The Timpson Lyceum · General Topology (Point Set Topology)

Set Theoretical Relations – Lecture 3 (Topology)

Overview

This lecture continues Chapter 0, the toolbox of set-theoretic language used throughout topology. It introduces indexed collections of sets — a precise way to attach an entire set to each element of some "indexing" set — and the notation for taking unions and intersections over such a family. From there it builds nested collections, disjoint and mutually disjoint families, and proves the Union Lemma, a deceptively simple result that is used constantly once topology proper begins. The lecture closes with set difference, Cartesian products, the distributive laws relating $\cap$, $\cup$, $\times$, and $\setminus$, and finally relations and equivalence relations. The throughline: in topology we manipulate sets directly, so fluency with this notation is the real payoff.

Key Concepts

Standard number sets and interval notation.

• $\varnothing$ empty set; $\mathbb{N} = \{0,1,2,\dots\}$; $\mathbb{Z}$ integers (with $\mathbb{Z}^+$, $\mathbb{Z}^-$); $\mathbb{Q}$ rationals (numbers expressible as $a/b$ with $a,b\in\mathbb{Z}$); $\mathbb{R}$ reals.
• Open interval $(a,b)=\{x\in\mathbb{R} : a < x < b\}$; closed $[a,b]=\{x\in\mathbb{R} : a \le x \le b\}$; half-open mix-and-match. Infinite endpoints: $[a,\infty)=\{x\in\mathbb{R}: x\ge a\}$; $(-\infty,b)=\{x\in\mathbb{R}: x<b\}$; $(-\infty,\infty)=\mathbb{R}$. Writing $\pm$ together, or "$<\infty$", is nonsense notation.

Indexed collection of sets (Def 0.1). Let $D$ be a set such that for every $d\in D$ there exists a set $S_d$. The collection $\mathcal{C}=\{S_d : d\in D\}$ is a collection of sets indexed by $D$; $D$ is the indexing set. So $D$ is a set of objects and $\mathcal{C}$ is a set of sets, one per element of $D$. Example: let $D$ be the towns of Arizona; for each town $d$ let $S_d$ be the set of barbers working in $d$. Then $\mathcal{C}$ is the collection of those barber-sets. The subscript is only a name — indexing by $0,1,2,\dots$ (as with $A_0, A_1, A_2$) is the same idea.

Special case — consecutive positive integers. When $D = \{m, m+1, \dots, n\}$, write $\{S_d\}_{d=m}^{n}$ as shorthand for $S_m, S_{m+1}, \dots, S_n$.

Nested collection (Def 0.2). A collection $\{S_d\}_{d=m}^{n}$ is nested provided each set contains the next: $S_{k+1} \subseteq S_k \quad\text{for } k = m, \dots, n-1.$ This ordering only makes sense when the indexing set is consecutive positive integers. Example: $S_n=\left[-\tfrac1n, \tfrac1n\right]$ gives $[-1,1]\supseteq[-\tfrac12,\tfrac12]\supseteq[-\tfrac13,\tfrac13]\supseteq\cdots$, zeroing in on $0$.

Union and intersection of an indexed family.

• $U\cup V=\{x : x\in U \text{ or } x\in V\}$; $\ \ U\cap V=\{x : x\in U \text{ and } x\in V\}$.
• $\displaystyle\bigcup_{d\in D} S_d = \{x : \exists\, d\in D \text{ such that } x\in S_d\}$ — $x$ need only appear in at least one $S_d$.
• $\displaystyle\bigcap_{d\in D} S_d = \{x : \forall\, d\in D,\ x\in S_d\}$ — $x$ must appear in every $S_d$.
• For $S_n=[-\tfrac1n,\tfrac1n]$: $\bigcup_{n} S_n = [-1,1]$ and $\bigcap_{n} S_n = \{0\}$.

Disjoint and mutually disjoint. Sets $A,B$ are disjoint if $A\cap B=\varnothing$. An indexed collection is mutually disjoint if for any two distinct indices $m\ne n$, $S_m\cap S_n=\varnothing$ — no two sets share an element. (E.g. "students who are exactly $d$ years old" is mutually disjoint; "at least $d$ years old" is not.)

Set difference. $A\setminus B = \{x : x\in A \text{ and } x\notin B\}$. Note $A\setminus B \ne B\setminus A$. If $A\cap B=\varnothing$ then $A\setminus B = A$; if $A\subseteq B$ then $A\setminus B=\varnothing$.

