$ $\left | A \right |\times\left |B \right |=\left | A \times B \right |$. $\max \left (\left | A \right |,\left |B \right | \right)$ $\max \left (\left | A \right |,\left |B \right | \right) = \left\{\begin{matrix} \left | A \right |, \\ \left | B \right |, \end{matrix}\right. \begin{matrix} B \preceq A \\ A \preceq B \end{matrix}$ . $\min \left (\left | A \right |,\left |B \right | \right)$ $\min \left (\left | A \right |,\left |B \right | \right) = \left\{\begin{matrix} \left | A \right |, \\ \left | B \right |, \end{matrix}\right. \begin{matrix} A \preceq B \\ B \preceq A \end{matrix}$ .

### Axioms of Set Theory

The most typical axiomatization of set theory is that of Zermelo-Fraenkel with the axiom of choice (ZFC). We will here list these axioms:
 Extensionality Sets A and B are equivalent (symbolized $A=B$), iff, for all x, $x \in A$ iff $x \in B$. Regularity If A is a non-empty set, then, for some element x of A, $A \cap x=\emptyset$. Schema of Specification If, for any x, we can determine whether or not Px is true of x (P being some predicate), then we can always construct the set $\left\{\begin{matrix} x \end{matrix}\right|\left.\begin{matrix} Px \end{matrix}\right\}$. Pairing If A and B are sets, then there exists a set C such that $A \in C$ and $B \in C$. Union If A is a set, then there exists a set B such that, if $Y \in A$ and $x \in Y$ then $x \in B$. More generally, the union of any collection of sets exists. Schema of Replacement If $f$ is a definable function, and A is a set in the domain of $f$, then, $f(A)$ is a set. Infinity There exists a set X such that $\emptyset \in X$, and, if $y \in X$ then $\left \{y \right \} \in X$. That is, $X=\left \{\emptyset, \left \{\emptyset\right \}, \left \{\left \{\emptyset\right \}\right \},...\right \}$. In particular, X contains infinitely many elements. Power Set If A is a set, then $\mathcal{P}(A)$ is also a set. Choice If $A_{1}$, $A_{2}$, $A_{3}$, ... is a sequence of non-empty sets, then there exists a set $X=\left \{x_{1},x_{2},x_{3},...\right \}$, such that $x_{n} \in A_{n}$.

### Some Elementary Theorems

We will now discuss a few theorems. Theorems stated without proof are beyond the level of our discussion or are too involved. The reader is directed to a more technical discussion if more detail is desired.
Theorem: If $A \subseteq B$ then $A \preceq B$.
Proof: Let $f(x)=x$. Then $f$ is injective from A to B.

Theorem: If $f: A \rightarrowtail B$, $C \subseteq A$, $D \subseteq A$ and $C \cap D=\emptyset$, then $f(C) \cap f(D)=\emptyset$.
Proof: Assume $C \cap D=\emptyset$ and $f(C) \cap f(D)=F\neq\emptyset$. Let $\phi \in F$ then, for some $c \in C$ and some $d \in D$, $f(c)=f(d)=\phi$. However, $f$ is injective, and therefore, if $f(c)=f(d)$, then $c=d$. Therefore, $c \in C \cap D$, and therefore $C \cap D \neq \emptyset$, which contradicts our assumption. Therefore $f(C) \cap f(D)=\emptyset$.
Corollary: If $f: A \rightarrowtail B$, then $f(C \cap D)=f(C) \cap f(D)$.
Corollary: If $f: A \rightarrowtail B$, then $f(C \cup D)=f(C) \cup f(D)$.

Theorem: $A \sim A$.
Proof: The identity function $f(x)=x$ is a bijection from A to A.

Theorem: If $A \sim B$ then $B \sim A$.
Proof: Let $f$ be a bijective function from A to B. As all bijective functions have inverses, $f^{-1}$ is a bijection from B to A.

Theorem: If $A \sim B$ and $B \sim C$ then $A \sim C$.
Proof: Let $f$ be a bijective function from A to B, and $g$ be a bijective function from B to C. Then the composition $g \circ f$ is a bijective function from A to C.

