# nLab closed subspace

This entry is about closed subsets of a topological space. For other notions of “closed space” see for instance closed manifold.

# Contents

## Idea

A subset $C$ of a topological space (or more generally a convergence space) $X$ is closed if its complement is an open subset, or equivalently if it contains all its limit points. When equipped with the subspace topology, we may call $C$ (or its inclusion $C \hookrightarrow X$) a closed subspace. More abstractly, a subspace $A$ of a space $X$ is closed if the inclusion map $A \hookrightarrow X$ is a closed map.

The collection of closed subsets of a space $X$ is closed under arbitrary intersections. If $A \subseteq X$, then the intersection of all closed subsets containing $A$ is the smallest closed subset that contains $A$, called the topological closure of $A$, and variously denoted $Cl(A)$, $Cl_X(A)$, $\bar{A}$, $\overline{A}$, etc. It follows that $A \subseteq B$ implies $Cl(A) \subseteq Cl(B)$ and $Cl(Cl(A)) = Cl(A)$, so that $A \mapsto Cl(A)$ forms a Moore closure operator on the power set $P(X)$.

Since closed subsets are closed with respect to finite unions, we have $Cl(A \cup B) = Cl(A) \cup Cl(B)$.

A topological closure operator is a Moore closure operator $Cl: P(X) \to P(X)$ that preserves finite unions ($Cl(0) = 0$ and $Cl(A \cup B) = Cl(A) \cup Cl(B)$). It is easy to see that all such closure operators come from a topology whose closed sets are the fixed points of $Cl$.

(There is a lot more to say, about convergence spaces, smooth spaces, schemes, etc.)

## Definition

###### Definition

(closed subsets)

Let $(X,\tau)$ be a topological space.

1. A subset $S \subset X$ is called a closed subset if its complement $X \setminus S$ is an open subset:

$\left( S \subset X\,\, \text{is closed} \right) \phantom{AA} \Leftrightarrow \phantom{AA} \left( X\setminus S \, \subset X \,\, \text{is open} \right) \,.$

graphics grabbed from Vickers 89

2. If a singleton subset $\{x\} \subset X$ is closed, one says that $x$ is a closed point of $X$.

3. Given any subset $S \subset X$, then its topological closure $Cl(S)$ is the smallest closed subset containing $S$:

$Cl(S) \;\coloneqq\; \underset{ {C \subset X\, \text{closed} } \atop {S \subset C } }{\cap} \left( C \right) \,.$
4. A subset $S \subset X$ such that $Cl(S) = X$ is called a dense subset of $(X,\tau)$.

## Properties

### Basic properties

###### Lemma

(alternative characterization of topological cosures)

Let $(X,\tau)$ be a topological space and let $S \subset X$ be a subset of its underlying set. Then a point $x \in X$ is contained in the topological closure $Cl(S)$ (def. ) precisely if every open neighbourhood $U_x \subset X$ of $x$ intersects $S$:

$\left( x \in Cl(S) \right) \phantom{AA} \Leftrightarrow \phantom{AA} \not\left( \underset{ {U \subset X \setminus S} \atop { U \subset X \, \text{open} } }{\exists} \left( x \in U \right) \right) \,.$
###### Proof

Due to de Morgan duality we may rephrase the definition of the topological closure as follows:

\begin{aligned} Cl(S) & \coloneqq \underset{ {S \subset C } \atop { C \subset X\,\text{closed} } }{\cap} \left(C \right) \\ & = \underset{ { U \subset X \setminus S } \atop {U \subset X\, \text{open}} }{\cap} \left( X \setminus U \right) \\ & = X \setminus \left( \underset{ {U \subset X \setminus S} \atop { U \subset X\, \text{open} }}{\cup} U \right) \end{aligned} \,.
###### Proposition

(closure of a finite union is the union of the closures)

For $I$ a finite set and $\{U_i \subset X\}_{i \in I}$ is a finite set of subsets of a topological space, then

$Cl(\underset{i \in I}{\cup}U_i) = \underset{i \in I}{\cup} Cl(U_i) \,.$
###### Proof

By lemma we use that a point is in the closure of a set precisely if every open neighbourhood of the point intersects the set.

