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

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

For $A \subseteq S \subseteq X$ the closure of $A$ in $S$ is $\overline{A}^{X} \cap S$ , while the interior only contains $\operatorname{int}^{X}(A) \cap S$ , with equality when $S$ is open; and a dense subset of $X$ traces to a dense subset of every open $S$

Statement

Let $(X, \mathcal{T})$ be a topological space (Topology on a set, open and closed sets, clopen sets, the closed-set axiomatisation, and the coarser/finer comparison), let $S \subseteq X$ carry the subspace topology $\mathcal{T}_S$ (Subspace topology: the traces of the open sets, its closed sets and its bases, the continuity of the inclusion, and the characteristic property of a map into a subspace) and let $A \subseteq S$ . Write $\overline{A}$ and $\operatorname{int}(A)$ for the closure and the interior of $A$ in $X$ , and $\operatorname{cl}_S(A)$ and $\operatorname{int}_S(A)$ for those taken in the space $(S, \mathcal{T}_S)$ (Interior, closure, boundary, exterior, derived set and isolated point in a topological space). Then:

Closure traces exactly. $\operatorname{cl}_S(A) \;=\; \overline{A} \cap S .$
Interior traces only one way. $\operatorname{int}(A) \subseteq S$ , so $\operatorname{int}(A) \cap S = \operatorname{int}(A)$ , and $\operatorname{int}(A) \;\subseteq\; \operatorname{int}_S(A) ,$ an inclusion that may be strict.
Equality for an open subspace. If $S \in \mathcal{T}$ then $\operatorname{int}_S(A) = \operatorname{int}(A)$ .
Density traces to open subspaces only. If $D \subseteq X$ is dense in $X$ (Dense, nowhere dense and codense subsets of a topological space, and the criterion by basic open sets) and $S \in \mathcal{T}$ , then $D \cap S$ is dense in $(S, \mathcal{T}_S)$ . Without the hypothesis $S \in \mathcal{T}$ this fails.

Both failures are witnessed inside the proof, in Sierpinski space (The discrete, indiscrete, cofinite, cocountable, particular-point and Sierpinski topologies): the unqualified forms of claims 2 and 3 and of claim 4 are false, and the counterexamples are two lines each rather than deferred.

Facts & Assumptions

Given: A topological space $(X,\mathcal{T})$ , a subset $S \subseteq X$ with its subspace topology $\mathcal{T}_S = \{\, U \cap S : U \in \mathcal{T} \,\}$ , and a subset $A \subseteq S$ . Also Sierpinski space $E = \{a,b\}$ with $a \ne b$ and $\mathcal{T}_E = \{\varnothing, \{b\}, E\}$ .

[A1]

$\mathcal{T}_S$ is a topology on $S$ , and $C \subseteq S$ is closed in $S$ if and only if $C = F \cap S$ for some closed $F \subseteq X$ (Subspace topology: the traces of the open sets, its closed sets and its bases, the continuity of the inclusion, and the characteristic property of a map into a subspace).

[A2]

If $S \in \mathcal{T}$ then a subset of $S$ is open in $S$ if and only if it is open in $X$ (Subspace topology: the traces of the open sets, its closed sets and its bases, the continuity of the inclusion, and the characteristic property of a map into a subspace).

[L1]

$\operatorname{int}(A)$ is the largest open subset of $A$ and $\overline{A}$ is the smallest closed superset of $A$ ; both are taken in whichever space is named (Interior, closure, boundary, exterior, derived set and isolated point in a topological space).

[L2]

$D$ is dense in a space exactly when $D$ meets every nonempty open subset of that space (Dense, nowhere dense and codense subsets of a topological space, and the criterion by basic open sets).

[L3]

In Sierpinski space $E$ the open sets are $\varnothing$ , $\{b\}$ and $E$ , so the closed sets are $E$ , $\{a\}$ and $\varnothing$ (The discrete, indiscrete, cofinite, cocountable, particular-point and Sierpinski topologies, Topology on a set, open and closed sets, clopen sets, the closed-set axiomatisation, and the coarser/finer comparison).

Proof

technique · direct

1.1

$\overline{A} \cap S$ is closed in $S$ by [A1], since $\overline{A}$ is closed in $X$ , and it contains $A$ , since $A \subseteq \overline{A}$ and $A \subseteq S$ .

A1L1

1.2

$\operatorname{cl}_S(A) = F \cap S$ for some closed $F \subseteq X$ , by [A1] applied to the set $\operatorname{cl}_S(A)$ , which is closed in $S$ ; and $A \subseteq \operatorname{cl}_S(A) = F \cap S \subseteq F$ .