Theorem: If $a \notin A$ and $A \sim \mathbb{N}_{n}$ then $A \cup \left\{a \right\} \sim \mathbb{N}_{n+1}$.
Proof: Let $f$ be a bijective function from A to $\mathbb{N}_{n}$. Let $g(x)=\left\{\begin{matrix} f(x), \;\; \;\; \\ n+1, \;\; \;\; \end{matrix}\right. \begin{matrix} x \in A \\ x=a \end{matrix}$.
Then g is bijective from $A \cup \left\{a \right\}$ to $\mathbb{N}_{n+1}$.
Corollary: If $B \cap A=\emptyset$ and $B \sim \mathbb{N}_{m}$ and $A \sim \mathbb{N}_{n}$ then $A \cup B \sim \mathbb{N}_{n+m}$.

Theorem: If $a \in A$ and $A \sim \mathbb{N}_{n}$ then $A \setminus \left\{a \right\} \sim \mathbb{N}_{n-1}$.
Proof: Let $f$ be a bijective function from $\mathbb{N}_{n}$ to A. Suppose $a=f(m)$. Let $g(x)=\left\{\begin{matrix} f(x), \;\; \;\; \\ f(x+1), \;\; \;\; \end{matrix}\right. \begin{matrix} x < m \\ x \geq m \end{matrix}$.
Then g is bijective from $\mathbb{N}_{n-1}$ to $A \setminus \left\{a \right\}$.
Corollary: If $B \subseteq A$ and $B \sim \mathbb{N}_{m}$ and $A \sim \mathbb{N}_{n}$ then $A \setminus B \sim \mathbb{N}_{n-m}$.

Theorem: If $A \sim B$ and $a \in A$ and $a \in B$ then $A \setminus \left\{a \right\} \sim B \setminus \left\{a \right\}$.
Proof: Let $f$ be a bijective function from A to B. Suppose $f(c)=a$. Define $g(x)=\left\{\begin{matrix} a, \\ c, \\ x, \;\; \end{matrix}\right. \begin{matrix} x=c \;\;\;\; \;\; \\ x=a \;\;\;\; \;\; \\ x \neq a \wedge x \neq c \end{matrix}$.
Clearly g is bijective (in fact, it is its own inverse) and $g(A)=A$. Also, $(f \circ g)(a)=a$, and $f \circ g$ is bijective. Therefore, $f \circ g$ is a bijection from $A \setminus \left\{a \right\}$ to $B \setminus \left\{a \right\}$.
Corollary: If $A \sim B$, and $C \sim \mathbb{N}_{n}$ for some $n \in \mathbb{N}$, where $C \subseteq A$ and $C \subseteq B$, then $A \setminus C \sim B \setminus C$.
Proof: Apply the theorem above $n$ times. Note that the set subtracted must be finite. The theorem is inapplicable for infinite sets. In fact, as a counterexample for infinite sets: $\mathbb{Z} \setminus \mathbb{N} \nsim \mathbb{W} \setminus \mathbb{N}$ even though $\mathbb{Z} \sim \mathbb{W}$.

Theorem: $\mathbb{N}_{n} \sim \mathbb{N}_{m}$ if and only if ${n = m}$.
Proof: Suppose that ${n \neq m}$ and $\mathbb{N}_{n} \sim \mathbb{N}_{m}$. Without loss of generality, $n < m$. By a theorem proved above, $\mathbb{N}_{n} \setminus \mathbb{N}_{n} \sim \mathbb{N}_{m} \setminus \mathbb{N}_{n}$. But $\mathbb{N}_{n} \setminus \mathbb{N}_{n} = \emptyset$. However, $m \in \mathbb{N}_{m} \setminus \mathbb{N}_{n}$, so $\mathbb{N}_{m} \setminus \mathbb{N}_{n} \neq \emptyset$. Clearly, then, it is impossible that $\mathbb{N}_{n} \setminus \mathbb{N}_{n} \sim \mathbb{N}_{m} \setminus \mathbb{N}_{n}$, and so our assumption that ${n \neq m}$ must be false. Conversely, if ${n=m}$ then $\mathbb{N}_{n}= \mathbb{N}_{m}$, and thus $\mathbb{N}_{n} \sim \mathbb{N}_{m}$.