Hence in one direction

$\underset{i \in I}{\cup} Cl(U_i) \subset Cl(\underset{i \in I}{\cup}U_i)$

because if every neighbourhood of a point intersects all the $U_i$, then every neighbourhood intersects their union.

The other direction

$Cl(\underset{i \in I}{\cup}U_i) \subset \underset{i \in I}{\cup} Cl(U_i)$

is equivalent by de Morgan duality to

$X \setminus \underset{i \in I}{\cup} Cl(U_i) \subset X \setminus Cl(\underset{i \in I}{\cup}U_i)$

On left now we have the point for which there exists for each $i \in I$ a neighbourhood $U_{x,i}$ which does not intersect $U_i$. Since $I$ is finite, the intersection $\underset{i \in I}{\cap} U_{x,i}$ is still an open neighbourhood of $x$, and such that it intersects none of the $U_i$, hence such that it does not intersect their union. This implis that the given point is contained in the set on the right.

### In metric spaces

###### Proposition

Using classical logic then:

Let $(X,d)$ be a metric space, regarded as a topological space via its metric topology, and let $V \subset X$ be a subset. Then the following are equivalent:

1. $V \subset X$ is a closed subspace.

2. For every sequence $x_i \in V \subset X$ with elements in $V$, which converges as a sequence in $X$ it also converges in $V$.

###### Proof

First assume that $V \subset X$ is closed and that $x_i \overset{i \to \infty}{\longrightarrow} x_{\infty}$ for some $x_\infty \in X$. We need to show that then $x_\infty \in V$. Suppose it were not, then $x_\infty \in X\backslash V$. Since by definition this complement $X \backslash V$ is an open subset, it follows that there exists a real number $\epsilon \gt 0$ such that the open ball around $x$ of radius $\epsilon$ is still contained in the complement: $B^\circ_x(\epsilon) \subset X \backslash V$. But since the sequence is assumed to converge in $X$, this means that there exists $N_\epsilon$ such that all $x_{i \gt N_{\epsilon}}$ are in $B^\circ_x(\epsilon)$, hence in $X\backslash V$. This contradicts the assumption that all $x_i$ are in $V$, and hence we have proved by contradiction that $x_\infty \in V$.

Conversely, assume that for all sequences in $V$ that converge to some $x_\infty \in X$ then $x_\infty \in V \subset W$. We need to show that then $V$ is closed, hence that $X \backslash V \subset X$ is an open subset, hence that for every $x \in X \backslash V$ we may find a real number $\epsilon \gt 0$ such that the open ball $B^\circ_x(\epsilon)$ around $x$ of radius $\epsilon$ is still contained in $X \backslash V$. Suppose on the contrary that such $\epsilon$ did not exist. This would mean that for each $k \in \mathbb{N}$ with $k \geq 1$ then the intersection $B^\circ_x(1/k) \cap V$ is non-empty. Hence then we could choose points $x_k \in B^\circ_x(1/k) \cap V$ in these intersections. These would form a sequence which clearly converges to
the original $x$, and so by assumption we would conclude that $x \in V$, which violates the assumption that $x \in X \backslash V$. Hence we proved by contradiction $X \backslash V$ is in fact open.

