LemmaStatement: AI-adaptedProof: AI-generatedSession-authored (Fable 5 assisted)precheck passjudge pass (z-ai/glm-5.2)audited 2026-07-27

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Interior commutes with finite intersections and closure with finite unions, while the two reverse combinations are inclusions only and both fail for infinite families; the space is the disjoint union of interior, boundary and exterior

Statement

Let $(X, \mathcal{T})$ be a topological space, with interior, closure, boundary and exterior as in Interior, closure, boundary, exterior, derived set and isolated point in a topological space. Then:

Monotonicity. $A \subseteq B$ implies $\operatorname{int}(A) \subseteq \operatorname{int}(B)$ and $\overline{A} \subseteq \overline{B}$ .
The two identities. For all $A, B \subseteq X$ , $\operatorname{int}(A \cap B) = \operatorname{int}(A) \cap \operatorname{int}(B), \qquad \overline{A \cup B} = \overline{A} \cup \overline{B} .$
The two reverse combinations are inclusions only. For all $A, B \subseteq X$ , $\operatorname{int}(A) \cup \operatorname{int}(B) \subseteq \operatorname{int}(A \cup B), \qquad \overline{A \cap B} \subseteq \overline{A} \cap \overline{B},$ and both inclusions are strict for $A = \{p\}$ and $B = X \setminus \{p\}$ in the cofinite topology on an infinite set $X$ with $p \in X$ (The discrete, indiscrete, cofinite, cocountable, particular-point and Sierpinski topologies).
Both identities of claim 2 fail for infinite families. In the same space, with $I := X \setminus \{p\}$ , $\bigcup_{x \in I} \overline{\{x\}} = I \subsetneq X = \overline{\bigcup_{x \in I} \{x\}}, \qquad \operatorname{int}\Big(\bigcap_{x \in I} (X \setminus \{x\})\Big) = \varnothing \subsetneq \{p\} = \bigcap_{x \in I} \operatorname{int}(X \setminus \{x\}) .$
Trichotomy of position. For every $A \subseteq X$ the three sets $\operatorname{int}(A)$ , $\partial A$ and $\operatorname{ext}(A)$ are pairwise disjoint and their union is $X$ .

Facts & Assumptions

Given: A topological space $(X,\mathcal{T})$ and subsets $A, B \subseteq X$ ; and, for claims 3 and 4, an infinite set $X$ carrying the cofinite topology, a point $p \in X$ and the index set $I := X \setminus \{p\}$ .

[A1]

$\operatorname{int}(A)$ is the largest open subset of $A$ and $\overline{A}$ is the smallest closed superset of $A$ ; $\operatorname{int}(A) \subseteq A \subseteq \overline{A}$ ; $\operatorname{ext}(A) = X \setminus \overline{A}$ and $\partial A = \overline{A} \setminus \operatorname{int}(A)$ (Interior, closure, boundary, exterior, derived set and isolated point in a topological space).

[L1]

A topology is closed under binary intersections (T3) and its closed sets under binary unions (C3) (Topology on a set, open and closed sets, clopen sets, the closed-set axiomatisation, and the coarser/finer comparison).

[L2]

In the cofinite topology on $X$ the open sets are $\varnothing$ together with the sets of finite complement, and the closed sets are $X$ together with the finite subsets of $X$ (The discrete, indiscrete, cofinite, cocountable, particular-point and Sierpinski topologies).

[L3]

A subset of a finite set is finite, and a union of two finite sets is finite (The discrete, indiscrete, cofinite, cocountable, particular-point and Sierpinski topologies); "infinite" means "not finite" (Finite, countably infinite, countable, uncountable).

Proof

technique · direct

1.1

Claim 1: if $A \subseteq B$ then every open $U \subseteq A$ satisfies $U \subseteq B$ , so $\operatorname{int}(A) \subseteq \operatorname{int}(B)$ ; and $\overline{B}$ is a closed set containing $B \supseteq A$ , so $\overline{A} \subseteq \overline{B}$ .

1.2

$\operatorname{int}(A) \cap \operatorname{int}(B)$ is open by (T3) and is contained in $A \cap B$ , so it is contained in $\operatorname{int}(A \cap B)$ ; and $\overline{A} \cup \overline{B}$ is closed by (C3) and contains $A \cup B$ , so it contains $\overline{A \cup B}$ .

A1L1

1.3

Since $X$ is infinite, $X \setminus \{p\}$ is infinite: were it finite, $X = \{p\} \cup (X \setminus \{p\})$ would be a union of two finite sets and hence finite. In particular $I \ne \varnothing$ .

givenL3

1.4

Claim 5: $\operatorname{int}(A) \subseteq \overline{A}$ by [A1], so the three sets $\operatorname{int}(A)$ , $\partial A = \overline{A} \setminus \operatorname{int}(A)$ and $\operatorname{ext}(A) = X \setminus \overline{A}$ are pairwise disjoint, the first two inside $\overline{A}$ and the third outside it; and their union is $\operatorname{int}(A) \cup (\overline{A} \setminus \operatorname{int}(A)) \cup (X \setminus \overline{A}) = \overline{A} \cup (X \setminus \overline{A}) = X$ .