Theorem: For any sets $A$ and $B$, there is an injection from $A$ to $B$ if and only if there is a surjection from $B$ to $A$.
Proof: Suppose $f: A \rightarrowtail B$. Then, let $a \in A$ be any element in $A$. Now, define $g(x)=\left\{\begin{matrix} f^{-1}(x), \;\; \;\; \\ a, \;\; \;\; \end{matrix}\right. \begin{matrix} x \in f(B) \\ x \notin f(B) \end{matrix}$
Clearly $g$ is surjective from $B$ to $A$.
Now suppose $f: A \twoheadrightarrow B$. That means that, for every $b\in B$ there is at least one $a \in A$ such that $f(a)=b$. Now define $g(x)=a$, where $a$ is arbitrarily chosen for any $x$, from among those $a \in A$ such that $f(a)=x$. Clearly $g$ is injective from $B$ to $A$.
Corollary: "There is a surjection from $A$ to $B$ " $\Leftrightarrow B \preceq A$.

Theorem: For any sets $A$ and $B$, there exists either an injection from $A$ to $B$ or a surjection from $A$ to $B$.
Proof (Informal): If there is no function from $A$ to $B$ that is either an injection or an injection, then, for any function $f$, there are $a_1,a_2 \in A$ such that $a_1 \neq a_2$ and $f(a_1)=f(a_2)$ and a $b \in B$ such that $b \notin f(A)$. But clearly we can construct a new function such that $f(a_2)=b$. Moreover, we can continue this for an arbitrary set of $\left\{\begin{matrix} b \end{matrix}\right|\left.\begin{matrix} b \in B \wedge b \notin f(A) \end{matrix}\right\}$ and/or $\left\{\begin{matrix} a \end{matrix}\right|\left.\begin{matrix} a \in A \wedge \exists a' \in A, a' \neq a, f(a)=f(a') \end{matrix}\right\}$. Thus, we can always exhaust at least one set, as each set has a set of larger cardinality, and we can continue this up to any cardinality. Thus, we can exhaust either the domain (which would make the function injective) or the co-domain (which would make the function surjective).
Corollary: For any sets A and B, either $A \preceq B$ or $B \preceq A$.
Corollary: For any sets A and B, exactly one of the following is true:
• $A \prec B$
• $B \prec A$
• $A \sim B$

Theorem: If $A \preceq B$ and $B \preceq A$ then $A \sim B$.
Proof: Let $f: A \rightarrowtail B$ and $g: B \rightarrowtail A$ and $h=g \circ f$. Let us introduce the notation $h^{k}(x)$ defined by $h^{0}(x)=x$ and $h^{n+1}(x)=h(h^{n}(x))$. We now let $C=A \setminus g(B)$ and $X=h^{0}(C) \cup h^{1}(C) \cup h^{2}(C) \cup ...$.
We then define $k(x)=\left\{\begin{matrix} f(x), \\ g^{-1}(x), \end{matrix}\right. \begin{matrix} x \in X \;\;\;\; \\ x \notin X \;\;\;\; \end{matrix}$.
We can then verify that $k$ is bijective from $A$ to $B$:
• Let $b \in B$. If $b \in f(X)$, then, for some $a \in X \subseteq A$, $k(a)=b$. Otherwise, we can find an $a$ such that $a=g(b)$. By construction, $a \notin C$, and $b \notin f(X)$, and since $f(h^{n}(X)) \subseteq f(X)$, we find $a \notin X$ and so $k(a)=b$. Thus, for any $b \in B$, $b=k(a)$ for some $a\in A$. Therefore $k$ is surjective.
• Assume, for some $c \in X$ and some $d \notin X$, $f(c)=g^{-1}(d)$, or $d=g(f(c))$. Therefore, for some $n \in \mathbb{W}$, $c \in h^{n}(C)$. But then $d \in g(f(h^{n}(C)))=h^{n+1}(C) \subseteq X$. But that contradicts our assumption that $d\notin X$, and thus, if $k(c)=k(d)$, then $c=d$. Therefore, $k$ is injective.

