LemmaStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedjudge pass (gpt-6-sol)audited 2026-09-27

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Irreducibility via nonempty open subsets, connectedness and open subspaces

Statement

Let $X$ be a topological space (Topology on a set, open and closed sets, clopen sets, the closed-set axiomatisation, and the coarser/finer comparison), irreducibility and irreducible subsets being as in Irreducible topological spaces and irreducible subsets in the subspace topology. Then:

$X$ is irreducible if and only if $X \neq \emptyset$ and every two nonempty open subsets of $X$ have nonempty intersection;
$X$ is irreducible if and only if $X \neq \emptyset$ and every nonempty open subset of $X$ is dense in $X$ (Dense, nowhere dense and codense subsets of a topological space, and the criterion by basic open sets);
if $X$ is irreducible then $X$ is connected (Separation of a topological space, connected and disconnected spaces, clopen sets, and connected subsets);
if $X$ is irreducible and $U \subseteq X$ is a nonempty open subspace, then $U$ is irreducible, hence connected (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);
the empty space is not irreducible, and the one-point space is irreducible.

Facts & Assumptions

[F1]

$X$ is irreducible when $X \neq \emptyset$ and, whenever $X = F_{1} \cup F_{2}$ with $F_{1}, F_{2} \subseteq X$ closed, one has $X = F_{1}$ or $X = F_{2}$ (Irreducible topological spaces and irreducible subsets in the subspace topology).

[F2]

A subset $C$ of a subspace $S \subseteq X$ is closed in $S$ exactly when $C = F \cap S$ for a closed $F \subseteq X$ , and the open subsets of $S$ are the traces $U \cap S$ of the open subsets $U \subseteq X$ (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).

[F3]

A subset $A \subseteq X$ is dense in $X$ if and only if $U \cap A \neq \emptyset$ for every nonempty open $U \subseteq X$ (Dense, nowhere dense and codense subsets of a topological space, and the criterion by basic open sets).

[F4]

A separation of $X$ is a pair $(U, V)$ of open, nonempty, disjoint subsets with $U \cup V = X$ , and $X$ is connected when no separation exists (Separation of a topological space, connected and disconnected spaces, clopen sets, and connected subsets).

Proof

Given: A topological space $X$ , its closed and open subsets, and the irreducibility notion of [F1].

1.1

I prove the equivalence of clause 1. Assume first that $X$ is irreducible, and let $U_{1}, U_{2} \subseteq X$ be nonempty open subsets. If $U_{1} \cap U_{2} = \emptyset$ , then $X = (X ∖ U_{1}) \cup (X ∖ U_{2})$ is a union of two closed subsets, and neither equals $X$ because $U_{1}$ and $U_{2}$ are nonempty; this contradicts irreducibility [F1]. Hence $U_{1} \cap U_{2} \neq \emptyset$ , and $X \neq \emptyset$ is part of [F1]. Conversely, assume $X \neq \emptyset$ and that every two nonempty open subsets meet, and let $X = F_{1} \cup F_{2}$ with $F_{1}, F_{2}$ closed. If $F_{1} \neq X$ and $F_{2} \neq X$ , then $U_{i} : = X ∖ F_{i}$ are nonempty open subsets with $U_{1} \cap U_{2} = X ∖ (F_{1} \cup F_{2}) = \emptyset$ , a contradiction; hence $X = F_{1}$ or $X = F_{2}$ , and $X$ is irreducible by [F1].

1.2

The empty space is not irreducible, because irreducibility requires nonemptiness by [F1]. A one-point space $X = {*}$ is irreducible: its only subsets are $\emptyset$ and $X$ , so a union $X = F_{1} \cup F_{2}$ of closed subsets forces one of them to be $X$ , and $X \neq \emptyset$ ; this is clause 5.

2.1

By the density criterion of [F3], a subset $A \subseteq X$ is dense exactly when $U \cap A \neq \emptyset$ for every nonempty open $U \subseteq X$ . Hence, for nonempty $X$ , the assertion that every nonempty open subset is dense says precisely that for all nonempty open $A, U \subseteq X$ one has $U \cap A \neq \emptyset$ , which is the intersection condition of [step 1.1]; with the nonemptiness clause this proves clause 2.

F3step 1.1

2.2

Let $X$ be irreducible and suppose that $(U, V)$ is a separation of $X$ as in [F4]. Then $U$ and $V$ are nonempty open subsets with $U \cap V = \emptyset$ , contradicting the intersection condition of [step 1.1]. Hence no separation exists and $X$ is connected.

F4step 1.1

3.1

Let $X$ be irreducible and let $U \subseteq X$ be a nonempty open subspace. Let $W_{1}, W_{2} \subseteq U$ be nonempty open subsets of the subspace $U$ ; by [F2] there are open $V_{i} \subseteq X$ with $W_{i} = U \cap V_{i}$ , so each $W_{i}$ is open in $X$ , being the intersection of two open subsets of $X$ , and nonempty by assumption. By [step 1.1] applied in $X$ we get $W_{1} \cap W_{2} \neq \emptyset$ ; since $W_{1}, W_{2} \subseteq U$ were arbitrary nonempty open subsets of the subspace $U$ , the criterion of [step 1.1] applied in the space $U$ , which is nonempty, shows that $U$ is irreducible, and [step 2.2] applied in $U$ shows that $U$ is connected. This is clause 4.

F2step 2.2step 1.1

4.1

Clauses 1 and 2 are [step 1.1] and [step 2.1], clause 3 is [step 2.2], clause 4 is [step 3.1] and clause 5 is [step 1.2]; the proof is complete. ∎

step 2.2step 1.2step 2.1step 3.1step 1.1

Depends on

Used by

Constant sheaves on irreducible spaces are flasque and acyclic Lemma

Dependency tree · two levels

13 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

The Stacks Project, Topology (standard reference, not scraped)