The Timpson Lyceum · Geometry (2024)

Set Theory and Proofs – Lecture 3 (Geometry)

Overview

This lecture introduces set theory as the foundation on which geometry will be built, layering on top of the propositional and predicate logic developed earlier. Rather than an axiomatic construction, the goal is intuition: what a set is, how to write one, the relations between sets (membership, subset, equality), the common named number sets, set-builder notation, and the core set operations. The lecture closes with worked proofs drawn from an algebra review, showing how to argue rigorously from a tiny set of assumptions about the integers. Together these give you the vocabulary and the proof habits the rest of the course presupposes.

Key Concepts

Sets vs. lists

• A set is an unordered, unique collection: order does not matter and duplicates do not matter. Written with curly braces: $\{1,2,3\}$.
• Because duplicates and order are irrelevant, $\{1,1,2,3\} = \{3,2,1\} = \{1,2,3\}$ — all the same three-element set.
• A list is an ordered, non-unique collection, written with parentheses: $(1,2)$. Here order and repetition matter, so $(1,2) \neq (2,1)$ and $(1,1,2) \neq (1,2)$. Lists model points such as $(x,y)$ in the plane.

Membership ($\in$) — the first set relation (a two-term predicate returning true/false).

• $x \in A$ means "$x$ is an element of $A$"; $x \notin A$ is its negation.
• Watch nesting: $\{1\} \notin \{\{1\},2,3\}$ as the number $1$, but $\{1\} \in \{\{1\},2,3\}$ as the object $\{1\}$. The set $\{1\}$ and the number $1$ are different objects.
• Elements need not be numbers — any objects may be collected into a set.

Subset ($\subseteq$) — defined using only membership: $A \subseteq B \iff \big(\,\forall x:\ x \in A \implies x \in B\,\big)$ Every element of $A$ must also lie in $B$. Note $\{1,2,3\} \subseteq \{1,2,3\}$ (a set is a subset of itself), and $\{\{1\}\} \not\subseteq \{1,3\}$ because $\{1\}$ is not an element of $\{1,3\}$.

Set equality ($=$) — equality is extended beyond logic's "identical symbols" rule: $A = B \iff (A \subseteq B \ \text{and}\ B \subseteq A)$ i.e. each set contains exactly the elements of the other. This is precisely why order and duplicates do not matter.

Proper subset ($\subset$ / $\subsetneq$) — analogous to $<$ versus $\le$: $A \subsetneq B \iff (A \subseteq B \ \text{and}\ A \neq B)$ $A$ is contained in $B$ but is strictly smaller.

Common named sets

• $\varnothing$ — the empty set, the set containing nothing.
• $\mathbb{N}$ — the natural numbers $\{0,1,2,3,\dots\}$ (zero included).
• $\mathbb{Z}$ — the integers $\{\dots,-2,-1,0,1,2,\dots\}$.
• $\mathbb{Q}$ — the rationals, all fractions of integers.
• $\mathbb{R}$ — the reals, every number on the number line (e.g. $\pi$, $\sqrt{2}$).
• These nest: $\varnothing \subseteq \mathbb{N} \subseteq \mathbb{Z} \subseteq \mathbb{Q} \subseteq \mathbb{R}$.

Set-builder notation — $\{\,x \mid P(x)\,\}$, read "the set of all $x$ such that $P(x)$," where $P$ is a predicate.

• Evens: $\{\,n \mid n \in \mathbb{Z} \ \text{and}\ 2 \mid n\,\}$.
• Rationals: $\mathbb{Q} = \left\{\,\tfrac{y}{x} \;\middle|\; y \in \mathbb{Z},\ x \in \mathbb{Z},\ x \neq 0\,\right\}$.
• A parabola: $\{\,(x,y) \mid x \in \mathbb{R},\ y \in \mathbb{R},\ y = x^2\,\}$.

Set operations (relations return true/false; operations take sets and return a set).

