A topological concept dual to compactness

Question

We say that a subset A in a topological space X is anti-compact if every covering of A by closed sets has a finite subcover. Clearly if X is Hausdorff then all anti-compact subsets of X are finite. What is the structure of anti-compact sets in non-Hausdorff spaces? Is there any reference?

Answer 1

When looking for a dual concept we should be careful not to be tricked by a shallow symmetry. I will not comment on your definition of anti-compactness. Instead I would like to explain what the "true" dual to compactness is.

The notion dual to compactness is overtness. The concept has appeared in different forms in various approaches to topology. It is not easy to grasp for a classical topologist, and therefore some resistance and ignoring is to be expected. If you think not, consider this:

Theorem: All spaces are overt.

Even without knowing what "overt" means, the theorem has convinced you that it must be a useless concept. Anyhow, let me define the concept. Given a space $X$ let $\mathcal{O}(X)$ be its topology, equipped with the Scott topology. Homeomorphically, for a nice enough $X$, the space $\mathcal{O}(X)$ corresponds to the space of continuous maps $\mathcal{C}(X, \Sigma)$, equipped with the compact-open topology. Here $\Sigma = \{\bot, \top\}$ is the Sierpinski space. Define the map $A_X : \mathcal{O}(X) \to \Sigma$ by $$A_X(U) = \begin{cases} \top & \text{if $U = X$}\\ \bot & \text{else} \end{cases}$$ In essence, $A_X$ is the universal quantifier, for it tells us whether every point of $x \in X$ is in $U$. It would be logical to actually write $$A_X(U) = (\forall x \in X \,.\, x \in U).$$ Now we have:

Theorem: A space $X$ is compact if, and only if, $A_X$ is continuous.

Let us dualize. Define the existential quantifier $E_X : \mathcal{O} \to \Sigma$ by $$E_X(U) = \begin{cases} \bot & \text{if $U = \emptyset$}\\ \top & \text{else} \end{cases}$$ or in logical notation $$E_X(U) = (\exists x \in X \,.\, x \in U).$$ Is this just shallow symmetry? Well, certainly we have a symmetry between universal and existential quantifier, and it is cool that there is a direct connection between compactness and logic.

Definition: A space $X$ is overt when $E_X$ is continuous.

As I already stated, all spaces are overt in classical topology. That's why it is very hard to discover overtness (and this is an example of a monoculture being unable to make progress for a long time). But overtness has been independently discovered in point-free topology, in computable topology, and in constructive topology. Because there it is not a vacuous notion. For instance, speaking somewhat vaguely, in computable topology a subspace $S \subseteq X$ is computably overt when it is semidecidable whether $U \in \mathcal{O}(X)$ intersects $S$. Dually, a subspace $S$ is computably compact when it is semidecidable whether $U \in \mathcal{O}(X)$ covers $S$. It is not hard to come up with subspaces which are not computably overt, and subspaces which are compact but not computably compact: take the closed interval $[-\alpha, \alpha] \subseteq \mathbb{R}$ where $\alpha$ is a non-computable real.

Here is a further category-theoretic observation that makes the duality more convincing. I am going to skip over technicalities and side conditions, you can read about Paul Taylor's Abstract Stone Duality to get them all and more. The assignment $X \mapsto \mathcal{O}(X)$ which takes a space to its topology is a contravariant functor from spaces to frames. A continuous map $f : X \to Y$ is mapped to the inverse image map $\mathcal{O}(f) : \mathcal{O}(Y) \to \mathcal{O}(X)$, defined by $\mathcal{O}(f)(U) = f^{-1}(U)$. There is a unique map $t_X : X \to 1$ to the singleton space. This map is taken to a map $$\mathcal{O}(t_X) : \mathcal{O}(1) \to \mathcal{O}(X)$$ where we observe that $\mathcal{O}(1)$ is just $\Sigma$, so we have $$\mathcal{O}(t_X) : \Sigma \to \mathcal{O}(X)$$ in the category of frames. Since frames are posets, we can ask for left and right adjoint of $\mathcal{O}(t_X)$, i.e., Galois connections. And it turns out that:

• the right adjoint exists (and is $A_X$) if, and only if, $X$ is compact
• the left adjoint exists (and is $E_X$) if, and only if, $X$ is overt

[NB: in the category of frames morphisms are supposed to preserve all joins and finite meets, which gives a punch to existence of adjoints.] The characterization of quantifiers in terms of adjoints works very generally, and is in fact the basis for category-theoretic treatment of predicate calculus. Thus not only we have a duality between compactness and overtness, but also (again) a connection with logic.

