Sets: characterized by elements, equal by extensionality, and why unrestricted formation fails

On this page
  1. Goal and scope
  2. Givens and symbols
  3. Rules and sources
  4. Main derivation
  5. Substeps and lemmas
  6. Reverse audits
  7. Error controls and negative controls
  8. Result and limitations
  9. Verification and source binding

← Guide index · Original MKO

derivation-mko-set-01 · production-20261019-adopted-96

Goal and scope

In the universe ℤ, let A={x∈ℤ ∣ x²=4}, B={−2,2}, C={2,−2,2} and D={2}. Prove A=B and C=B; prove D≠B and distinguish the element 2 from the set {2}. Finally prove that in any domain U with a membership relation there is no R∈U with 'y∈R ⇔ y∉y' for all y∈U, so not every condition forms a set.

Follows the MKO: within a specified universe a set is characterized by its elements, and sets with the same elements are equal. The first part works in the universe ℤ, where all sets are sets of integers; the second part, Russell's argument, uses pure logic and holds for every domain with a membership relation, without choosing ZF, ZFC or any other axiom system. It does not treat the general theory of set operations and does not change the canonical MKO or its English edition.

Givens and symbols

  • set_extensionality
  • set_builder
  • subset_def
  • zero_product
  • distributive
  • numeric_evaluation
  • G1 — Universe and types: the first part works in ℤ, and the sets discussed are sets of integers; in x∈S the left side is an integer and the right side a set of integers. List notation: x∈{a,b,c} ⇔ x=a ∨ x=b ∨ x=c, and likewise with two elements or one.
  • G2 — A={x∈ℤ ∣ x²=4}, B={−2,2}, C={2,−2,2} and D={2}.
  • G3 — Setting for Russell's argument: U is any domain of objects with a binary relation ∈ (for y,z∈U, y∈z is a proposition); no axioms of set theory are assumed; classical logic is used.
  • x\in A — x is an element of A
  • \{a,b,c\} — list notation: x∈{a,b,c} ⇔ x=a∨x=b∨x=c
  • \{x\in\mathbb{Z}\mid P(x)\} — the set carved out of ℤ by the condition P
  • A=B — set equality: exactly the same elements
  • U — any domain with a membership relation (for Russell's argument)

Rules and sources

  • set_extensionality · prerequisite — Set equality: A=B if and only if A⊆B and B⊆A, i.e. they have the same elements.
  • set_builder · prerequisite — Set-builder notation: x∈{y∈A ∣ P(y)} if and only if x∈A and P(x).
  • subset_def · prerequisite — Subset: A⊆B means every element of A is an element of B, i.e. ∀x (x∈A⇒x∈B).
  • zero_product · prerequisite — Zero-product property: in ℤ, ℚ, ℝ, ℂ (or any integral domain), uv=0 ⇔ u=0 or v=0.
  • distributive · prerequisite — Distributivity: u(v+w)=uv+uw; with commutativity also (v+w)u=vu+wu.
  • numeric_evaluation · prerequisite — Exact evaluation of an explicit integer or rational expression; independently recomputable.
  • and_intro_elim · prerequisite — Conjunction: from P and Q infer P∧Q; from P∧Q infer P and infer Q.
  • eq_congruence · prerequisite — Congruence of equality: if u=v, applying the same defined operation f to both sides gives f(u)=f(v). Operating on one side only is not an instance of this rule.
  • add_inverse · prerequisite — Additive inverse: every u has −u with u+(−u)=0; u−v means u+(−v).
  • add_assoc · prerequisite — Associativity of addition: (u+v)+w=u+(v+w).
  • add_zero · prerequisite — Additive identity: u+0=u.
  • neg_mul · prerequisite — Signs and multiplication: (−u)v=−(uv) and (−u)(−v)=uv.
  • eq_replace · prerequisite — Replacement of equals: an expression inside a statement may be replaced by an equal expression without changing whether the statement is true.
  • logic_chain · prerequisite — Chaining: from P⇒Q and Q⇒R infer P⇒R; likewise for ⇔.
  • forall_intro · prerequisite — ∀-introduction (generalization): if a statement is proved for an arbitrary element x, with no extra assumption on x, it holds for every x in the domain.
  • contradiction · prerequisite — Proof by contradiction: if assuming ¬P leads to a contradiction (a statement and its negation both hold), then P holds.
  • exists_elim · prerequisite — ∃-elimination: from ∃x∈D, P(x) we may name some element a with P(a) by a fresh name; reasoning with a is allowed as long as the final conclusion does not mention a.
  • forall_elim · prerequisite — ∀-elimination (instantiation): if ∀x∈D, P(x) holds and a∈D, then P(a) holds.
  • case_analysis · prerequisite — Proof by cases: if P∨Q holds and both P⇒R and Q⇒R are proved, then R holds. The cases must cover every possibility.
  • modus_ponens · prerequisite — Modus ponens: from P and P⇒Q infer Q.
  • iff_intro · prerequisite — Biconditional: P⇔Q holds exactly when both P⇒Q and Q⇒P hold.
  • mko_set · cited — Within a specified universe or type, a set A is characterized by which objects are its elements; sets with exactly the same elements are regarded as equal. · mko-set
  • list_set · prerequisite — List notation: x∈{a,b,c} ⇔ x=a ∨ x=b ∨ x=c; likewise with two elements or one.
  • or_laws · prerequisite — Commutativity, associativity and idempotence of disjunction: P∨Q ⇔ Q∨P; (P∨Q)∨R ⇔ P∨(Q∨R); P∨P ⇔ P.
  • excluded_middle · prerequisite — Excluded middle (classical logic): for every proposition P, P∨¬P.
  • typed_membership · prerequisite — Typed membership: in the context of G1, x∈S is a well-formed proposition only when x is an integer and S is a set of integers; a set of integers is not itself an integer.
  • L_C · lemma — For every x∈ℤ: x∈C ⇔ x∈B.

