irreducible closed subspace (changes) in nLab

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Definition

Given a topological space XX, a closed subspace FF of XX is irreducible if it is inhabited and not the union of two closed proper (i.e. smaller) subspaces. In other words, FF is irreducible if whenever F 1F_1 and F 2F_2 are closed subsets of XX such that

F=F 1∪F 2 F = F_1 \cup F_2

then F 1=FF_1 = F or F 2=FF_2 = F.

Proposition

Let (X,τ)(X, \tau) be a topological space, and let P∈τ⊂P(X)P \in \tau \subset P(X) be a proper open subset, so that the complement F≔X\PF \coloneqq X\backslash P is an inhabited closed subspace. Then FF is irreducible in the sense of def. 1 precisely if whenever U 1,U 2∈τU_1,U_2 \in \tau are open subsets with U 1∩U 2⊂PU_1 \cap U_2 \subset P then U 1⊂PU_1 \subset P or U 2⊂PU_2 \subset P:

X\Pirreducible⇔(∀U 1,U 2∈τ((U 1∩U 2⊂P)⇒(U 1⊂PorU 2⊂P))) X \backslash P \,\text{irreducible} \;\Leftrightarrow\; \left( \underset{U_1, U_2 \in \tau}{\forall } \left( \left( U_1 \cap U_2 \subset P \right) \;\Rightarrow\; \left(U_1 \subset P \;\text{or}\; U_2 \subset P\right) \right) \right)

Proof

Every closed subset F i⊂FF_i \subset F may be exhibited as the complement

F i=F\U i F_i = F \backslash U_i

for some open subset U i∈τU_i \in \tau. Observe that under this identification the condition that U 1∩U 2⊂PU_1 \cap U_2 \subset P is equivalent to the condition that F 1∪F 2=FF_1 \cup F_2 = F, because it is equivalent to the equation labeled (⋆)(\star) in the following sequence of equations:

F 1∪F 2 =(F\U 1)∪(F\U 2) =(X\(P∪U 1))∪(X\P∪U 2) =X\(P∪(U 1∩U 2)) =(⋆)X\P =F. \begin{aligned} F_1 \cup F_2 & = (F \backslash U_1) \cup (F \backslash U_2) \\ & = \left( X \backslash (P \cup U_1) \right) \cup \left( X \backslash P \cup U_2 \right) \\ & = X \backslash ( P \cup (U_1 \cap U_2) ) \\ & \stackrel{(\star)}{=} X \backslash P \\ & = F \end{aligned} \,.

Similarly, the condition that U i⊂PU_i \subset P is equivalent to the condition that F i=FF_i = F , because it is quivalent equivalent to the equality(⋆)(\star) in the following sequence of equalities:

F i =F\U i =X\(P∪U i) =(⋆)X\P =F. \begin{aligned} F_i &= F \backslash U_i \\ & = X \backslash ( P \cup U_i ) \\ & \stackrel{(\star)}{=} X \backslash P \\ & = F \end{aligned} \,.

Under these equivalences, the two conditions are manifestly the same.

Proposition

For (X,τ)(X,\tau) a topological space, then there is a bijection between the irreducible closed subspaces of (X,τ)(X,\tau) and the frame homomorphisms from τ X\tau_X to τ *\tau_\ast, given by

Hom Frame(τ X,τ *) ⟶≃ IrrClSub(X) ϕ ↦ X\U ∅(ϕ) \array{ Hom_{Frame}(\tau_X, \tau_\ast) &\underoverset{\simeq}{}{\longrightarrow}& IrrClSub(X) \\ \phi &\mapsto& X \backslash U_\emptyset(\phi) }

where U ∅(ϕ)U_\emptyset(\phi) is the union of all elements U∈τ xU \in \tau_x such that ϕ(U)=∅\phi(U) = \emptyset:

U ∅(ϕ)≔∪U∈τ Xϕ(U)=∅U. U_{\emptyset}(\phi) \coloneqq \underset{{U \in \tau_X} \atop {\phi(U) = \emptyset} }{\cup} U \,.

Proof

First we need to show that the function is well defined in that given a frame homomorphism ϕ:τ X→τ *\phi \colon \tau_X \to \tau_\ast then X\U ∅(ϕ)X \backslash U_\emptyset(\phi) is indeed an irreducible closed subspace.