A couple of theorems about compactness which dualize using overtness, where we note that the dual of Hausdorff (closed diagonal) is discrete (open diagonal):

Theorem:

• A closed subspace of a compact space is compact.
• An open subspace of an overt space is overt.

Theorem: (assuming the exponential $Y^X$ exists)

• If $X$ is compact and $Y$ is discrete then $Y^X$ is discrete.

• If $X$ is overt and $Y$ is Hausdorff then $Y^X$ is Hausdorff.

Answer 2

Andrej Bauer has given an excellent answer showing how overtness is dual to compactness, but to understand this idea more deeply and convince classical mathematicians that there is something in it, we have to break the symmetry.

Overt sub spaces are more interesting than overt spaces. We define such a ``subspace'' by a predicate $\lozenge$ on the open subset lattice that preserves unions, so $$ \lozenge \bigcup U_i \Longrightarrow \exists i.\lozenge U_i, $$ and it is easy to see that Andrej's $E_X$ has this property.

Think of $\lozenge$ like a Geiger counter: if it detects radioactivity in some neighbourhood $U=\bigcup U_i$ then it will do so in one or more of the sub-neighbourhoods $U_i$ into which we divide $U$.

If the ambient space has a basis then we only need to know the value of $\lozenge$ on basic open subspaces.

For example, in Cantor space, which computer scientists think of as consisting of infinite sequences of 0s and 1s, the basic open sets are named by finite sequences. The one called $s$ is the union of those called $s0$ and $s1$, so $$ \lozenge s \Longrightarrow \lozenge(s0) \lor \lozenge(s1). $$ This is an important property in Process Algebra.

In $R^n$, the basic opens are balls $B(x,r)$. It turns out to be reasonable to write $$ d(x) < r \quad\mbox{for}\quad \lozenge B(x,r) $$ because it satisfies $$ d(x) < r \iff \exists r'.d(x) < r' < r $$ $$ d(x) < r \Longrightarrow \exists y.d(y) < \epsilon\;\land\;d(x,y) < r \Longrightarrow d(x) < r+\epsilon. $$ Then we say that $a$ is an accumulation point of $\lozenge$ or $d$ if $d(a)=0$ and find that $d(x)$ is the distance of $x$ from the nearest accumulation point.

This formulation relates overt to located subspaces in Bishop's Constructive Analysis.

An example of almost these properties is the Newton-Raphson algorithm.

Recall that, for any function $f:R^n\to R^n$ that has a continuous invertible derivative, this algorithm defines a sequence $(x_n)$ by $ x_{n+1} = x_n + g(x_n) $ where $ g(x) = \big(\dot f(x)\big)^{-1}\cdot\big(y-f(x)\big) $.

We define $\Delta(x) < r$ to be the conjunction of the three conditions

• $\dot f(x)$ is invertible,
• ${|g(x)|} < r/2$, and
• $ \forall x' x''\in\ \overline B(x,r). {|\dot f(x)^{-1}\cdot\big(\dot f(x')-\dot f(x'')\big)|} \;\leq\;{|x'-x''|}/r$,

so if these conditions fail we define $\Delta(x)=\infty$.

Then we obtain a function $d$ with the above properties by $$ d(x) < r \quad\mbox{if}\quad \exists y s. \Delta(y) < s \land d(x,y) < r-s. $$

These ideas are developed in my draft paper Overt subspaces of $R^n$. I claim that a $\lozenge$ operator gives evidence of the existence of the solution of a problem, the solutions themselves being its accumulation points.

The results that Andrej mentioned are in my (published) paper A Lambda Calculus for Real Analysis, motivated by the Intermediate Value Theorem. (Andrej played a major part in the existence and writing of this paper.)

