Open and closed sets: classifying all sixteen subsets of a four-point topology

Goal and scope

Let X={1,2,3,4} and τ={∅,{1},{1,2},{3,4},{1,3,4},X}. First verify that τ is a topology on X, then list all closed sets and sort the sixteen subsets of X into four classes: clopen, open only, closed only, and neither; finally prove that in every topological space ∅ and X are clopen, and that the union of two closed sets is closed.

Works inside the MKO's scope: the underlying set X and topology τ are fixed (assumption a1), and every complement is taken relative to the same X (assumption a2). The general results proved are only '∅ and X are always clopen' and 'the union of two closed sets is closed'; the dual statement for intersections of arbitrarily many closed sets is mentioned but not proved separately. The canonical MKO and its English edition are not changed.

Givens and symbols

• set_operations_def
• set_extensionality
• subset_def
• finite_counting_basics
• nat_induction

• G1 — The underlying set X={1,2,3,4} (MKO assumption a1).
• G2 — The family τ={∅,{1},{1,2},{3,4},{1,3,4},X}; every member is a subset of X.
• G3 — Every complement is the complement X∖C relative to X, and every candidate set is first confirmed to be a subset of X (MKO assumption a2).

• X — the underlying set {1,2,3,4}
• \tau — the topology on X (the family of open sets)
• \mathcal{C} — the family of closed sets {C⊆X ∣ X∖C∈τ}
• X\setminus C — the complement of C relative to X
• \mathcal{P}(X) — all subsets of X

Rules and sources

• set_operations_def · prerequisite — Set operations: x∈A∪B ⇔ x∈A or x∈B; x∈A∩B ⇔ x∈A and x∈B; x∈A∖B ⇔ x∈A and x∉B; in a universe U, Aᶜ=U∖A.
• set_extensionality · prerequisite — Set equality: A=B if and only if A⊆B and B⊆A, i.e. they have the same elements.
• subset_def · prerequisite — Subset: A⊆B means every element of A is an element of B, i.e. ∀x (x∈A⇒x∈B).
• finite_counting_basics · prerequisite — Finite counting: for disjoint finite sets A, B, |A∪B|=|A|+|B|; |A×B|=|A|·|B|; there is a bijection between finite sets ⇔ they have the same number of elements.
• nat_induction · prerequisite — Principle of induction: for a family P(n) starting at n₀, if P(n₀) holds and P(n)⇒P(n+1) for every n≥n₀, then P(n) holds for every n≥n₀.
• topology_def · prerequisite — Topology: a family τ of open subsets of X containing ∅ and X, closed under arbitrary unions and finite intersections.
• def_unfold · prerequisite — Unfolding a definition: replacing a defined term by its definition (or folding it back) does not change whether a statement is true.
• de_morgan_logic · prerequisite — De Morgan's laws for logic: ¬(P∧Q) ⇔ ¬P∨¬Q and ¬(P∨Q) ⇔ ¬P∧¬Q.
• injective_surjective_def · prerequisite — Injective: f(a)=f(a′) ⇒ a=a′. Surjective: every b∈B has some a∈A with f(a)=b. Bijective: both; then there is a unique inverse f⁻¹ with f⁻¹∘f=id_A and f∘f⁻¹=id_B.
• numeric_evaluation · prerequisite — Exact evaluation of an explicit integer or rational expression; independently recomputable.
• 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.
• forall_elim · prerequisite — ∀-elimination (instantiation): if ∀x∈D, P(x) holds and a∈D, then P(a) holds.
• quantifier_negation · prerequisite — Negating quantifiers: ¬(∀x, P(x)) is equivalent to ∃x, ¬P(x), and ¬(∃x, P(x)) is equivalent to ∀x, ¬P(x).
• iff_intro · prerequisite — Biconditional: P⇔Q holds exactly when both P⇒Q and Q⇒P hold.
• mko_open_closed_def · cited — In a topological space (X,τ), a subset U⊆X is open exactly when U∈τ, and C⊆X is closed exactly when X∖C∈τ. Open and closed are not mutually exclusive; ∅ and X are always both open and closed. · mko-open-closed-set
• mko_topological_space_def · cited — A topology on X is a family τ⊆P(X) with ∅∈τ, X∈τ, the union of any subfamily of τ in τ, and the intersection of any finite subfamily of τ in τ; the union of the empty family is ∅ and its intersection is read as X. · mko-topological-space
• L_finite_family · lemma — Let τ be a finite family of subsets of X. If ∅∈τ and τ is closed under the union of two members, then the union of every subfamily of τ is in τ; if X∈τ and τ is closed under the intersection of two members, then the intersection of every subfamily of τ (the empty one read as X) is in τ.
• L_double_complement · lemma — For every C⊆X, X∖(X∖C)=C.
• L_demorgan · lemma — For A,B⊆X, X∖(A∪B)=(X∖A)∩(X∖B).