• Union: $A \cup B = \{\,x \mid x \in A \ \text{or}\ x \in B\,\}$.
• Intersection: $A \cap B = \{\,x \mid x \in A \ \text{and}\ x \in B\,\}$.
• Set difference: $A \setminus B = \{\,x \mid x \in A \ \text{and}\ x \notin B\,\}$.
• Cartesian (cross) product: $A \times B = \{\,(x,y) \mid x \in A,\ y \in B\,\}$, all ordered pairs with first coordinate from $A$, second from $B$.

Properties of operations

• Commutative: $A \cup B = B \cup A$ and $A \cap B = B \cap A$. Difference is not commutative: $A \setminus B \neq B \setminus A$.
• Associative: $(A \cup B) \cup C = A \cup (B \cup C)$ and $(A \cap B) \cap C = A \cap (B \cap C)$, which justifies writing $A \cup B \cup C$ without parentheses.
• Identities with $\varnothing$: $A \cup \varnothing = A$ (like adding $0$) and $A \cap \varnothing = \varnothing$ (like multiplying by $0$).
• Distributive (mirroring logic's $\lor$/$\land$): $A \cup (B \cap C) = (A \cup B) \cap (A \cup C)$ and $A \cap (B \cup C) = (A \cap B) \cup (A \cap C)$.

Proof setup (integer axioms). The proofs use two cancellation assumptions about integers, plus ordinary arithmetic and the two laws of identity ($\forall x:\ x=x$; equals may be substituted for equals):

• (A1) $a + b = a + c \implies b = c$.
• (A2) if $a \neq 0$, then $a \cdot b = a \cdot c \implies b = c$.

To prove an implication "if $P$ then $Q$," assume the antecedent $P$ and derive $Q$ (a false antecedent is vacuously true). A biconditional "$P$ iff $Q$" is proved in two directions: $P \implies Q$ and $Q \implies P$.

Worked Example

Claim. If $3x + 4 = 19$, then $x = 5$. (Subtraction and division are not yet available — only A1, A2, and arithmetic.)

Assume $3x + 4 = 19$. Rewrite $19 = 15 + 4$ and apply A1 to cancel the common $+4$; then rewrite $15 = 3\cdot 5$ and apply A2 (since $3 \neq 0$) to cancel the common factor $3$:

A subtler case. Given $r(7b - 5) = r$ with $b$ an integer, prove $r = 0$. From $r(7b-5) = r = r\cdot 1$, either $r = 0$ or $r \neq 0$. If $r \neq 0$, A2 cancels $r$ to give $7b - 5 = 1$, i.e. $7b = 6$, forcing $b = \tfrac{6}{7} \notin \mathbb{Z}$ — impossible since $b$ is an integer. Hence the assumption $r \neq 0$ fails, so $r = 0$ (and $r=0$ indeed satisfies the equation). $\blacksquare$

Summary

• A set is unordered with no duplicates ($\{1,1,2\}=\{1,2\}$); a list like $(x,y)$ is ordered and allows repeats.
• The three foundational relations build on one another: membership $\in$ defines $\subseteq$, and $\subseteq$ defines equality ($A=B \iff A\subseteq B \land B\subseteq A$) and proper subset ($A\subsetneq B \iff A\subseteq B \land A\neq B$).
• Memorize the nested number sets $\varnothing \subseteq \mathbb{N} \subseteq \mathbb{Z} \subseteq \mathbb{Q} \subseteq \mathbb{R}$ and read set-builder notation $\{x \mid P(x)\}$ fluently.
• The operations $\cup$, $\cap$, $\setminus$, and $\times$ take sets to sets; $\cup$ and $\cap$ are commutative and associative and distribute over each other, but $\setminus$ is not commutative.
• $\varnothing$ behaves like $0$: $A\cup\varnothing = A$ and $A\cap\varnothing = \varnothing$.
• To prove "if $P$ then $Q$," assume $P$ and derive $Q$; for a biconditional, prove both directions. Working only from cancellation axioms A1/A2 trains you to justify every step rigorously.