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Definition

Definition

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

$F = F_1 \cup F_2$

then $F_1 = F$ or $F_2 = F$.

Equivalently this may be expressed in terms of open subsets:

Proposition

Let $(X, \tau)$ be a topological space, and let $P \in \tau \subset P(X)$ be a proper open subset, so that the complement $F \coloneqq X\backslash P$ is an inhabited closed subspace. Then $F$ is irreducible in the sense of def. 1 precisely if whenever $U_1,U_2 \in \tau$ are open subsets with $U_1 \cap U_2 \subset P$ then $U_1 \subset P$ or $U_2 \subset P$:

$X \backslash P \,\text{irreducible} \;\Leftrightarrow\; \left( \underset{U_1, U_2 \in \tau}{\forall } \left( \left( U_1 \cap U_2 \subset P \right) \;\Rightarrow\; \left(U_1 \subset P \;\text{or}\; U_2 \subset P\right) \right) \right)$
Proof

Every closed subset $F_i \subset F$ may be exhibited as the complement

$F_i = F \backslash U_i$

for some open subset $U_i \in \tau$. Observe that under this identification the condition that $U_1 \cap U_2 \subset P$ is equivalent to the condition that $F_1 \cup F_2 = F$, because it is equivalent to the equation labeled $(\star)$ in the following sequence of equations:

\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 $U_i \subset P$ is equivalent to the condition that $F_i = F$, because it is equivalent to the equality $(\star)$ in the following sequence of equalities:

\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.

Yet another equivalent characterization is in terms of frame homomorphisms:

In the following we write

$\ast \coloneqq (\{1\}, \tau_\ast = \left\{ \emptyset, \{1\}\right\})$

for the point, regarded, uniquely, as a topological space, the point space.

Proposition

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

$\array{ Hom_{Frame}(\tau_X, \tau_\ast) &\underoverset{\simeq}{}{\longrightarrow}& IrrClSub(X) \\ \phi &\mapsto& X \backslash U_\emptyset(\phi) }$

where $U_\emptyset(\phi)$ is the union of all elements $U \in \tau_x$ such that $\phi(U) = \emptyset$:

$U_{\emptyset}(\phi) \coloneqq \underset{{U \in \tau_X} \atop {\phi(U) = \emptyset} }{\cup} U \,.$

Proof

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

To that end observe that:

$(\ast)$ If there are two elements $U_1, U_2 \in \tau_X$ with $U_1 \cap U_2 \subset U_{\emptyset}(\phi)$ then $U_1 \subset U_{\emptyset}(\phi)$ or $U_2 \subset U_{\emptyset}(\phi)$.

This is because

\begin{aligned} \phi(U_1) \cap \phi(U_2) & = \phi(U_1 \cap U_2) \\ & \subset \phi(U_{\emptyset}(\emptyset)) \\ & = \emptyset \end{aligned} \,,

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

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

Conversely, given an irreducible closed subset $X \backslash U_0$, define $\phi$ by

$\phi \;\colon\; U \mapsto \left\{ \array{ \emptyset & \vert \, \text{if} \, U \subset U_0 \\ \{1\} & \vert \, \text{otherwise} } \right. \,.$

This does preserve

1. arbitrary unions

because $\phi(\underset{i}{\cup} U_i) = \emptyset$ precisely if $\underset{i}{\cup}U_i \subset U_0$ which is the case precisely if all $U_i \subset U_0$, which means that all $\phi(U_i) = \emptyset$ and because $\underset{i}{\cup}\emptyset = \emptyset$;

while $\phi(\underset{i}{\cup}U_1) = \{1\}$ as soon as one of the $U_i$ is not contained in $U_0$, which means that one of the $\phi(U_i) = \{1\}$ which means that $\underset{i}{\cup} \phi(U_i) = \{1\}$;

2. finite intersections

because if $U_1 \cap U_2 \subset U_0$, then by $(\ast)$ $U_1 \in U_0$ or $U_2 \in U_0$, whence $\phi(U_1) = \emptyset$ or $\phi(U_2) = \emptyset$, whence with $\phi(U_1 \cap U_2) = \emptyset$ also $\phi(U_1) \cap \phi(U_2) = \emptyset$;

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

Hence this is indeed a frame homomorphism $\tau_X \to \tau_\ast$.

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

Remark

Note that frame homomorphisms from $\tau_X$ to $\tau_\ast$ are the same thing as continuous maps from the locale $\mathcal{O}_\ast$ to the locale $\mathcal{O}_X$ (that is, in the other direction, regarding the frames of opens as instead being locales). Thus, the irreducible closed subspaces correspond to what should be the points of the space, if we regard it as much as possible as a locale, that is the points of the locale $\mathcal{O}_X$. At a more elementary level, these are also the same thing as the completely prime filters in the frame $\tau_X$.

Properties

Note that the closure of (the singleton set on) any point/element of $X$ is an irreducible closed subspace. $X$ is sober if and only if every irreducible closed subspace is the closure of a unique point of $X$. In general, the irreducible closed subspaces of $X$ correspond to the points of the topological locale $\Omega(X)$, which are (by definition) the completely prime filters on the frame of open subspaces of $X$. Specifically, given an irreducibly closed subspace, the filter of open subspaces that contain it is completely prime; conversely, given a completely prime filter of open subspaces, the closure of its intersection is irreducible.

The theory of irreducible closed subspaces is not useful in constructive mathematics; instead, one must use the completely prime filters directly. While one might hope that the irreducibly open subsets (that is those that satisfy the conditions of Proposition 1) might be more tractible constructively, they are in fact no better. (We should probably put in the classical proof and see where it goes wrong.)

References

Last revised on April 24, 2017 at 15:37:23. See the history of this page for a list of all contributions to it.