Ground Zero Mathematics

Previously, we discussed predicate logic and claimed that it is ubiquitous in mathematics. In this post, we will apply said logic to a naïve notion of set theory.

In a naïve sense, a set is a collection of objects. Formally, we use the Zermelo-Fraenkel axioms to model symbols that behave like what we would expect from collections of objects. In particular, we want to define several properties about sets.

Definition 1. For any set $K$ , we call the set $L$ a subset of $K$ , denoted $L \subseteq K$ , to mean the following universally quantified proposition:

$\forall x \in L\quad (x \in K).$

This forms the basis of the element-chasing technique in proving subset relations. We illustrate an example below.

Example 1. Define the following sets:

$\begin{aligned} K &:= \{(p,q) : p \in \{\mathrm T, \mathrm F\} \wedge q \in \{\mathrm T, \mathrm F\}\}, \\ L &:= \{(p,q) \in K : p \wedge q\}, \\ M &:= \{(p,q) \in K : p \to q\}. \end{aligned}$

We claim that $L \subseteq M$ . The definitions of $K,L,M$ are legitimate under axiomatic set theory by the axiom schema of specification. Furthermore, we will rigorously define the elements $(p,q)$ of $K$ later on when discussing cartesian products. Fundamentally, it means that

$(p,q) = (p',q')\quad \iff \quad p=p'\quad \wedge \quad q=q'.$

Proof. Fix $(p, q) \in L$ . Then $(p,q) \in K$ and $p \wedge q$ . This means that $p$ and $q$ . Therefore, $p \to q$ . Since $(p,q) \in K$ and $p \to q$ , we must have $(p, q) \in M$ . Therefore, $L \subseteq M$ .

Definition 2. Let $L, M \subseteq K$ . We write $L = M$ to mean the following universally quantified proposition:

$\forall x \in K\quad (x \in L\quad \iff \quad x \in M).$

More technically, this is formulated precisely using the axiom of extensionality in axiomatic set theory.

This tells us the sets formed by the axiom schema of specification must be unique.

Theorem 1. Let $\phi$ be a predicate on a set $K$ . Then the set defined by

$\{x \in K : \phi(x)\}$

is unique.

Proof. Suppose $L, M$ are subsets of $K$ where $x \in L$ means that $\phi(x)$ , and $x \in M$ means that $\phi(x)$ . Then

$x \in L \quad \iff \quad \phi(x) \quad \iff \quad x \in M.$

By the axiom of extensionality, $L = M$ .

This technique also allows us to prove $L = M$ via two-way subset inclusion.

Theorem 2. Let $L,M \subseteq K$ . Then $L = M$ if and only if $L \subseteq M$ and $M \subseteq L$ .

Proof. Suppose $L = M$ . We first prove that $L \subseteq M$ . Fix $x \in L$ . Since $L \subseteq K$ , $x \in K$ . Since $L = M$ , we have

$x \in L\quad \iff \quad x \in M.$

This implies that

$x \in L \quad \Rightarrow \quad x \in M.$

Since $x \in L$ , by modus ponens, we conclude that $x \in M$ . Therefore, $L \subseteq M$ .

The proof that $M \subseteq L$ is left as an exercise. (The argument is, quite literally, symmetric.)

Now, on the other hand, suppose $L \subseteq M$ and $M \subseteq L$ . Fix $x \in K$ . We need to prove the following two propositions:

$x \in L \Rightarrow x \in M$ .
$x \in M \Rightarrow x \in L$ .

The first statement follows from $L \subseteq M$ . The second statement follows from $M \subseteq L$ . Therefore,

$x \in L \quad \iff \quad x \in M.$

This implies that $L = M$ , as required.

The power of this proving technique arises when our calculations are not reversible. We will elaborate on this when discussing sets in detail.

We can even define the complement, intersection, and union of sets rigorously.

Definition 3. Let $K$ be a set and $L, M \subseteq K$ . The complement of $L$ , denoted $\overline L$ , is defined by

$\overline L := \{x \in K : x \notin L\}.$

The intersection of $L$ and $M$ , denoted $L \cap M$ , is defined by

$L \cap M := \{x \in K : x \in L \wedge x \in M\}.$

The union of $L$ and $M$ , denoted $L \cup M$ , is defined by

$L \cup M := \{x \in K : x \in L \lor x \in M\}.$

The union is also conceptualised as a consequence of the axiom of union.

Notice that $\overline{\cdot}, \cap, \lor$ correspond to the logical notions $\neg, \wedge, \lor$ respectively.

In particular, we can discuss the notion of the empty set.

Theorem 3. For sets $L, M \subseteq K$ , $L \cap \overline L = M \cap \overline M$ .

Proof. Fix $x \in K$ . Then

$x \in L \cap \overline L \quad \iff \quad x \in L \wedge x \in \overline L\quad \iff \quad x \in L \wedge \neg(x \in L) = \mathrm F.$

Therefore, by a similar argument

$x \in L \cap \overline L \quad \iff \quad \mathrm F \quad \iff \quad x \in M \cap \overline M.$

This allows us to define the empty set as $\emptyset = K \cap \overline K$ for some (in fact, any) set $K$ .

We can even discuss cartesian products, but we will supplement this detail later on. Basically, we want to formalise the notion of ordered pairs. We’ll describe the crucial property that we need in the next post—nested quantifiers.

—Joel Kindiak, 27 Nov 24, 2226H

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Ground Zero Mathematics

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