A1L1

1.3

$\operatorname{int}(A)$ is open in $X$ and satisfies $\operatorname{int}(A) \subseteq A \subseteq S$ , so $\operatorname{int}(A) = \operatorname{int}(A) \cap S$ is a trace of an open set of $X$ and hence lies in $\mathcal{T}_S$ .

givenL1

1.4

In $E$ , put $S_0 := \{a\}$ , $A_0 := \{a\}$ and $D_0 := \{b\}$ . Then $\operatorname{int}_{S_0}(A_0) = S_0 = \{a\}$ , since $S_0$ is open in $S_0$ and $S_0 \subseteq A_0$ ; and the interior of $A_0$ in $E$ is $\varnothing$ , since by [L3] the only open subset of $\{a\}$ in $E$ is $\varnothing$ . So the inclusion of claim 2 is strict for this pair.

L1L3

1.5

Assume $S \in \mathcal{T}$ . Then $\operatorname{int}_S(A)$ , being open in $S$ , is open in $X$ by [A2], and it is contained in $A$ ; so $\operatorname{int}_S(A) \subseteq \operatorname{int}(A)$ by [L1].

A2L1

1.6

Assume $S \in \mathcal{T}$ and that $D$ is dense in $X$ , and let $W$ be a nonempty open subset of $S$ . By [A2] the set $W$ is open in $X$ , so $W \cap D \ne \varnothing$ by [L2]; and $W \subseteq S$ gives $W \cap D = W \cap (D \cap S)$ .

A2L2

2.1

In $E$ with the sets of step 1.4: the closure of $D_0$ in $E$ is $E$ , since by [L3] the only closed superset of $\{b\}$ is $E$ , so $D_0$ is dense in $E$ ; and $D_0 \cap S_0 = \varnothing$ , which is not dense in the nonempty space $S_0$ , because $S_0$ is a nonempty open subset of $S_0$ that $\varnothing$ does not meet.

L1L2L3

2.2

$\operatorname{cl}_S(A) \subseteq \overline{A} \cap S$ : by step 1.1 the set $\overline{A} \cap S$ is a closed subset of $S$ containing $A$ , and $\operatorname{cl}_S(A)$ is the smallest such.

step 1.1L1

2.3

$\overline{A} \cap S \subseteq \operatorname{cl}_S(A)$ : with $F$ as in step 1.2 one has $A \subseteq F$ with $F$ closed in $X$ , so $\overline{A} \subseteq F$ by [L1], whence $\overline{A} \cap S \subseteq F \cap S = \operatorname{cl}_S(A)$ .

step 1.2L1

2.4

$\operatorname{int}(A) \subseteq \operatorname{int}_S(A)$ : by step 1.3 the set $\operatorname{int}(A)$ is open in $S$ and contained in $A$ , and $\operatorname{int}_S(A)$ is the largest such.

step 1.3L1

3.1

Steps 2.2 and 2.3 give $\operatorname{cl}_S(A) = \overline{A} \cap S$ , which is claim 1.

step 2.2step 2.3

3.2

Step 1.3 gives $\operatorname{int}(A) \cap S = \operatorname{int}(A)$ , step 2.4 gives the inclusion, and step 1.4 exhibits a case where the inclusion is strict; this is claim 2.

step 1.3step 2.4step 1.4

3.3

Steps 2.4 and 1.5 give $\operatorname{int}_S(A) = \operatorname{int}(A)$ when $S \in \mathcal{T}$ , which is claim 3.

step 2.4step 1.5

4.1

By step 1.6 the set $D \cap S$ meets every nonempty open subset of $S$ , hence is dense in $(S,\mathcal{T}_S)$ by [L2]; and step 2.1 shows that the conclusion fails for a subspace that is not open. This is claim 4, and with steps 3.1, 3.2 and 3.3 all four claims are proved.

step 1.6step 2.1step 3.1step 3.2step 3.3L2∎

Remarks

The same two failures occur in $\mathbb{R}$ , and there they are the familiar ones. With the usual topology, $S = [0,1]$ and $A = [0,1]$ give $\operatorname{int}_S(A) = [0,1]$ while $\operatorname{int}(A) = (0,1)$ ; and $\mathbb{Q}$ is dense in $\mathbb{R}$ while its trace on the subspace of irrationals is empty, so a dense set need not trace to a dense set of a subspace that is not open. Sierpinski space is used in the proof only because it needs no real-number machinery.
Why closure behaves better than interior. Claim 1 holds for every $S$ , with no hypothesis, because the closed sets of a subspace are exactly the traces of the closed sets and tracing preserves the "smallest superset" that defines a closure. The interior is a largest subset, and tracing does not preserve that: a set can be open in $S$ without being the trace of any open set of $X$ that is contained in $A$ , which is exactly what step 1.4 exhibits.
Claim 4 is what makes "has a countable dense subset" behave the way it does. The property passes to open subspaces by claim 4, and it does not pass to arbitrary subspaces; the witness for the failure is worked on the companion page, where an uncountable discrete subspace is exhibited inside a space with a countable dense subset.

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 43 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

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