### Relation to interior subspaces

###### Lemma

Let $(X,\tau)$ be a topological space and let $S \subset X$ be a subset. Then the topological interior $Int(S)$ of $S$ equals the complement of the topological closure $Cl(X\backslash S)$ of the complement of $S$:

$Int(S) = X \backslash Cl\left( X \backslash S \right) \,.$
###### Proof

By taking complements once more, the statement is equivalent to

$X \backslash Int(S) = Cl( X \backslash S ) \,.$

Now we compute:

\begin{aligned} X \backslash Int(S) & = X \backslash \left( \underset{{U \, open} \atop {U \subset S}}{\cup}U \right) \\ & = \underset{U \subset S}{\cap} X \backslash U \\ & = \underset{{C\, closed} \atop {C \supset X \backslash S}}{\cap} C \\ & = Cl(X \backslash S) \end{aligned}

### Relation to compact subspaces

The relation of closed subspaces to compact subspaces is expressed by the following statements

### Kuratowski’s closure-complement problem

This mildly amusing curiosity asks how many set-theoretic operations on a topological space $X$ are derivable from closure $C$ and complementation $\neg$ and applying finite composition. The answer is that at most 14 operations are so derivable (and there are examples showing this number is achievable). As the proofs below indicate, this bare fact has little to do with topology; it has more to do with general Moore closures and how they interact with complements (using classical logic).

Let $P(X)$ denote the power set (ordered by inclusion) and $M$ the monoid of endofunctions $P(X) \to P(X)$ with order defined pointwise. Then $C^2 = C$ and $\neg^2 = 1$ in $M$, with $C$ order-preserving and $\neg$ order-reversing. Also

• $I \coloneqq \neg C \neg$ is the interior operation, with $I \leq Id$.
###### Proposition

$C \neg C \neg$ is idempotent.

###### Proof

$I (C \neg C) \leq (C \neg C)$, i.e., $\neg C \neg C \neg C \leq C \neg C$. Applying the order-preserving operation $C$ to both sides together with the fact that $C^2 = C$, this gives

$C\neg C \neg C \neg C \leq C C \neg C = C \neg C.$

Since $I C \leq C$, we have also $\neg C \neg C \leq C$. Applying the order-reversing operation $C \neg C$ to both sides, we obtain

$C \neg C = C \neg C C \leq C \neg C (\neg C \neg C).$

Combining the two displayed inequalities gives $C \neg C = C \neg C \neg C \neg C$, and then multiplying this on the right by $\neg$, the proposition follows.

###### Proposition

Let $K$ be the monoid presented by two generators $C, \neg$ and subject to the relations $C^2 = C$, $\neg^2 = 1$, and $C \neg C \neg C \neg C = C \neg C$. Then $K$, called the Kuratowski monoid, has at most 14 elements.

###### Proof

We may apply an obvious reduction algorithm on the set of words in two letters $C, \neg$, in which a word is reduced by replacing any substring $C C$ by $C$ and any substring $\neg\neg$ by an empty substring, so that any word which cannot be further reduced must be alternating in $C, \neg$. This leads to a list of 14 words

$1, \qquad \neg, \qquad C, \qquad \neg C, \qquad C \neg, \qquad \neg C \neg, \qquad C \neg C,$
$\,$
$\neg C \neg C, \qquad C \neg C \neg, \qquad \neg C \neg C \neg, \qquad C \neg C \neg C, \qquad \neg C \neg C \neg C, \qquad C \neg C \neg C \neg, \qquad \neg C \neg C \neg C \neg$

with any further alternating words reducible by replacing a substring $C \neg C \neg C \neg C$ by $C \neg C$. Thus each element in the monoid $K$ is represented by one of these 14 words.

These 14 words actually name distinct set-theoretic operations $P(X) \to P(X)$ for a judicious choice of space $X$; as a corollary, the Kuratowski monoid $K$ has exactly 14 elements. For instance (courtesy of Wikipedia), taking $X = \mathbb{R}$ with its standard topology, the orbit of the element $(0, 1) \cup (1, 2) \cup \{3\} \cup ([4, 5] \cap \mathbb{Q})$ under the monoid action consists of 14 distinct elements.