Cartesian product. $A\times B = \{(x,y) : x\in A,\ y\in B\}$ — a set of ordered pairs, not an interval. Extends to $n$-tuples $A_1\times\cdots\times A_n$. $\mathbb{R}\times\mathbb{R}=\mathbb{R}^2$ is the coordinate plane; $\mathbb{R}^3$ is 3-space.

Distributive set identities (each provable by translating $\cap,\cup$ into the logical "and/or" and using propositional logic, since "$x\in A$" is a proposition):

• $A\cap(B\cup C) = (A\cap B)\cup(A\cap C)$
• $A\cup(B\cap C) = (A\cup B)\cap(A\cup C)$
• $A\times(B\cup C) = (A\times B)\cup(A\times C)$, and likewise with $\cap$ and $\setminus$
• $A\setminus(B\cup C) = (A\setminus B)\cap(A\setminus C)$
• $A\setminus(B\cap C) = (A\setminus B)\cup(A\setminus C)$

Relation. A relation is just a set of ordered pairs. We write $x \mathbin{R} y \iff (x,y)\in R$. If $R\subseteq X\times Y$, then $R$ is a relation from $X$ to $Y$; if $R\subseteq X\times X$, then $R$ is a relation on $X$.

Equivalence relation. A relation $R$ on $X$ is an equivalence relation if for all $a,b,c\in X$:

• Reflexive: $a \mathbin{R} a$;
• Symmetric: $a \mathbin{R} b \implies b \mathbin{R} a$;
• Transitive: $a \mathbin{R} b$ and $b \mathbin{R} c \implies a \mathbin{R} c$.

Examples that are equivalence relations: $=$ on $\mathbb{R}$, "has the same mother as," congruence of triangles. Non-examples: "is an ancestor of" (transitive only), "loves" (none), "has a common parent as" (reflexive and symmetric but not transitive). Reflexivity is relative to the domain $X$: $R=\{(1,1),(2,2)\}$ is reflexive on $\{1,2\}$ but not on $\{1,2,3\}$.

Worked Example

The Union Lemma. Let $X$ be a set and $\mathcal{C}$ a collection of subsets of $X$. If for every $x\in X$ there exists some $U_x\in\mathcal{C}$ with $x\in U_x$, then $X = \bigcup_{x\in X} U_x.$

Proof. Assume that for every $x\in X$ there is some $U_x\in\mathcal{C}$ with $x\in U_x$. We show mutual containment.

$(\subseteq)$ Let $a\in\bigcup_{x\in X} U_x$. Then $a$ lies in at least one of the united sets, so there is some $x'\in X$ with $a\in U_{x'}$. Since each $U_{x'}\subseteq X$, we get $a\in X$. Hence $\bigcup_{x\in X} U_x \subseteq X$.

$(\supseteq)$ Let $a\in X$. By assumption there exists $U_a\in\mathcal{C}$ with $a\in U_a$. Since $U_a$ is one of the sets in the union, $a\in\bigcup_{x\in X} U_x$. Hence $X\subseteq \bigcup_{x\in X} U_x$.

A distributive proof, symbolically first. To verify $A\cap(B\cup C)=(A\cap B)\cup(A\cap C)$, translate to logic, where $\alpha:\ x\in A$, $\beta:\ x\in B$, $\gamma:\ x\in C$, and apply the distributive law $\alpha\wedge(\beta\vee\gamma)\equiv(\alpha\wedge\beta)\vee(\alpha\wedge\gamma)$:

Summary

• An indexed collection $\{S_d\}_{d\in D}$ attaches one set $S_d$ to each $d$ in an indexing set $D$; the subscript is just a name. Consecutive-integer indexing gives the shorthand $\{S_d\}_{d=m}^{n}$.
• A family is nested when $S_{k+1}\subseteq S_k$; $\bigcup_{d} S_d$ collects elements in some $S_d$, while $\bigcap_{d} S_d$ keeps elements in every $S_d$.
• Sets are disjoint if their intersection is empty; a family is mutually disjoint if every pair of distinct members is disjoint.
• The Union Lemma: if every point of $X$ lies in some chosen subset $U_x$, then $X$ is the union of those subsets — proved by showing each side is a subset of the other.
• $A\setminus B$, the Cartesian product $A\times B$ of ordered pairs, and the distributive identities ($\cap,\cup,\times,\setminus$) all reduce to propositional logic on the membership statements.
• A relation is a set of ordered pairs; a relation on $X$ is an equivalence relation iff it is reflexive, symmetric, and transitive (reflexivity judged relative to the chosen domain $X$).