Theorem: For any set A, $A \prec \mathcal{P}(A)$.
Proof: Clearly $f(x)=\left\{x \right\}$ is an injection from A to $\mathcal{P}(A)$. Let $f: A \rightarrow \mathcal{P}(A)$ be any function. Define $S= \left\{\begin{matrix} x \end{matrix}\right|\left.\begin{matrix} x \notin f(x) \end{matrix}\right\}$. Clearly, then, $x \in S \Leftrightarrow x \notin f(x)$. Assume that, for some $k \in A$, $S=f(k)$. In that case, $k \in f(k) \Leftrightarrow k \notin f(k)$, which is clearly impossible. Thus, our assumption fails, and there is no k such that $k \in A$, $S=f(k)$. Therefore, as $S \in \mathcal{P}(A)$, $f$ is not surjective, and thus not bijective.
Corollary: For any set A, $A \prec \mathcal{P}(A) \prec \mathcal{P}(\mathcal{P}(A)) \prec ...$.
Corollary: $\mathbb{N} \prec \mathcal{P}(\mathbb{N}) \prec \mathcal{P}(\mathcal{P}(\mathbb{N})) \prec ...$.

Theorem: $\mathbb{N} \sim \mathbb{Z}$.
Proof: Let $f(n)=\left\{\begin{matrix} \frac{n-1}{2} \;\; n \textrm{ odd} \\ -\frac{n}{2} \;\; n \textrm{ even} \end{matrix}\right.$. Then $f$ can be seen to be bijective from $\mathbb{N}$ to $\mathbb{Z}$.

Theorem: $\mathbb{N} \sim \mathbb{Q}$.
Proof: As $\mathbb{N} \subset \mathbb{Q}$, we have $\mathbb{N} \preceq \mathbb{Q}$. Let us now take $f \left( \frac{a}{b} \right)= 2^{\textrm{sgn}(\frac{a}{b})}3^{a}5^{b}$. Clearly $f(\mathbb{Q}) \subset \mathbb{N}$, and so $f$ is an injection from $\mathbb{Q}$ to $\mathbb{N}$, and thus $\mathbb{Q} \preceq \mathbb{N}$. Therefore $\mathbb{N} \sim \mathbb{Q}$.

Theorem: $\mathbb{N} \sim \mathbb{A}$.
Proof: As $\mathbb{N} \subset \mathbb{A}$, we have $\mathbb{N} \preceq \mathbb{A}$. Let us now take $f \left( x \right)$ to be defined as $f(x)=p_{1}^{k}p_{2}^{a_{0}}p_{3}^{a_{1}}...p_{n+2}^{a_{n}}$ where $x$ is the kth real solution of $0=\sum_{k=0}^{n} a_{k} x^{k}$, sorted according to increasing absolute magnitude, and $p_{k}$ is the kth prime number. Then, $f(\mathbb{A}) \subset \mathbb{Q} \sim \mathbb{N}$, and $f$ is injective. Thus $\mathbb{A} \preceq \mathbb{N}$, and thus $\mathbb{N} \sim \mathbb{A}$.

Theorem: $(0,1) \sim \mathbb{R}$.
Proof: The function $f=\frac{2 x-1}{x-x^{2}}$ is bijective from $(0,1)$ to $\mathbb{R}$.

Theorem: $(0,1) \sim [0,1]$.
Proof: Let $f(x)=\left\{\begin{matrix} \frac{1}{2}, \\ \frac{1}{2^{n-2}}, \\ x, \;\; \end{matrix}\right. \begin{matrix} x=0 \;\;\;\;\;\;\;\;\;\;\;\;\; \\ x=\frac{1}{2^{n}}, n \in \mathbb{W} \\ x \neq 0, x \neq \frac{1}{2^{n}} \;\; \end{matrix}$. Then f is bijective from $[0,1]$ to $(0,1)$.