To that end observe that:

(*)(\ast) If there are two elements U 1,U 2∈τ XU_1, U_2 \in \tau_X with U 1∩U 2⊂U ∅(ϕ)U_1 \cap U_2 \subset U_{\emptyset}(\phi) then U 1⊂U ∅(ϕ)U_1 \subset U_{\emptyset}(\phi) or U 2⊂U ∅(ϕ)U_2 \subset U_{\emptyset}(\phi).

This is because

ϕ(U 1)∩ϕ(U 2) =ϕ(U 1)∩ϕ(U 2) ⊂ϕ(U ∅(∅)) =∅, \begin{aligned} \phi(U_1 \phi(U_1) \cap U_2) \phi(U_2) & = \phi(U_1) \phi(U_1 \cap \phi(U_2) U_2) \\ & \subset \phi(U_{\emptyset}) \phi(U_{\emptyset}(\emptyset)) \\ & = \emptyset \end{aligned} \,,

(where the first equality holds because ϕ\phi preserves finite intersections, the inclusion holds because ϕ\phi respects inclusions, and the second equality holds because ϕ\phi preserves arbitrary unions). But in τ *={∅,{1}}\tau_\ast = \{\emptyset, \{1\}\} the intersection of two open subsets is empty precisely if at least one of them is empty, hence ϕ(U 1)=∅\phi(U_1) = \emptyset or ϕ(U 2)=∅\phi(U_2) = \emptyset. But this means that U 1⊂U ∅(ϕ)U_1 \subset U_{\emptyset}(\phi) or U 2⊂U ∅(ϕ)U_2 \subset U_{\emptyset}(\phi), as claimed.

Now according to prop. 1 the condition (*)(\ast) identifies the complement X\U ∅(ϕ)X \backslash U_{\emptyset}(\phi) as an irreducible closed subspace of (X,τ)(X,\tau).

Conversely, given an irreducible closed subset X\U 0X \backslash U_0, define ϕ\phi by

ϕ:U↦{∅ |ifU⊂U 0 {1} |otherwise. \phi \;\colon\; U \mapsto \left\{ \array{ \emptyset & \vert \, \text{if} \, U \subset U_0 \\ \{1\} & \vert \, \text{otherwise} } \right. \,.

This does preserve

1. arbitrary unions

   because ϕ(∪iU i)={∅0} \phi(\underset{i}{\cup} U_i) = \{0\} \emptyset precisely if ∪iU i⊂U 0\underset{i}{\cup}U_i \subset U_0 which is the case precisely if all U i⊂U 0U_i \subset U_0, which means that all ϕ(U i)=∅\phi(U_i) = \emptyset and because∪i∅=∅\underset{i}{\cup}\emptyset = \emptyset;

   while ϕ(∪iU 1)={1}\phi(\underset{i}{\cup}U_1) = \{1\} as soon as one of the U iU_i is not contained in U 0U_0, which means that one of the ϕ(U i)={1}\phi(U_i) = \{1\} which means that ∪iϕ(U i)={1}\underset{i}{\cup} \phi(U_i) = \{1\};

2. finite intersections, intersections

   because if U 1∩U 2 ∈ ⊂U 0 U_1 \cap U_2 \in \subset U_0, then by (*)(\ast) U 1∈U 0U_1 \in U_0 or U 2∈U 0U_2 \in U_0, whence ϕ(U 1)=∅\phi(U_1) = \emptyset or ϕ(U 2)=∅\phi(U_2) = \emptyset, whence with ϕ(U 1∩U 2)=∅\phi(U_1 \cap U_2) = \emptyset also ϕ(U 1)∩ϕ(U 2)=∅\phi(U_1) \cap \phi(U_2) = \emptyset;

   while if U 1∩U 2U_1 \cap U_2 is not contained in U 0U_0 then neither U 1U_1 nor U 2U_2 is contained in U 0U_0 and hence with ϕ(U 1∩U 2)={1}\phi(U_1 \cap U_2) = \{1\} also ϕ(U 1)∩ϕ(U 2)={1}∩{1}={1}\phi(U_1) \cap \phi(U_2) = \{1\} \cap \{1\} = \{1\}.

Hence this is indeed a frame homomorphism τ X→τ *\tau_X \to \tau_\ast.

Clearly Finally, it is clear that these two operations are inverse to each other.