Main derivation

\tau=\{\varnothing,\{1\},\{1,2\},\{3,4\},\{1,3,4\},X\}

(X,\tau)\ \text{is a topological space}

C\subseteq X\ \text{closed}\iff X\setminus C\in\tau

C\ \text{closed}\iff C=X\setminus U\ \text{for some}\ U\in\tau

\{X\setminus U\mid U\in\tau\}

\mathcal{C}=\{X,\{2,3,4\},\{3,4\},\{1,2\},\{2\},\varnothing\}

\tau,\ \mathcal{C}

\tau\cap\mathcal{C}=\{\varnothing,\{1,2\},\{3,4\},X\}

\tau,\ \mathcal{C}

\tau\setminus\mathcal{C}=\{\{1\},\{1,3,4\}\}

\tau,\ \mathcal{C}

\mathcal{C}\setminus\tau=\{\{2\},\{2,3,4\}\}

|\mathcal{P}(X)|=2^4=16,\ |\tau\cup\mathcal{C}|=6+6-4=8

\mathcal{P}(X)\setminus(\tau\cup\mathcal{C})=\{\{3\},\{4\},\{1,3\},\{1,4\},\{2,3\},\{2,4\},\{1,2,3\},\{1,2,4\}\}

(Y,\sigma)\ \text{any topological space}

\varnothing\ \text{and}\ Y\ \text{are clopen in}\ (Y,\sigma)

C_1,C_2\in\mathcal{C}

X\setminus(C_1\cup C_2)=(X\setminus C_1)\cap(X\setminus C_2)\in\tau,\ \text{so}\ C_1\cup C_2\in\mathcal{C}

\{2\},\ \{3,4\}\in\mathcal{C}

\{2\}\cup\{3,4\}=\{2,3,4\}\in\mathcal{C},\quad X\setminus\{2,3,4\}=\{1\}=\{1,3,4\}\cap\{1,2\}

\text{claim:}\ C\ \text{closed}\iff C\notin\tau

\{1,2\}\in\tau\cap\mathcal{C},\ \{3\}\notin\tau\cup\mathcal{C}\ \Rightarrow\ \text{claim false in both directions}

Substeps and lemmas

F\subseteq\tau,\ |F|=k

\textstyle\bigcup F\in\tau\ \text{for every}\ k

F\subseteq\tau,\ |F|=k

\textstyle\bigcap F\in\tau\ \text{for every}\ k\ (\bigcap\varnothing=X)

U,V\in\tau

U\cup V\in\tau,\ U\cap V\in\tau

x\in X\setminus(X\setminus C)

x\in C

\{3\}

\{3\}\notin\tau,\ X\setminus\{3\}=\{1,2,4\}\notin\tau

x\in X\setminus(A\cup B)

x\in(X\setminus A)\cap(X\setminus B)

x\in(X\setminus A)\cap(X\setminus B)

x\in X\setminus(A\cup B);\ \text{hence}\ X\setminus(A\cup B)=(X\setminus A)\cap(X\setminus B)