Theorem: $[0,1) \sim [0,1]$.
Proof: Let $f(x)=\left\{\begin{matrix} \frac{1}{2^{n-1}}, \\ x, \;\; \end{matrix}\right. \begin{matrix} x=\frac{1}{2^{n}}, n \in \mathbb{W} \\ x \neq \frac{1}{2^{n}} \;\; \end{matrix}$. Then f is bijective from $[0,1]$ to $[0,1)$.

Theorem: $[0,1) \sim (0,1]$.
Proof: Let $f(x)= 1-x$. Then f is bijective from $(0,1]$ to $[0,1)$.

Theorem: $(a,b) \sim (0,1)$.
Proof: Let $f(x)= \frac{x-a}{b-a}$. Then f is bijective from $(a,b)$ to $(0,1)$. Relations of the form $[a,b) \sim [0,1)$ can be identically proven.

Theorem: $(a,\infty) \sim (0,1)$.
Proof: Let $f(x)=\frac{|x-a|}{|x-a|+1}$. Then f is bijective from $(a,\infty)$ to $(0,1)$. This function can also be used to prove $[a,\infty) \sim [0,1)$, $(-\infty, a) \sim (0,1)$ and $(-\infty, a] \sim [0,1)$.

Theorem: If $a \in (0,1)$, then $[0,a)\cup (a,1]=[0,1]\setminus \left \{a \right \} \sim [0,1]$.
Proof: Let $f(x)=\left\{\begin{matrix} a^{n}, \\ x, \;\; \end{matrix}\right. \begin{matrix} x=a^{n+1}, n \in \mathbb{N} \\ x \neq a^{n+1} \;\;\;\; \;\; \end{matrix}$.
Then f is bijective from $[0,1]\setminus \left \{a \right \}$ to $[0,1)$.

Theorem: $\mathbb{R} \sim \mathbb{R} \setminus \mathbb{A}$
Proof: Let $P= \left\{\begin{matrix} \frac{1}{p} \end{matrix}\right|\left.\begin{matrix} p \in \mathbb{P} \end{matrix}\right\}$. Clearly $P \sim \mathbb{N}$, as $P \sim \mathbb{P}$ and $\mathbb{P} \sim \mathbb{N}$. Moreover, as $\mathbb{N} \sim \mathbb{A}$, we have $P \sim \mathbb{A}$, and since $(0,1) \sim \mathbb{R}$, we just need to show that $(0,1) \sim (0,1) \setminus P$. Let
$f(x)=\left\{\begin{matrix} \frac{1}{p_k^{n+1}}, \\ x, \;\; \end{matrix}\right. \begin{matrix} x=\frac{1}{p_k^{n}}, n,k \in \mathbb{N} \\ x \neq \frac{1}{p_k^{n}} \;\;\;\;\;\;\;\;\;\;\; \end{matrix}$.
Then $f$ is bijective from $(0,1)$ to $(0,1) \setminus P$. Therefore, it follows that $\mathbb{R} \sim \mathbb{R} \setminus \mathbb{A}$.

Theorem: $\mathcal{P}(\mathbb{N}_{n})\sim \mathbb{N}_{2^{n}}$.
Proof: We use the fact that every number has a unique binary representation. Let $A \in \mathcal{P}(\mathbb{N}_{n})$. We then let $f(A)=1+\sum_{x \in A} 2^{x-1}$ and $f(\emptyset)=1$. Then f is bijective from $\mathcal{P}(\mathbb{N}_{n})$ to $\mathbb{N}_{2^{n}}$.

Theorem: If $A \sim B$ then $\mathcal{P}(A) \sim \mathcal{P}(B)$.
Proof: Let $f$ be a bijective function from A to B. Let $X \in \mathcal{P}(A)$. Then $Y= f(X)$ will be a bijective function from $\mathcal{P}(A)$ to $\mathcal{P}(B)$. However, by Easton's Theorem, the converse is only true if we assume the generalized continuum hypothesis, which is the proposition that, if $X$ is an infinite set, there is no set $Y$ such that $X \prec Y \prec \mathcal{P}(X)$.