PS in answer to the comments below. In the classical use of set theory in mathematics, subsets and predicates are interchangeable. In this case you may replace my $\lozenge U$ throughout by $U\in{\mathcal M}$, where $\mathcal M$ is a family of open subspaces that is closed under joins.

However, besides its being a clumsy notation, if you think of an overt subset as a set of sets of points you are deliberately going in the wrong direction to understand this idea.

My contention is that an overt subspace consists of the discoverable solutions of a problem. In the example the problem is to invert a differentiable function, but I believe these ideas are much more widely applicable in analysis. Such a problem is expressed syntactically and the discovery of a solution is computation.

Set theory (speaking of the set of all solutions) at best puts the cart before the horse but actually contributes nothing to solving the problem. To call set theory "classical" mathematics is a perverse use of the word compared to its meanings in other disciplines, which would naturally describe the Newton-Raphson algorithm as classical analysis.

My claim is that we start with the formula in analysis expressing the problem, turn it into a formula in logic (predicate) that defines the overt subspace and then potentially use methods of computational logic to extract (successive approximations to) a solution.

Of course there is no magic wand and in practice finding accumulation points of overt subspaces on ${\mathbb R}^n$ will require methods that are equivalent to those that are already used in numerical analysis. What we have gained is a conceptual division of the process into a pure mathematical part (a new concept in general topology that is dual to and usable like compactness) and a computational part.

Answer 3

Are you perhaps looking for overtness?

Answer 4

It suffices to consider the $T_0$ spaces as other cases can be handled by replicating points.

Let $X$ be an "anti-compact" $T_0$ space. It is worthwhile to look at the specialization order on the points of $X$ which is defined by $y \leq_X x$ iff $y \in \overline{\{x\}}$. Since the closed sets $\overline{\{x\}}$ cover $X$, the specialization order has finitely many maximal elements $x_1,\dots,x_n$ and every element of $X$ lies below one of these maximal elements. This criterion is also sufficient. Given a closed cover $\mathcal{C}$, we can certainly find $C_1,\dots,C_n \in \mathcal{C}$ such that $x_1 \in C_1,\dots,x_n \in C_n$ and then $X = C_1 \cup\cdots\cup C_n$.

Given a partial order $(X,{\leq_X})$ the finest topology that induces this specialization preorder that has ${\leq_X}$ as a specialization order is the Alexandrov topology, where all upper sets are open. Any coarser topology where the lower sets $\{y \in X : y \leq x\}$ are all closed will generate the same specialization preorder. This can be used to generate examples.

Since the original question is a about "anti-compact" subsets $A$ of a (not necessarily $T_0$) space $X$. It follows from the above that $A \subseteq X$ is "anti-compact" if and only if there are finitely many points $x_1,\dots,x_n \in A$ such that $A \subseteq \overline{\{x_1,\dots,x_n\}}$.

Also note that the continuous image of an "anti-compact" space is "anti-compact". Indeed if $f:X \to Y$ is continuous and $X = \overline{\{x_1,\dots,x_n\}}$ then $\overline{\{x_i\}} \subseteq f^{-1}(\overline{\{f(x_i)\}})$, so $f(\overline{\{x_i\}}) \subseteq \overline{\{f(x_i)\}}$ and therefore $\{f(x_1),\dots,f(x_n)\} \subseteq f(X) \subseteq \overline{\{f(x_1),\dots,f(x_n)\}}$.

Answer 5

Since my comment on point-free topology was upvoted, let me give a better description of what anti-compactness means in point-free topology and why this notion corresponds to the notion of the cardinality of a topological space. The anti-compactness number in a topological space $X$ is the smallest cardinal $\lambda$ such that every closed cover of $X$ has a subcover of cardinality less than $\lambda$. Clearly the anti-compactness number of a $T_{1}$ space $X$ is $|X|^{+}$, so the anti-compactness number measures the cardinality of $T_{1}$-spaces. In this answer, I shall generalize the notion of the anti-compactness number to point-free topology. The reader is encouraged to read Picado and Pultr's book Frames and Locales: Topology Without Points for more thorough explanations.