Main derivation

S01 · definition

Before

\text{a set}\ S\ \text{in the universe}\ \mathbb{Z}

After

A=B\iff\forall x\in\mathbb{Z}\,(x\in A\iff x\in B)

Conditions: none

Rule: mko_set · Uses: G1

Leaf basis: mko_set + set_extensionality

By the MKO, within a specified universe a set is characterized by which objects are its elements, and sets with exactly the same elements are equal. The universe is ℤ (G1), so proving two sets equal means proving x∈A ⇔ x∈B for an arbitrary integer x, and proving them unequal means finding one integer in one set but not the other.

S02 · equivalent

Before

x\in A

After

x^2=4

Conditions: x\in\mathbb{Z}

Rule: set_builder · Uses: G2

Leaf basis: set_builder + and_intro_elim

By set-builder notation, x∈A ⇔ x∈ℤ and x²=4. In the universe ℤ, x∈ℤ already holds, so x∈A ⇔ x²=4. A is carved out of the existing set ℤ by a condition, which is a restricted form of set formation.

S03 · equivalent

Before

x^2=4

After

x^2-4=0

Conditions: none

Rule: eq_congruence · Uses: S02

Leaf basis: eq_congruence + add_inverse + add_assoc + add_zero

Add −4 to both sides; backwards, add 4 to both sides.

S04 · equivalent

Before

x^2-4=0

After

(x-2)(x+2)=0

Conditions: none

Rule: distributive · Uses: S03

Leaf basis: distributive + neg_mul + add_inverse + eq_replace

By distributivity (x−2)(x+2)=x²+2x−2x−4=x²−4 for every integer; replacing equals does not change whether the equation holds.

S05 · equivalent

Before

(x-2)(x+2)=0

After

x-2=0\ \lor\ x+2=0

Conditions: none

Rule: zero_product · Uses: S04

Leaf basis: zero_product

ℤ is an integral domain, so by the zero-product property a product of two integers is 0 exactly when at least one factor is 0.

S06 · equivalent

Before

x-2=0\ \lor\ x+2=0

After

x=2\ \lor\ x=-2

Conditions: none

Rule: eq_congruence · Uses: S05

Leaf basis: eq_congruence + add_inverse + numeric_evaluation

Add 2 to both sides of the first disjunct and −2 to both sides of the second; each is a two-way equivalence, so the disjunction before and after is true or false together.

S07 · equivalent

Before

x=2\ \lor\ x=-2

After

x\in B

Conditions: none

Rule: list_set · Uses: S06, G1, G2

Leaf basis: or_laws + list_set

By commutativity of disjunction rewrite as x=−2 ∨ x=2; by list notation (G1) this is exactly x∈B.