Reverse audits

• S11 · premise_audit · S04, S07.1, quantifier_negation — Here each direction is refuted separately by a counterexample from the computations above; a counterexample to either one direction alone already suffices to refute the universal equivalence.
• S10 · premise_audit · S09, S03, forall_elim — This is one instance of the general result and a cross-check; it cannot prove the general result in reverse.
• S09.2 · premise_audit · S09.1, set_extensionality, set_operations_def — A set equality needs both inclusions, and both are proved.
• S09.1 · premise_audit · set_operations_def, de_morgan_logic — This sub-step proves one inclusion only; the other is S09.2.
• S09 · premise_audit · L_demorgan, mko_topological_space_def, mko_open_closed_def — It needs closure under finite intersections; nothing is claimed for unions of infinitely many closed sets (in general topologies they may fail to be closed).
• S08 · premise_audit · mko_topological_space_def, mko_open_closed_def, forall_intro — Only two axioms of a topology and the definition of closed are used; it is not claimed that any other set must be clopen (for example {1} is not).
• S07.1 · premise_audit · G2, mko_open_closed_def — Both verdicts use only the definition and the complete list of τ; the complement is computed in X.
• S07 · premise_audit · S04, S05, S06, finite_counting_basics — The count uses inclusion and exclusion for the union (recomputable as N2, N3, N4). The eight listed sets appear in neither list, consistent with the count.
• S06 · premise_audit · S03, G2 — 'Not open' comes from comparing with each of the six members of τ.
• S05 · premise_audit · S03, set_operations_def — Each 'not closed' rests on the concrete check that the complement is not in τ.
• S04 · premise_audit · G2, S03, set_operations_def — The intersection needs both complete lists; the conclusion is an equality of families.
• S03 · premise_audit · S02, set_operations_def, L_double_complement — It needs the equivalence of S02 and the complete list of τ; every complement is computed inside X.
• S02.1 · inverse_equivalence · set_operations_def, set_extensionality, de_morgan_logic — Each move is a logical equivalence (De Morgan, with x∈X known), so both directions hold. If C were not a subset of X, the last move would give only x∈C∩X and the formula would fail.
• S02 · inverse_equivalence · L_double_complement, mko_open_closed_def — Both directions have reasons, using only the double complement and the definition. The complement must be relative to the same X, or the double complement formula fails.
• S01.3 · premise_audit · G2, set_operations_def — The four nontrivial members form comb(4,2)=6 pairs, all listed (N1), with none omitted.
• S01.2 · premise_audit · nat_induction, set_operations_def — It needs X∈τ as the base for the empty intersection, and closure under pairwise intersections.
• S01.1 · premise_audit · nat_induction, set_operations_def, finite_counting_basics — The induction needs every subfamily to have finitely many members, which comes from τ being finite. The base case needs ∅∈τ; the empty family may not be skipped.
• S01 · premise_audit · G2, L_finite_family, mko_topological_space_def — It needs τ finite, containing ∅ and X, and the pairwise check of S01.3. The lemma holds only for finite families; for an infinite family, closure under pairwise unions is not enough (see error control three).

Error controls and negative controls

Closed does not mean 'not open'

The misconception identified in the MKO is reading 'closed' as 'not open'. Here {1,2} is both open and closed, and {3} is neither (S11). Closedness can only be decided by the definition: is the complement in τ?

Complements must be taken in the same X

If the complement of {1,2} is wrongly taken in {1,2,3}, one gets {3}∉τ and wrongly concludes {1,2} is not closed; the correct X∖{1,2}={3,4}∈τ. If the complement of {3,4} is taken in {1,2,3,4,5}, one gets {1,2,5}, not even a subset of X, so comparing with τ is meaningless (MKO assumption a2).

Pairwise closure suffices only for finite families

S01 could check only pairwise unions because τ is finite (L_finite_family). Counterexample: the family of all finite subsets of ℕ together with ℕ itself contains ∅ and ℕ and is closed under pairwise unions and intersections, but the union of all {2k} is the set of even numbers, which is infinite and not equal to ℕ, so it is not in the family and the family is not a topology.

• {3} is not open, so {3} is closed → reject — X∖{3}={1,2,4}∉τ, so {3} is not closed; it is neither open nor closed.
• {1,2} is closed, so it is not open → reject — {1,2}∈τ, so it is both open and closed.
• In {1,2,3,4,5} the complement of {3,4} is {1,2,5}∉τ, so {3,4} is not closed → reject — The complement must be taken in X: X∖{3,4}={1,2}∈τ, so {3,4} is closed.
• The closed sets are exactly the 10 subsets not in τ → reject — There are only 6 closed sets (S03), and 4 of them are in τ; of the 10 subsets not in τ, 8 are not closed.

Result and limitations

τ is a topology on X={1,2,3,4}; the closed family is {X,{2,3,4},{3,4},{1,2},{2},∅}. The sixteen subsets split into: clopen ∅, {1,2}, {3,4}, X; open only {1}, {1,3,4}; closed only {2}, {2,3,4}; the other eight are neither open nor closed. In general, ∅ and X are clopen in every topological space, and the union of two closed sets is closed.

• The dual statement in MKO paragraph 2, that an intersection of arbitrarily many closed sets is closed, follows from De Morgan's law for arbitrary families; this guide proves only the case of the union of two closed sets and does not write sub-steps for arbitrary families.

Verification and source binding

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