Theorem: $\mathbb{N}_{m} \times \mathbb{N}_{n} \sim \mathbb{N}_{mn}$.
Proof: Let $(a,b)\in \mathbb{N}_{m} \times \mathbb{N}_{n}$. Let $f((a,b))= (a-1)n+b$, then f is bijective from $\mathbb{N}_{m} \times \mathbb{N}_{n}$ to $\mathbb{N}_{mn}$.

Theorem: $\mathcal{P}(\mathbb{N}) \sim (0,1)$.
Proof: We use the fact that every number has a binary representation. However, there are equivalent representations for numbers with finite binary representations. For instance, $\frac{3}{4}=0.11_{2}=0.101111..._{2}$. Let $A\in \mathcal{P}(\mathbb{N})$. We the define
$f(A)=\left\{\begin{matrix} \frac{1}{8}+\frac{1}{4}\sum_{k \in A} 2^{-k}, \\ \frac{1}{2}+\frac{1}{4}\sum_{k \in A} 2^{-k}, \end{matrix}\right. \begin{matrix} A \prec \mathbb{N} \\ A \sim \mathbb{N} \end{matrix}$.
Then we can see that $f$ is injective from $\mathcal{P}(\mathbb{N})$ to $(0,1)$. We then define $g(x)= \left\{\begin{matrix} n \end{matrix}\right|\left.\begin{matrix} n\in \mathbb{N} \wedge \left \lfloor 2^n x \right \rfloor \equiv 1 \pmod 2 \end{matrix}\right\}$. Then we can see that $g$ is injective from $(0,1)$ to $\mathcal{P}(\mathbb{N})$. Thus, as $\mathcal{P}(\mathbb{N}) \preceq (0,1)$ and $(0,1) \preceq \mathcal{P}(\mathbb{N})$, we find that $(0,1) \sim \mathcal{P}(\mathbb{N})$.
Corollary: $\mathcal{P}(\mathbb{N}) \sim \mathbb{R}$.

Theorem: $\mathbb{N} \times \mathbb{N} \sim \mathbb{N}$.
Proof: The function $f(n)=(n,1)$ is injective from $\mathbb{N}$ to $\mathbb{N} \times \mathbb{N}$. Therefore $\mathbb{N} \preceq \mathbb{N} \times \mathbb{N}$. The function $f((a,b))=2^{a}3^{b}$ is injective from $\mathbb{N} \times \mathbb{N}$ to $\mathbb{N}$. Therefore $\mathbb{N} \times \mathbb{N} \preceq \mathbb{N}$. Therefore $\mathbb{N} \times \mathbb{N} \sim \mathbb{N}$.

Theorem: $[0,1] \times [0,1] \sim [0,1]$.
Proof: We use the fact that ever number has a unique binary representation. Let $a,b \in [0,1]$, and $a=\sum_{k=1}^{\infty} a_{k} 2^{-k}$ and $b=\sum_{k=1}^{\infty} b_{k} 2^{-k}$, where $a_k , b_k \in \left \{ 0,1 \right \}$. Then, let $f((a,b))=\sum_{k=1}^{\infty} \left ( 2 a_k+b_k \right ) 4^{-k}$ and then we can see that $f$ is bijective from $[0,1] \times [0,1]$ to $[0,1]$.
Corollary: $\mathbb{R} \times \mathbb{R} \sim \mathbb{R}$.

Theorem: If $\mathbb{N} \preceq A$ or $\mathbb{N} \preceq B$, then $|A|+|B|=|A|\times|B|=\max(|A|,|B|)$.

Theorem: $\mathbb{S}_{m}+\mathbb{S}_{n}=\mathbb{S}_{m+n}$ and $\mathbb{S}_{m}\times\mathbb{S}_{n}=\mathbb{S}_{mn}$.

Theorem: The following proposition is undecidable (cannot be proved or disproved) in ZFC: "There does not exist a set $X \subseteq \mathbb{R}$ such that $\mathbb{N} \prec X \prec \mathbb{R}$."