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The Timpson Lyceum · General Topology (Point Set Topology)

Axiomatic Set Theory & Intro to Topology – Lecture 2 (Topology)

Overview

This lecture completes the axiomatic development of set theory and then opens the study of topology. We finish walking through the axioms of ZFC — building up from membership and the empty set to unions, set-builder notation, power sets, infinity, choice, and regularity — and we see why each axiom is needed and why it cannot simply be defined away. The recurring lesson is that "calling something a set does not make it a set": every construction must be justified, or paradoxes like Russell's appear. The lecture closes by introducing topology as the study of proximity (which points stay near which) and reviewing the number systems and interval notation we will lean on for examples.

Key Concepts

The two primitives. All of set theory is built from exactly two undefined notions: set and the membership relation ∈\in. Everything else is defined from these.

Axiom 1 (membership is a proposition). x∈yx \in y is a proposition if and only if xx and yy are sets. This blocks Russell's paradox: if we try to form U={x:x∉x}U = \{x : x \notin x\} (the set of all sets that do not contain themselves), then asking U∈UU \in U is contradictory — if U∈UU \in U then U∉UU \notin U, and vice-versa. Since "U∈UU \in U" is not a proposition, UU was never a set. Calling a collection a set does not make it one, just as calling 66 prime does not make it prime.

Axiom 2 (empty set). There exists a set with no elements. One can prove it is unique, which justifies giving it the name ∅\varnothing.

Axiom 3 (pairing). If xx and yy are sets, then {x,y}\{x, y\} is a set whose members are precisely xx and yy.

Axiom 4 (union). For every set xx there exists a set mm whose elements are exactly the elements of the elements of xx: ∀x ∃m ∀y (y∈m  ⟺  ∃s (s∈x∧y∈s)).\forall x\, \exists m\, \forall y\,\big(y \in m \iff \exists s\,(s \in x \wedge y \in s)\big). We write m=⋃xm = \bigcup x. For example ⋃{{1,2},{3,4}}={1,2,3,4}\bigcup \{\{1,2\},\{3,4\}\} = \{1,2,3,4\}. Note ⋃x≠x\bigcup x \neq x: here xx has 22 elements but ⋃x\bigcup x has 44. The familiar infix union is the special case a∪b=⋃{a,b}a \cup b = \bigcup\{a,b\}.

Predicate. A Boolean-valued (proposition-valued) function: feed it an object and it returns true or false, e.g. P(x)≡P(x) \equiv "xx is even."

Relation. A predicate of two arguments — given two objects it returns true or false (e.g. x<yx < y, or "xx loves yy").

Functional relation. A relation RR is functional iff for every xx there is one and only one yy with x R yx\,R\,y. Example: x R y  ⟺  y=2xx\,R\,y \iff y = 2x is functional (each xx has a unique double); x R y  ⟺  x<yx\,R\,y \iff x < y is not functional (one xx relates to many yy). For a functional relation we write R(x)=yR(x) = y.

Image. For a set MM and functional relation RR, the image of RR on MM is the collection of all yy such that some x∈Mx \in M has x R yx\,R\,y. With R(x)=2xR(x)=2x and M={12,2,7}M=\{\tfrac12, 2, 7\}, the image is {1,4,14}\{1, 4, 14\}. This is exactly the "range" of an ordinary function over its domain.

Axiom 5 (replacement / image is a set). If RR is functional and MM is a set, then the image of RR on MM is a set.

Set-builder notation (a theorem, not an axiom). For any predicate PP and any set MM, the collection {y∈M:P(y)}\{y \in M : P(y)\} is a set. Crucially we must restrict to an existing set MM — the unrestricted form {x:P(x)}\{x : P(x)\} is what produced Russell's paradox. The proof realizes {y∈M:P(y)}\{y \in M : P(y)\} as the image of a cleverly built functional relation, so Axiom 5 forces it to be a set (sketched below).

Axiom 6 (power set). For any set MM there exists a set P(M)\mathcal{P}(M) whose elements are exactly the subsets of MM. This cannot be obtained from set-builder notation (there is no pre-existing larger set to filter), so it must be an axiom. Example: P({a,b})={∅,{a},{b},{a,b}}\mathcal{P}(\{a,b\}) = \{\varnothing, \{a\}, \{b\}, \{a,b\}\}. A set of nn elements has 2n2^n subsets — P\mathcal{P} of a 1010-element set has 210=10242^{10} = 1024 elements.

