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The Timpson Lyceum · General Topology (Point Set Topology)

Axiomatic Set Theory – Lecture 1 (Topology)

Axiomatic Set Theory – Lecture 1

Overview

This lecture opens the topology course by building its foundation: axiomatic set theory. Topology is introduced as the study of proximity — the structural relation it preserves (homeomorphism) cares only about which points are "close," not about distances, just as a rubber band can be stretched without separating neighboring points. Because every mathematical object will ultimately be defined as a set, we first lay down a rigorous, axiom-based theory of sets to avoid the contradictions of naive set theory. The lecture covers the first several of nine axioms, the core definitions (subset, equality, ordered pair), and the construction of set-builder notation.

Key Concepts

Topology as the study of proximity. Every field of mathematics has its own "sameness" relation: congruence in geometry, isomorphism in algebra. In topology the sameness relation is homeomorphism — two spaces are the same if one can be deformed into the other (stretched like a rubber band, without tearing or gluing) so that proximity is preserved. Distances may change freely; only closeness matters. The familiar notion of a continuous function from R\mathbb{R} to R\mathbb{R} is just one special case (the standard topology on R\mathbb{R}) of this more general idea.

Primitives. The theory leaves two notions undefined:

Axiom 1 (membership is a proposition). The expression x∈yx \in y is a proposition (true or false, never both) if and only if xx and yy are both sets: ∀x ∀y ((x∈y)⊻¬(x∈y))\forall x\, \forall y \,\big( (x \in y) \veebar \neg(x \in y) \big) This rules out defining objects that aren't sets. It is motivated by Russell's Paradox: let U={ x:x∉x }U = \{\, x : x \notin x \,\}, the set of all sets that don't contain themselves. Asking whether U∈UU \in U gives U∈U  ⟺  U∉UU \in U \iff U \notin U — a contradiction. The resolution: "U∈UU \in U" is neither true nor false, so it isn't a proposition, so by Axiom 1, UU is not a set. You cannot conjure a set into existence merely by writing it down.

Basic definitions (all built from ∈\in):

Axiom 2 (empty set exists). ∃x ∀y (y∉x)\exists x\, \forall y\,(y \notin x). A theorem then proves this set is unique, licensing the definition of the constant ∅\varnothing, the empty set.

Axiom 3 (pairing). If XX and YY are sets, then {X,Y}\{X, Y\} — the set containing precisely XX and YY — is a set: ∀X ∀Y ∃M ∀u (u∈M  ⟺  (u=X ∨ u=Y))\forall X\, \forall Y\, \exists M\, \forall u\,\big(u \in M \iff (u = X \,\vee\, u = Y)\big) This is what introduces the brace notation. A singleton is defined as {X}:={X,X}\{X\} := \{X, X\}. So far only one- and two-element sets are defined.

Ordered pairs as sets. Order is not built in, so it must be encoded: (X,Y):={ {X},  {X,Y} }(X, Y) := \{\, \{X\},\; \{X, Y\} \,\} The left element is recoverable (it is the one belonging to both inner sets / the member of the other), so the information is preserved: (X,Y)=(a,b)  ⟺  X=a∧Y=b(X,Y) = (a,b) \iff X = a \wedge Y = b. Triples and longer tuples are defined recursively, e.g. (X,Y,Z):=((X,Y), Z)(X, Y, Z) := \big((X, Y),\, Z\big).

Axiom 4 (union). If XX is a set, then ⋃X\bigcup X — whose elements are the elements of the elements of XX — is a set: ∀X ∃M ∀y (y∈M  ⟺  ∃u (u∈X∧y∈u))\forall X\, \exists M\, \forall y\,\big(y \in M \iff \exists u\,(u \in X \wedge y \in u)\big) This is used to define the familiar multi-element brace notation: {a,b,c}:=⋃{ {a,b},{c} }\{a,b,c\} := \bigcup\{\,\{a,b\},\{c\}\,\}, extended recursively to {a1,…,an}\{a_1, \dots, a_n\} for n≥3n \ge 3.

Functional relations. A relation RR is functional if every input is related to exactly one output: ∀x ∃! y (x R y)\forall x\, \exists!\, y\,(x \,R\, y). Example: the doubling relation, where xx maps to 2x2x.

Axiom 5 (replacement / image). If MM is a set and RR is a functional relation, then the image Im(R,M)\mathrm{Im}(R, M) — all yy such that some x∈Mx \in M has x R yx \,R\, y — is a set.

Set-builder notation (theorem from Axiom 5). For a predicate PP and a set MM, the collection { y∈M:P(y) }\{\, y \in M : P(y) \,\} is a set. Crucially, the element must be drawn from an existing set MM — you can never write { y:P(y) }\{\, y : P(y)\,\} unrestricted, which is exactly the move that produced Russell's Paradox.

Worked Example

Claim (pairing is unordered): {X,Y}={Y,X}\{X, Y\} = \{Y, X\}.

To prove two sets equal, show each is a subset of the other (set equality), and to show A⊆BA \subseteq B, take an arbitrary a∈Aa \in A and prove a∈Ba \in B.

set equality:X=Y  ⟺  (X⊆Y)∧(Y⊆X)take arbitrary a∈{X,Y}:a∈{X,Y}  ⟺  (a=X ∨ a=Y)commutativity of ∨:(a=X ∨ a=Y)  ⟺  (a=Y ∨ a=X)by Axiom 3 (pairing):(a=Y ∨ a=X)  ⟺  a∈{Y,X}therefore:{X,Y}⊆{Y,X}  and (same steps)  {Y,X}⊆{X,Y}conclude:{X,Y}={Y,X}■\begin{aligned} &\text{set equality:} && X = Y \iff (X \subseteq Y) \wedge (Y \subseteq X) \\ &\text{take arbitrary } a \in \{X, Y\}: && a \in \{X,Y\} \iff (a = X \,\vee\, a = Y) \\ &\text{commutativity of } \vee: && (a = X \,\vee\, a = Y) \iff (a = Y \,\vee\, a = X) \\ &\text{by Axiom 3 (pairing):} && (a = Y \,\vee\, a = X) \iff a \in \{Y, X\} \\ &\text{therefore:} && \{X,Y\} \subseteq \{Y,X\} \;\text{and (same steps)}\; \{Y,X\} \subseteq \{X,Y\} \\ &\text{conclude:} && \{X, Y\} = \{Y, X\} \quad\blacksquare \end{aligned}

Every step is a logical equivalence (  ⟺  \iff), so the containment holds in both directions and the two sets are equal — confirming that listing order does not distinguish a set.

Summary