TheoremStatement: AI-adaptedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (z-ai/glm-5.2)verified 2026-07-29 (claude-fable-5)

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Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
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For a topological space the following agree: no separation exists, the only clopen subsets are $\varnothing$ and $X$ , and every continuous map to the two-point discrete space is constant

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) and let $\mathbf{2} = \{0,1\}$ carry the discrete topology (The discrete, indiscrete, cofinite, cocountable, particular-point and Sierpinski topologies). Call a map $\chi : X \to \mathbf{2}$ constant when $\chi(x) = \chi(y)$ for all $x, y \in X$ .

1. The following four conditions are equivalent.

(a) $X$ is connected: no separation of $X$ exists (Separation of a topological space, connected and disconnected spaces, clopen sets, and connected subsets).
(b) The only clopen subsets of $X$ are $\varnothing$ and $X$ .
(c) Every continuous map $\chi : X \to \mathbf{2}$ is constant (Continuity of a map of topological spaces at a point and globally).
(d) The only subsets of $X$ with empty boundary are $\varnothing$ and $X$ (Interior, closure, boundary, exterior, derived set and isolated point in a topological space).

2. For $A \subseteq X$ with the subspace topology (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), $A$ is a connected subset of $X$ if and only if the only subsets of $A$ that are clopen in $(A, \mathcal{T}_A)$ are $\varnothing$ and $A$ , if and only if every continuous map $A \to \mathbf{2}$ is constant.

Claim 2 is claim 1 applied to the space $(A, \mathcal{T}_A)$ and is stated separately because it is the form used in every later proof on this page: a connected set is tested by showing that a continuous two-valued function on it cannot take both values.

Facts & Assumptions

Given: A topological space $(X, \mathcal{T})$ and the two-point discrete space $\mathbf{2} = \{0,1\}$ .

[A1]

A separation of $X$ is a pair $(U,V)$ of open, nonempty, disjoint sets with $U \cup V = X$ ; $X$ is connected when none exists; a subset carries the subspace topology and is connected when it is connected as a space (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).

[A2]

A set is closed exactly when its complement is open, clopen when it is both open and closed; $\varnothing$ and $X$ are clopen (Topology on a set, open and closed sets, clopen sets, the closed-set axiomatisation, and the coarser/finer comparison).

[A3]

Every subset of $\mathbf{2}$ is open, hence also closed; the subsets of $\mathbf{2}$ are $\varnothing$ , $\{0\}$ , $\{1\}$ and $\mathbf{2}$ (The discrete, indiscrete, cofinite, cocountable, particular-point and Sierpinski topologies).

[A4]

A map is continuous exactly when the preimage of every open set is open, equivalently exactly when the preimage of every closed set is closed (For a map of spaces the following agree: continuity at every point, preimages of open sets open, preimages of closed sets closed, preimages of subbasic open sets open, and $f(\overline{A}) \subseteq \overline{f(A)}$ , clauses (b) and (c), and Continuity of a map of topological spaces at a point and globally).

[A5]

$\operatorname{int}(A) \subseteq A \subseteq \overline{A}$ and $\partial A = \overline{A} \setminus \operatorname{int}(A)$ ; $A$ is open exactly when $A = \operatorname{int}(A)$ and closed exactly when $A = \overline{A}$ (Interior, closure, boundary, exterior, derived set and isolated point in a topological space).

Proof

technique · direct

1.1

If $C \subseteq X$ is clopen with $C \ne \varnothing$ and $C \ne X$ , then $C$ and $X \setminus C$ are both open by [A2], both nonempty, disjoint, and their union is $X$ ; so $(C, X \setminus C)$ is a separation of $X$ .

A1A2

1.2

If $\chi : X \to \mathbf{2}$ is continuous, then $\chi^{-1}[\{0\}]$ and $\chi^{-1}[\{1\}]$ are clopen in $X$ , since $\{0\}$ and $\{1\}$ are both open and closed in $\mathbf{2}$ by [A3] and preimages of open sets are open and of closed sets closed by [A4].

A3A4

1.3

If $(U,V)$ is a separation of $X$ , then $\chi_{U,V} : X \to \mathbf{2}$ taking the value $0$ on $U$ and $1$ on $V$ is a well-defined function, because $U$ and $V$ are disjoint and cover $X$ by [A1]; it is continuous, because the preimages of $\varnothing$ , $\{0\}$ , $\{1\}$ , $\mathbf{2}$ are $\varnothing$ , $U$ , $V$ , $X$ , all open by [A1] and [A2], so [A4] and [A3] apply; and it is not constant, because $U$ and $V$ are nonempty.

A1A2A3A4

1.4

For $A \subseteq X$ the conditions $\partial A = \varnothing$ and " $A$ is clopen" agree: by [A5] $\partial A = \varnothing$ says $\overline{A} = \operatorname{int}(A)$ , which together with $\operatorname{int}(A) \subseteq A \subseteq \overline{A}$ forces $A = \operatorname{int}(A) = \overline{A}$ , that is $A$ open and closed; conversely if $A$ is clopen then $\operatorname{int}(A) = A = \overline{A}$ and $\partial A = \varnothing$ .

2.1

(a) implies (b): if (b) fails there is a clopen $C \notin \{\varnothing, X\}$ , and step 1.1 turns it into a separation of $X$ , so (a) fails.

step 1.1

2.2

(b) implies (c): let $\chi : X \to \mathbf{2}$ be continuous; by step 1.2 the set $\chi^{-1}[\{0\}]$ is clopen, hence by (b) it is $\varnothing$ or $X$ ; in the first case $\chi$ takes only the value $1$ and in the second only the value $0$ , so $\chi$ is constant.

step 1.2

2.3

(c) implies (a): if (a) fails there is a separation $(U,V)$ of $X$ , and step 1.3 produces a continuous $\chi_{U,V} : X \to \mathbf{2}$ that is not constant, so (c) fails.

step 1.3

2.4

(b) and (d) are the same condition, by step 1.4 applied to each subset of $X$ .

step 1.4

3.1

Steps 2.1, 2.2 and 2.3 give (a) implies (b) implies (c) implies (a), so (a), (b) and (c) are equivalent, and step 2.4 adjoins (d); this is claim 1.

step 2.1step 2.2step 2.3step 2.4

4.1

Claim 2 is claim 1 applied to the topological space $(A, \mathcal{T}_A)$ , whose connectedness is by [A1] the definition of $A$ being a connected subset of $X$ .

step 3.1A1∎

Remarks

Which clause is used where. Clause (c) is the workhorse: it converts a connectedness claim into a statement about functions, so it composes with continuous maps and with unions, which is what makes the theorems after it short. Clause (b) is the one to use when a candidate clopen set is already in hand. Clause (d) is stated because a boundary computation is often the quickest route in a concrete space.
Why $\mathbf{2}$ and not an arbitrary discrete space. Any discrete space with at least two points would serve for clause (c), since a non-constant map into it composes with a retraction onto two of its points. Fixing $\mathbf{2}$ avoids having to say which two, and every use below needs no more.
The empty space satisfies all four clauses. Its only subset is $\varnothing = X$ , which is clopen and has empty boundary; the unique map $\varnothing \to \mathbf{2}$ is constant vacuously; and no separation exists, since a separation needs a nonempty piece. So the convention of Separation of a topological space, connected and disconnected spaces, clopen sets, and connected subsets is consistent with every clause here rather than being an exception to them.

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 49 results over 18 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

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

For a topological space the following agree: no separation exists, the only clopen subsets are ∅\varnothing∅ and XXX, and every continuous map to the two-point discrete space is constant