###### Remark

At most 7 operations are possible with interior and closure, corresponding to the covariant Kuratowski operations. Thus there is a 7-element submonoid $K_{cov} \hookrightarrow K$. Spaces $X$ for which the topological action $K_{cov} \to Set(P(X), P(X))$ is not injective are of some structural interest; for instance, the spaces for which $C I C = I C$ are the extremally disconnected spaces, whereas spaces for which $I C = C$ are those where the open sets are equivalence classes for some equivalence relation (partition spaces). Those for which $I = C$ are discrete spaces.

###### Remark

A more manifestly topological consideration is what happens when we throw joins (or meets) into the $I, C$ mix. Briefly, at most 13 subsets can be obtained by starting with a subset $A \in P(X)$ and generating new subsets by taking closures, interiors, and unions; the order structure of these 13 subsets coincides with the free cocompletion of the finite ordered monoid $K_{cov}$ with respect to nonempty joins. Here we must use distributivity of $C$ over joins.

For some details on these remarks (and quite a bit more), see Gardner and Jackson, 2008 and Sherman 2004. An example of a non-topological Moore closure where the 14 operations are all distinct is given here.

## Generalizations

### Locales

In locale theory, every open $U$ in a locale $X$ defines a closed sublocale $\mathsf{C} U$ which is given by the closed nucleus

$j_{\mathsf{C} U}\colon V \mapsto U \cup V .$

The idea is that $\mathsf{C}U$ is the part of $X$ which does not involve $U$ (hence the notation $\mathsf{C}U$, or any other notation for a complement), and we may identify $V$ with $U \cup V$ when we are looking only away from $U$.

The sublocale $\mathsf{C}U$ is literally a complement of $U$ in the lattice of sublocales of $X$, i.e. $U\cap \mathsf{C}U = \emptyset$ and $U\cup \mathsf{C}U = X$ as sublocales. Moreover, if $X$ is a (sober) topological space regarded as a locale, then the locale $U$ is also spatial, and so is $\mathsf{C}U$, corresponding exactly to the topological closed set $X\setminus U$. (The fixed points of $j_{\mathsf{C}U}$ can be identified with the open sets containing $U$, which are bijectively related to the open subsets of $X\setminus U$.) Thus there is really only one notion of “closed subspace” whether we regard $X$ as a space or as a locale (at least as long as $X$ is sober).

### Constructive mathematics

In constructive mathematics, however, there are many possible inequivalent definitions of a closed subspace, including:

1. A subspace $C\subset X$ is closed if it is the complement of an open subspace, i.e. if $C = X\setminus U$ for some open subspace $U$;
2. A subspace $C\subset X$ is closed if its complement $X\setminus C$ is open;
3. A subspace $C\subset X$ is closed if it contains all its limit points, i.e. if for any $x\in X$ such that $U\cap C$ is inhabited for all neighborhoods $U$ of $x$, we have $x\in C$.

Definition (1) coincides with definition (2) or (3) only if excluded middle holds, since under (2) or (3) every subspace of a discrete space is closed, while under (1) the only closed subspaces are those that are complements, and if every proposition is a negation then the law of double negation follows. On the other hand, Definition (2) is clearly too strong, because even closed intervals $[a,b]$ in the real numbers can’t be proved constructively to satisfy it (though they do satisfy definitions (1) and (3)). Note also that Definition (1) always implies Definition (3), since if $C = X\setminus U$ and every open neighborhood of $x$ intersects $C$, then $x\notin U$ and thus $x\in C$. Thus it is not unreasonable (see also below) to define:

• $C\subset X$ is strongly closed if it is the complement of an open subspace.
• $C\subset X$ is weakly closed if it contains all its limit points.

This constructive variety of notions of closed subspace gives rise to a corresponding variety of notions of Hausdorff space when applied to the diagonal subspace.

Note also that neither of the sensible constructive definitions behaves quite like closed subspaces do classically. In particular, neither of them is apparently stable under finite unions (though the too-strong Definition (2) is). The situation is better for locales; see below.

