False statementConstruction: AI-adaptedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck passverified 2026-08-04 (gpt-5.6-sol-codex-subscription)

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FALSE: the intersection of two connected subspaces is connected

Statement

False claim: if $A$ and $B$ are connected subsets of a topological space $X$ (Separation of a topological space, connected and disconnected spaces, clopen sets, and connected subsets, 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) then $A \cap B$ is connected.

The corresponding statement for unions is true under a meeting hypothesis (A union of connected subspaces with a point in common is connected, and so is a union of a family in which every member meets a fixed connected member); there is no such repair for intersections, and the witness below has $A \cap B \neq \varnothing$ , so nonemptiness is not what is missing.

Witness. In $X = \mathbb{R}^2$ with the product topology (For $n \ge 1$ the product topology on $n$ copies of the usual topology of $\mathbb{R}$ is the metric topology of $d_\infty$ on $\mathbb{R}^n$ , and hence also of $d_1$ and $d_2$ , so $\mathbb{R}^n$ as a product and $\mathbb{R}^n$ as a metric space are one space, The product set $\prod_{i \in I} X_i$ of functions choosing a point in each factor, the projections, the box topology, and the product topology as the initial topology of the projections; the empty product is a one-point space) put

$S_0 := \{0\} \times [0,1], \qquad S_1 := \{1\} \times [0,1], \qquad T_0 := [0,1] \times \{0\}, \qquad T_1 := [0,1] \times \{1\},$

and let $A := T_0 \cup S_0 \cup S_1$ and $B := T_1 \cup S_0 \cup S_1$ : the three sides of the unit square other than the top, and the three other than the bottom. Both are connected, and $A \cap B = S_0 \cup S_1$ is disconnected, being two disjoint closed segments.

Facts & Assumptions

Given: $\mathbb{R}^2$ with the product topology and the sets $S_0, S_1, T_0, T_1, A, B$ above; subsets carry the subspace topology.

[A1]

$[0,1]$ is order-convex, hence a connected subset of $\mathbb{R}$ ; a subset of $\mathbb{R}$ is connected exactly when it is order-convex (The connected subspaces of $\mathbb{R}$ with its usual topology are exactly the order-convex subsets, the published characterisation transported by the identification of the two descriptions of "open in $\mathbb{R}$ ", Intervals of $\mathbb{R}$ : the nine order-convex forms, nondegeneracy, and length).

[A2]

A map into $\mathbb{R}^2$ is continuous exactly when both components are, and a constant map and the identity are continuous; a continuous image of a connected space is a connected subset of the target (A map into a product is continuous iff each of its components is; the projections are continuous and open; and each projection is surjective when every factor is nonempty, which for an infinite index set uses the Axiom of Choice, A continuous image of a connected space is connected, and connectedness is a topological property, Continuity may be checked on any open cover, and on any finite closed cover; composites of continuous maps are continuous, Metrizable space: a topological space whose topology is induced by some metric; metrizability is topological, the metric is not, $\mathbb{R}^n$ as the set of functions $n \to \mathbb{R}$ , and $d_1$ , $d_2$ , $d_\infty$ are metrics on it, The metric topology: a set is open when every one of its points has a ball around it inside the set; closed means open complement).

[A3]

A union of connected subsets each meeting a fixed connected subset, together with that subset, is connected (A union of connected subspaces with a point in common is connected, and so is a union of a family in which every member meets a fixed connected member, claim 2).

[A4]

A subset $E$ of a space is disconnected exactly when $E = E_1 \cup E_2$ with $E_1, E_2$ nonempty and separated, that is neither meeting the other's closure (A subspace $A \subseteq X$ is disconnected exactly when $A = A_1 \cup A_2$ with $A_1, A_2$ nonempty and separated in $X$ , which is the criterion this library already uses on the real line, Interior, closure, boundary, exterior, derived set and isolated point in a topological space).