2.1

Claim 1 applied to $A \cap B \subseteq A$ and $A \cap B \subseteq B$ gives $\operatorname{int}(A \cap B) \subseteq \operatorname{int}(A) \cap \operatorname{int}(B)$ and $\overline{A \cap B} \subseteq \overline{A} \cap \overline{B}$ ; applied to $A \subseteq A \cup B$ and $B \subseteq A \cup B$ it gives $\operatorname{int}(A) \cup \operatorname{int}(B) \subseteq \operatorname{int}(A \cup B)$ and $\overline{A} \cup \overline{B} \subseteq \overline{A \cup B}$ .

step 1.1

2.2

In the cofinite topology on the infinite $X$ : the set $\{p\}$ is closed, being finite, so $\overline{\{p\}} = \{p\}$ ; and $\{p\}$ is not open, since $X \setminus \{p\}$ is infinite by step 1.3, so the only open subset of $\{p\}$ is $\varnothing$ and $\operatorname{int}(\{p\}) = \varnothing$ .

step 1.3A1L2

2.3

In the same space $B := X \setminus \{p\}$ is open, its complement $\{p\}$ being finite, so $\operatorname{int}(B) = B$ ; and $B$ is infinite by step 1.3, so the only closed set containing $B$ is $X$ and $\overline{B} = X$ .

step 1.3A1L2

2.4

For $x \in I$ the singleton $\{x\}$ is finite, hence closed, so $\overline{\{x\}} = \{x\}$ and $\bigcup_{x \in I} \overline{\{x\}} = I$ ; meanwhile $\bigcup_{x \in I} \{x\} = I$ is infinite by step 1.3, so its closure is $X$ , and $p \in X \setminus I$ makes the inclusion strict.

step 1.3A1L2

3.1

Combining step 1.2 with step 2.1 proves claim 2, and the two inclusions of claim 3 are among those obtained in step 2.1.

step 1.2step 2.1

3.2

With $A := \{p\}$ and $B := X \setminus \{p\}$ : $\operatorname{int}(A) \cup \operatorname{int}(B) = \varnothing \cup B = X \setminus \{p\}$ , while $A \cup B = X$ is open and so $\operatorname{int}(A \cup B) = X$ ; the inclusion is therefore strict.

step 2.2step 2.3A1

3.3

With the same $A$ and $B$ : $A \cap B = \varnothing$ , which is closed, so $\overline{A \cap B} = \varnothing$ , while $\overline{A} \cap \overline{B} = \{p\} \cap X = \{p\}$ ; the inclusion is therefore strict, and claim 3 is proved.

step 2.2step 2.3A1L2

3.4

For $x \in I$ the set $X \setminus \{x\}$ is open, so $\operatorname{int}(X \setminus \{x\}) = X \setminus \{x\}$ and $\bigcap_{x \in I} \operatorname{int}(X \setminus \{x\}) = X \setminus I = \{p\}$ ; meanwhile $\bigcap_{x \in I} (X \setminus \{x\}) = \{p\}$ has empty interior by step 2.2, so the inclusion is strict and claim 4 is proved.

step 1.3step 2.2A1L2

4.1

Claims 1, 2, 3, 4 and 5 are established by step 1.1, step 3.1, steps 3.2 and 3.3, steps 2.4 and 3.4, and step 1.4 respectively.

step 1.1step 3.1step 3.2step 3.3step 2.4step 3.4step 1.4∎

Remarks

Why the witnesses are all in one space. The cofinite topology on an infinite set makes every finite set closed and every infinite set dense, so it separates the four combinations of $\operatorname{int}$ , $\overline{\ \cdot\ }$ , $\cup$ and $\cap$ with a single pair of sets and a single index set, the two families of claim 4 being the singletons and their complements. The same four failures occur in $\mathbb{R}$ with its usual topology, and the sharpest form of the first is on the companion page: the interiors of $\mathbb{Q}$ and of its complement are both empty while the interior of their union is everything (In $\mathbb{R}$ the interiors of $\mathbb{Q}$ and of its complement are both empty while the interior of their union is everything ↗).
Claim 2 does extend to any finite number of sets, by iterating it, but not to a family indexed by a set that merely happens to be finite in some other sense: the induction is on the number of sets and claim 4 shows where it stops.
The four inclusions of claims 2 and 3 are the only ones that hold in general. For an arbitrary family the surviving statements are $\bigcup_i \operatorname{int}(A_i) \subseteq \operatorname{int}(\bigcup_i A_i)$ , $\operatorname{int}(\bigcap_i A_i) \subseteq \bigcap_i \operatorname{int}(A_i)$ , $\bigcup_i \overline{A_i} \subseteq \overline{\bigcup_i A_i}$ and $\overline{\bigcap_i A_i} \subseteq \bigcap_i \overline{A_i}$ , each by monotonicity alone, and claim 4 shows that two of the four are already strict: the third for the family of singletons and the second for the family of their complements.

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 37 results over 17 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources

Interior (topology) (Wikipedia) (standard reference, not scraped)
Closure (topology) (Wikipedia) (standard reference, not scraped)
J. Munkres, Topology, 2nd ed., §17 (standard reference, not scraped)