Cross product. X×Y={(a,b):a∈X, b∈Y}X \times Y = \{(a,b) : a \in X,\ b \in Y\}. It is not commutative: X×Y≠Y×XX \times Y \neq Y \times X. (Strictly it is carved out of P(P(X∪Y))\mathcal{P}(\mathcal{P}(X \cup Y)), but in practice we use the naive form.)

Axiom 7 (infinity). There exists a set XX with ∅∈X\varnothing \in X and, for every y∈Xy \in X, also y∪{y}∈Xy \cup \{y\} \in X. This guarantees an infinite set and lets us define the natural numbers (von Neumann ordinals): 0=∅0 = \varnothing, and n+1=n∪{n}n+1 = n \cup \{n\}, so 0=∅,1={0},2={0,1},3={0,1,2},…0=\varnothing,\quad 1=\{0\},\quad 2=\{0,1\},\quad 3=\{0,1,2\},\dots Two payoffs: each nn has exactly nn elements (cardinality), and order is membership — x<yx < y means x∈yx \in y (so 2<32 < 3 because 2∈32 \in 3).

Disjoint and mutually disjoint. Two sets are disjoint if A∩B=∅A \cap B = \varnothing. A collection XX is mutually disjoint if any two distinct members of XX have empty intersection.

Axiom 8 (choice). If XX is a set whose members are non-empty and mutually disjoint, then there exists a set YY containing exactly one element from each member of XX. This matters only when we cannot name a representative — we know each set is non-empty but have no rule to pick from it (the "socks vs. shoes" picture: with shoes you can say "take the left one"; with identical socks there is no rule, and Choice supplies the selection set). It is the "C" in ZFC; dropping it gives ZF.

Axiom 9 (regularity / foundation). Every non-empty set XX contains an element yy with X∩y=∅X \cap y = \varnothing. This forbids pathologies like X={X}X = \{X\}: if a set equaled the set containing itself, intersecting would never reach ∅\varnothing. It is the guard against self-referential, ill-founded sets.

Independence of axioms. An axiom is independent of the others if there is an interpretation making the others true but it false; this proves it cannot be derived from them. Logically equivalent axiom systems exist (e.g. several equivalent axiomatizations of Euclidean geometry), and which set of axioms you adopt is what defines the field you work in — just as the parallel postulate distinguishes Euclidean, elliptic, and hyperbolic geometry.

What topology studies. Topology is the study of proximity — which points are "close" to (touching) which — preserved under continuous deformation. Picture a rubber band: stretching or shrinking keeps neighboring atoms neighbors (allowed), but breaking (tearing apart adjacent points) or gluing (welding distant points together) destroys proximity (not allowed). Under this sameness relation a coffee mug is equivalent to a donut: both have exactly one hole, and you can reshape one into the other by redistributing material without cutting or sealing a hole. Topology is sometimes called "rubber-sheet geometry."

Number systems and interval notation.

Worked Example

Claim (set-builder is a set). For any predicate PP and set MM, the collection E={y∈M:P(y)}E = \{y \in M : P(y)\} is a set.

Take the concrete case M={1,2,3,4,5,6}M = \{1,2,3,4,5,6\} with P(x)≡P(x) \equiv "xx is even", so we want E={2,4,6}E = \{2,4,6\}. If nothing in MM satisfies PP, define E=∅E = \varnothing (a set by Axiom 2). Otherwise fix some z∈Mz \in M with P(z)P(z) true (here z=2z = 2) and build the functional relation

xRy    ⟺    (P(x)∧y=x) ∨ (¬P(x)∧y=z),xRy    ⟺    (x even∧y=x) ∨ (x odd∧y=2).\begin{aligned} x \mathbin{R} y \;&\iff\; \big(P(x) \wedge y = x\big)\ \vee\ \big(\lnot P(x) \wedge y = z\big), \\ x \mathbin{R} y \;&\iff\; (x\text{ even} \wedge y = x)\ \vee\ (x\text{ odd} \wedge y = 2). \end{aligned}

Each x∈Mx \in M relates to exactly one yy (itself if even, else zz), so RR is functional. Tabulating:

1↦2,2↦2,3↦2,4↦4,5↦2,6↦6.\begin{aligned} 1 \mapsto 2,\quad 2 \mapsto 2,\quad 3 \mapsto 2,\quad 4 \mapsto 4,\quad 5 \mapsto 2,\quad 6 \mapsto 6. \end{aligned}

Every right-hand value is an element of MM satisfying PP, and every element of MM satisfying PP appears. So the image of RR on MM is exactly {2,4,6}=E\{2,4,6\} = E. By Axiom 5 that image is a set, hence EE is a set. ■\blacksquare

Summary