[A5]

A product of closed sets is closed in $\mathbb{R}^2$ , and $\{c\}$ and $[0,1]$ are closed in $\mathbb{R}$ ; a closed set equals its own closure (Interior, closure, boundary, exterior, derived set and isolated point in a topological space, The product set $\prod_{i \in I} X_i$ of functions choosing a point in each factor, the projections, the box topology, and the product topology as the initial topology of the projections; the empty product is a one-point space, Intervals of $\mathbb{R}$ : the nine order-convex forms, nondegeneracy, and length).

Refutation

technique · contradiction

1.1

Suppose, for contradiction, that the claim holds: the intersection of two connected subsets is connected.

assume-contra

1.2

Each of $S_0, S_1, T_0, T_1$ is a connected subset of $\mathbb{R}^2$ : for instance $S_0$ is the image of $[0,1]$ under $t \mapsto (0,t)$ , whose components are a constant map and the identity, hence continuous by [A2], and $[0,1]$ is connected by [A1]; the other three are the images of $t \mapsto (1,t)$ , $t \mapsto (t,0)$ and $t \mapsto (t,1)$ .

A1A2

1.3

Each of $S_0, S_1, T_0, T_1$ is closed in $\mathbb{R}^2$ by [A5], being a product of two closed subsets of $\mathbb{R}$ , so each equals its own closure.

1.4

$A \cap B = S_0 \cup S_1$ : each of $S_0, S_1$ lies in both $A$ and $B$ ; and a point of $T_0$ not in $S_0 \cup S_1$ has second coordinate $0$ and first coordinate strictly between $0$ and $1$ , so it lies in neither $T_1$ nor $S_0$ nor $S_1$ , hence not in $B$ ; symmetrically for $T_1$ .

given

2.1

$A = T_0 \cup S_0 \cup S_1$ is connected: $T_0$ is connected by step 1.2, and $S_0$ and $S_1$ are connected and meet $T_0$ , in $(0,0)$ and $(1,0)$ respectively; so [A3] applies with $T_0$ as the fixed connected set. Symmetrically $B = T_1 \cup S_0 \cup S_1$ is connected, $S_0$ and $S_1$ meeting $T_1$ in $(0,1)$ and $(1,1)$ .

step 1.2A3

2.2

$S_0$ and $S_1$ are nonempty, disjoint, and separated in $\mathbb{R}^2$ : by step 1.3 each is its own closure, and $S_0 \cap S_1 = \varnothing$ because a common point would have first coordinate both $0$ and $1$ . So $A \cap B$ is disconnected by [A4] and step 1.4.

step 1.3step 1.4A4

3.1

By step 2.1 both $A$ and $B$ are connected, so the supposed claim of step 1.1 makes $A \cap B$ connected, contradicting step 2.2. The claim is therefore false.

step 1.1step 2.1step 2.2discharge-contradiction∎

Remarks

Nonemptiness is not the missing hypothesis. In the witness $A \cap B = S_0 \cup S_1$ is nonempty, and it is even a union of two connected sets — they simply do not meet. Nor does convexity of the pieces help: each of $A$ and $B$ is a union of three straight segments.
Why unions behave and intersections do not. [A3] works because a point common to two connected sets welds them: a continuous two-valued function must agree on both. An intersection has no such welding point available, and indeed the intersection of two connected sets can be split as badly as one likes; taking longer chains of segments makes $A \cap B$ a union of any number of disjoint segments while keeping $A$ and $B$ connected.
Both witnesses are as simple as the plane allows. $A$ and $B$ are the boundary of the unit square with one side removed, in the two ways of doing so that leave the two vertical sides. Each is path-connected, hence connected by Every path-connected space is connected, and every path component lies inside a component and Paths, path-connected spaces and path components, so the failure has nothing to do with the pathologies of the zigzag curve elsewhere on this page.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · next 3 levels

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Sources

Connected space (Wikipedia) (standard reference, not scraped)