S08 · implies

Before

\text{for arbitrary}\ x\in\mathbb{Z}:\ x\in A\iff x\in B

After

A=B

Conditions: none

Rule: set_extensionality · Uses: S02, S03, S04, S05, S06, S07

Leaf basis: logic_chain + forall_intro + set_extensionality

S02 to S07 hold for an arbitrary integer x; chaining them gives x∈A ⇔ x∈B, and by generalization and the criterion for set equality, A=B. As a contrasting non-element: 3²=9≠4, so 3∉A; also 3≠−2 and 3≠2, so 3∉B, and the two sides agree.

S09 · implies

Before

\forall x\in\mathbb{Z}:\ x\in C\iff x\in B

After

C=B

Conditions: none

Rule: set_extensionality · Uses: L_C, G2

Leaf basis: set_extensionality + list_set + or_laws

By Lemma L_C (proved in S09.1 and S09.2), every integer x has x∈C ⇔ x∈B; by the criterion for set equality, C=B. C is written with three symbols in a different order, but a set depends only on its elements, so order and repetition do not create a new set.

S10 · implies

Before

-2\in B,\ -2\notin D

After

D\neq B

Conditions: none

Rule: contradiction · Uses: G1, G2, list_set

Leaf basis: contradiction + set_extensionality + list_set + numeric_evaluation

By list notation −2∈B (the first disjunct −2=−2 holds), while −2∈D ⇔ −2=2, and −2≠2, so −2∉D. If D=B the two would have the same elements and −2∈D would follow, a contradiction. So D={2}≠B.

S11 · implies

Before

2\ \text{versus}\ \{2\}

After

\{2\}\subseteq B\iff 2\in B;\quad \{2\}\in B\ \text{is not a true statement}

Conditions: none

Rule: subset_def · Uses: G1, G2, list_set, typed_membership

Leaf basis: subset_def + list_set + typed_membership

2 is an integer and {2}=D is a set of integers; they are not the same object. Their correct relation: by the definition of subset, {2}⊆B ⇔ for every y, y∈{2} ⇒ y∈B; since y∈{2} ⇔ y=2, this is equivalent to 2∈B. As 2∈B holds, {2}⊆B. On the other hand, by typed_membership the elements of B are integers and {2} is not an integer, so '{2}∈B' is not a true statement in this context (in the typed reading it is not even well formed). Confusing 2 with {2} leads to confusing ⊆ with ∈.

S12 · implies

Before

\exists R\in U\ \forall y\in U:\ (y\in R\iff y\notin y)

After

\text{fix such an}\ R\in U

Conditions: none

Rule: exists_elim · Uses: G3

Leaf basis: exists_elim

Turn to the second mistake identified in the MKO, that any condition defines a set without restriction. If the condition 'y∉y' defined a set in the domain U, there would be some R∈U whose elements are exactly the objects that are not elements of themselves. For a proof by contradiction, temporarily assume such an R exists and give it the fresh name R.

S13 · implies

Before

\forall y\in U:\ (y\in R\iff y\notin y)

After

R\in R\iff R\notin R

Conditions: R\in U

Rule: forall_elim · Uses: S12

Leaf basis: forall_elim

The property in S12 holds for every y∈U, and R itself is in U, so we may take y=R and get R∈R ⇔ R∉R.

S14 · case_split

Before

R\in R\iff R\notin R

After

R\in R\ \lor\ R\notin R

Conditions: none

Rule: case_analysis · Uses: S13, excluded_middle

Leaf basis: case_analysis + excluded_middle

By excluded middle, R∈R or R∉R. In S14.1 and S14.2 each case yields a contradiction from S13, and the two branches cover every possibility.

S15 · implies

Before

\text{the assumption of S12 leads to a contradiction}

After

\neg\exists R\in U\ \forall y\in U:\ (y\in R\iff y\notin y)

Conditions: none

Rule: contradiction · Uses: S12, S14, S14.1, S14.2

Leaf basis: contradiction + exists_elim

The temporary assumption of S12 leads to a contradiction in both cases, so by proof by contradiction no such R exists. This holds for every domain U and every membership relation and uses no axiom of set theory. Thus no object R∈U represents the condition y∉y for all y∈U: one cannot require every condition to be represented by a set within U. This does not rule out forming the relative subset R={y∈U | y∉y}⊆U by separation in ordinary set theory; that R is not an element of U. By contrast, S02 used the restricted form that carves elements out of the existing set ℤ.