### Locales in constructive mathematics

Of the “topological” definitions of closed subspace above, it is “strongly closed” (and the too-strong Definition (2)) that seem closest to the localic one. However, in the topological definitions we may not have $X = U \cup (X\setminus U)$ or $X = (X\setminus C) \cup C$ even as sets, whereas it remains true constructively that $X = U \cup \mathsf{C}U$ in the lattice of sublocales. In fact, we have the following:

###### Theorem

The following are equivalent:

1. The law of excluded middle.
2. Every closed sublocale of a spatial locale is spatial.
3. Every closed sublocale of a discrete locale is spatial.
###### Proof

We remarked above that (1)$\Rightarrow$(2), and of course (2)$\Rightarrow$(3). So assume (3). Every spatial sublocale is a union of its points, and in a discrete space points are open; thus if closed sublocales are spatial, they are also open. Since $X = U \cup \mathsf{C}U$ is constructively true, it follows that every open set is complemented in the open-set lattice of any discrete locale, which is to say that all powersets are Boolean algebras, i.e. excluded middle holds.

On the other hand, there is a localic notion of weakly closed sublocale that is closely related to topologically weakly closed subspaces (so that the above notion of “closed sublocale” — the formal complement of an open sublocale — could also be called strongly closed). It is the specialization of the notion of fiberwise closed sublocale? to locale maps $X\to 1$ into the terminal object.

###### Definition

A sublocale $C\subseteq X$ is weakly closed if it is not strongly dense in any larger sublocale of $X$. That is, if whenever $D$ is a sublocale of $X$ such that $C\subseteq D$ by a strongly dense inclusion, then $C=D$.

Since strong and weak denseness coincide classically, so do strong and weak closedness. And as we expect, strong closedness implies weak closedness, since strong density implies weak density. Moreover, both of them are better-behaved than the corresponding topological notions. For instance, strongly and weakly closed sublocales are both stable under finite unions (in the lattice of sublocales), even constructively.

Both strongly and weakly closed sublocales are “correct” notions; they are simply different. Some classical theorems about closed sublocales are constructively about strongly closed ones, while others (such as the theorem that any subgroup of a localic group is closed) are about weakly closed ones.

However, as we saw above, not every strongly closed sublocale (hence not every weakly closed sublocale either) can be spatial. But it is shown at strongly dense sublocale that a subspace of a space is localically strongly dense if and only if it is topologically strongly dense. This leads us to guess:

###### Conjecture

A subspace $C\subseteq X$ of a (sober) topological space $X$ is topologically weakly closed if and only if it is the spatial coreflection of a weakly closed sublocale.

In one direction this is easy: suppose $C$ is topologically weakly closed, and let $D$ be its localic weak closure. This is, by definition, the largest sublocale of $X$ in which $C$ is localically strongly dense. Now let $E$ be the spatial coreflection of $D$; since $C$ is spatial we have $C\subseteq E$, and since $C$ is localically strongly dense in $D$, it is also so in $E$, and hence topologically strongly dense. But $C$ is also topologically weakly closed, hence $C=E$ and is the spatial coreflection of the weakly closed sublocale $D$.

The other direction is harder.

## References

• C. Kuratowski, Sur l’opération $\bar{A}$ de l’analysis situs , Fund. Math. III (1922) pp.192-195. (pdf)

Weakly closed sublocales are discussed in

• Sketches of an Elephant, C1.1 and C1.2

• Peter Johnstone, A constructive ‘closed subgroup theorem’ for localic groups and groupoids, Cahiers de Topologie et Géométrie Différentielle Catégoriques (1989), Volume: 30, Issue: 1, page 3-23 link

• M. Jibladze and Peter Johnstone, The frame of fibrewise closed nuclei, Cahiers de Topologie et Géométrie Différentielle Catégoriques (1991), Volume: 32, Issue: 2, page 99-112, link

• Peter Johnstone, Fiberwise separation axioms for locales

Last revised on June 6, 2017 at 14:33:56. See the history of this page for a list of all contributions to it.