S16 · implies

Before

\text{conclusions of S08, S09, S10, S11, S15}

After

A=B=C,\ D\neq B,\ \{2\}\subseteq B,\ \neg\exists R\in U\ \forall y\in U:\ (y\in R\iff y\notin y)

Conditions: none

Rule: and_intro_elim · Uses: S08, S09, S10, S11, S15

Leaf basis: and_intro_elim

Together: A written by a condition, B written as a list and C written with a repetition are one and the same set; D={2} differs from B; the element 2 and the set {2} relate to B by ∈ and by ⊆ respectively; no R∈U satisfies y∈R iff y∉y for every y∈U.

Substeps and lemmas

S09.1 · equivalent

Before

x\in C

After

x=2\lor x=-2\lor x=2

Conditions: none

Rule: list_set · Uses: G2, list_set

Leaf basis: list_set

Unfold C={2,−2,2} by list notation: x∈C ⇔ x=2 ∨ x=−2 ∨ x=2.

S09.2 · equivalent

Before

x=2\lor x=-2\lor x=2

After

x=-2\lor x=2,\ \text{i.e.}\ x\in B

Conditions: none

Rule: or_laws · Uses: S09.1, or_laws

Leaf basis: or_laws + list_set

By commutativity and associativity of disjunction rewrite as (x=2∨x=2)∨x=−2; by idempotence x=2∨x=2 ⇔ x=2, giving x=2∨x=−2; commuting gives x=−2∨x=2, which by list notation is x∈B. S09.1 and S09.2 together prove Lemma L_C.

S14.1 · implies

Before

R\in R

After

R\notin R,\ \text{contradiction}

Conditions: none

Rule: modus_ponens · Uses: S13, S14

Leaf basis: modus_ponens + iff_intro

Case one: R∈R. The '⇒' direction of S13 gives R∉R, which together with R∈R is a contradiction.

S14.2 · implies

Before

R\notin R

After

R\in R,\ \text{contradiction}

Conditions: none

Rule: modus_ponens · Uses: S13, S14

Leaf basis: modus_ponens + iff_intro

Case two: R∉R. The '⇐' direction of S13 gives R∈R, which together with R∉R is a contradiction.

Reverse audits

  • S16 · sufficient_goal_reduction · S08, S09, S10, S11, S15, and_intro_elim — Goal G: the four conclusions the problem asks for. Sufficient condition H: S08, S09, S10, S11 and S15; H⇒G just lists them. H is proved forwards from G1 to G3; G is never used as a premise.
  • S15 · premise_audit · S12, S14, contradiction, exists_elim — The witness name R has no free occurrence in the conclusion, as exists_elim requires; nothing is claimed about which conditions form sets in any particular axiom system (the MKO chooses neither ZF nor ZFC).
  • S14.2 · premise_audit · S13, modus_ponens, iff_intro — Only the other direction of the biconditional is used.
  • S14.1 · premise_audit · S13, modus_ponens, iff_intro — Only one direction of the biconditional is used.
  • S14 · premise_audit · S13, excluded_middle, case_analysis — The two cases cover every possibility, by excluded middle.
  • S13 · premise_audit · S12, forall_elim — The instantiation needs R∈U, which is exactly what 'R is a set of the domain' means.
  • S12 · premise_audit · G3, exists_elim — This is a temporary assumption, used only inside the proof by contradiction S12 to S14; the witness R has no free occurrence in the final conclusion S15.
  • S11 · premise_audit · subset_def, list_set, typed_membership — {2}⊆B ⇔ 2∈B holds both ways by the definition of subset and list notation; '{2}∈B is not true' depends only on the type distinction and needs no further axiom.
  • S10 · premise_audit · list_set, set_extensionality, contradiction — One element in one set but not the other is enough to refute equality; it is not claimed that they share no element (2 is in both).
  • S09.2 · inverse_equivalence · or_laws, list_set — Commutativity, associativity and idempotence are all two-way equivalences.
  • S09.1 · inverse_equivalence · list_set — The definition of list notation is two-way.
  • S09 · premise_audit · L_C, set_extensionality — The equivalence is needed for every x; the lemma is proved for an arbitrary x.
  • S08 · premise_audit · S02, S07, logic_chain, forall_intro, set_extensionality — Every integer must be covered: x was arbitrary, with no extra assumption. x=3 is only an example and not part of the proof.
  • S07 · inverse_equivalence · or_laws, list_set — Commutativity and the definition of list notation are both two-way.
  • S06 · inverse_equivalence · eq_congruence, add_inverse, numeric_evaluation — Substituting back: 2−2=0 and (−2)+2=0.
  • S05 · inverse_equivalence · zero_product — Both directions are contained in zero_product.
  • S04 · inverse_equivalence · distributive, eq_replace — Replacement by an identity is two-way; N1 is an independent polynomial check.
  • S03 · inverse_equivalence · eq_congruence, add_inverse — Adding −4 and adding 4 undo each other, with no side condition in ℤ.
  • S02 · inverse_equivalence · set_builder, and_intro_elim — The set-builder definition is itself two-way, and dropping the conjunct x∈ℤ, which already holds, is two-way as well.
  • S01 · premise_audit · G1, mko_set, set_extensionality — This step only states the criterion and derives no new fact.

Error controls and negative controls

An element and the set containing only it

The MKO names the common mistake of confusing an element x with the set containing only x. 2 is an integer and {2} is a set of integers: 2∈B and {2}⊆B both hold and are equivalent (S11), but '{2}∈B' is not true and {2} is not equal to B (S10). On meeting notation such as 2 or {2}, first identify whether it denotes an integer element or a set of integers in this context, then decide between ∈ and ⊆.

Not every condition forms a set

Another mistake identified in the MKO is believing that any condition defines a set without restriction. S12 to S15 prove that for any domain U, no R∈U satisfies y∈R iff y∉y for every y∈U. This does not exclude forming the relative subset R={y∈U | y∉y}⊆U by separation in ordinary set theory; that R has R∉U. So set formation needs principles, for example carving elements out of a known set (as with {x∈ℤ ∣ x²=4} in S02); this guide does not choose a particular axiom system.

Order and repetition do not create new sets

C={2,−2,2} is written with three symbols, but by extensionality C=B={−2,2} (S09). Distinguishing order or multiplicity needs a different kind of object, such as the ordered triple (2,−2,2) or a multiset; those are not sets in the sense of this MKO.

  • {2,−2,2} has three elements, so it differs from {−2,2} → reject — x∈{2,−2,2} ⇔ x=−2∨x=2, the same elements as B, so they are equal by extensionality (S09).
  • 2 and {2} are the same thing → reject — 2 is an integer and {2} is a set of integers; 2∈B holds while '{2}∈B' is not true (S11).
  • For every condition P(y), {y ∣ P(y)} is a set → reject — Take P(y) to be y∉y; S12 to S15 derive a contradiction.
  • A is written with a condition and B as a list, so A≠B → reject — A set is determined by its elements, not by how it is written; S08 proves A=B.

Result and limitations

In the universe ℤ, A={x∈ℤ ∣ x²=4}, B={−2,2} and C={2,−2,2} have the same elements, so A=B=C; D={2}≠B because −2∈B and −2∉D. 2 is an integer and {2} is a set: 2∈B and {2}⊆B hold, while '{2}∈B' is not true. For any domain U with a membership relation, no R∈U satisfies y∈R iff y∉y for every y∈U. This rules out requiring every condition to be represented without restriction by a set within U; it does not prohibit the relative subset R={y∈U | y∉y}⊆U formed by separation in ordinary set theory (which has R∉U).

  • The typed context of the first part (typed_membership) is a premise of this guide, matching the MKO's 'specified universe or type'; in untyped set theory the relation between 2 and {2} depends on the chosen axioms and the coding of numbers, which this guide does not treat.
  • Russell's argument shows only that no internal R∈U represents y∉y for every y∈U, and does not rule out a relative subset R⊆U formed by separation; which conditions do form sets is fixed by a concrete axiom system (the MKO's third paragraph), which this guide does not choose.

Verification and source binding

9 declared checks for this guide; only the listed expressions are covered.

  • source commit: b28a5edbd46279b1db2a3f3f48bb3f447e18899f
  • adopted source SHA-256: 81bb053444cb9fae0c721ca4c93b2655272c5d7fc28922dbed489183ee255de8
  • original source SHA-256: 81bb053444cb9fae0c721ca4c93b2655272c5d7fc28922dbed489183ee255de8
  • Lean: NOT_RUN · independent AI: NOT_MEASURED

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