Suppose that $L$ is a frame. Then a subset $S\subseteq L$ is a sublocale if $S$ is closed under arbitrary meets and whenever $l\in L$ and $s\in S$, then $l\rightarrow s\in S$ as well. The notion of a sublocale is a point-free analogue of the notion of a subspace of a topological space. There are several different characterizations of the notion of a sublocale. It should be also noted that the sublocales are in a one-to-one correspondence with the congruences on a frame. The collection of all sublocales of a frame is a co-frame under the inclusion ordering. In particular, the collection of all sublocales of a frame is a complete lattice. In this complete lattice, the greatest lower bound of sublocales is simply the intersection of sublocales, and the least upper bound of sublocales is given by the following simple formula:

$$\bigvee_{i\in I}S_{i}=\{\bigwedge R|R\subseteq\bigcup_{i\in I}S_{i}\}.$$

If $L$ is a frame and $a\in L$, then the set $\uparrow a=\{y\in L|y\geq a\}$ is a sublocale of $L$. The sublocales of the form $\uparrow a$ are called closed sublocales. There is also a notion of an open sublocale of a frame.

It should be noted that the greatest lower bound of sublocales does not correspond very well with the intersection of sets. The greatest lower bound of all dense open sublocales of a frame is a dense sublocale since every frame has a smallest dense sublocale. However, the intersection of all dense open subspaces of a topological space with no isolated points is the empty set.

On the other hand, the least upper bound of sublocales of a frame corresponds quite nicely to the union of subsets of a set. If $(X,\mathcal{T})$ is a topological space and $R\subseteq X$, then define an equivalence relation $E_{R}$ on $\mathcal{T}$ by letting $(U,V)\in E_{R}$ iff $U,V\in\mathcal{T}$ and $U\cap R=V\cap R$. Then each $E_{R}$ is a congruence on $\mathcal{T}$. Let $\widetilde{R}$ be the collection of all maximum elements from each equivalence class in $E_{R}$. Then $\widetilde{R}$ is the sublocale of $\mathcal{T}$ that corresponds to the subspace $R$ and $E_{R}$ is the congruence on the frame $\mathcal{T}$ that corresponds to the subspace $R$. Furthermore, if $R_{i}\subseteq X$ for $i\in I$, then

$$\widetilde{\bigcup_{i\in I}R_{i}}=\bigvee_{i\in I}\widetilde{R_{i}}$$

(here the least upper bound is taken in the lattice of sublocales of $\mathcal{T}$), and

$$E_{\bigcup_{i\in I}R_{i}}=\bigcap_{i\in I}E_{R_{i}}.$$

We define the anti-compactness number of a frame $L$ to be the least cardinal $\lambda$ such that whenever $\mathcal{C}$ is a collection of closed sublocales of $L$ with $\bigvee\mathcal{C}=1$, then there is some $\mathcal{D}\subseteq\mathcal{C}$ with $|\mathcal{D}|<\lambda$ and $\bigvee\mathcal{D}=1$.

If $(X,\mathcal{T})$ is a $T_{1}$-space, then the anti-compactness number of the frame $\mathcal{T}$ is $|X|^{+}$. Therefore the anti-compactness number of a frame truly measures the cardinality of the corresponding topological space.

Also, the anti-compactness number of a complete Boolean algebra is its saturation (as defined in the Handbook of Boolean Algebras).

I have not found anything published that has attempted to generalize the notion of the cardinality of a space to point-free topology, and there may be similar ways or even better ways to generalize the notion of the cardinality to point-free topology (in fact, sometimes concepts in general topology generalize to point free topology in several natural ways). On the other hand, unlike cardinality, most other well known notions from general topology have been generalized quite nicely and easily to point-free topology. I will probably come back later and add more detail to this answer. Perhaps, I should write a paper on this issue.

Answer 6

When one considers the more general notion of a closure systems instead of topological space, then a modification of anti-compactness of a lattice corresponds to precisely compactness in complete lattices. Here we modify anti-compactness by only requiring the collection to cover a dense subset of your space.

Before I begin, I should mention that the notion that having a finite subset whose union is dense is a generalization of compactness that has been studied before and is an interesting notion: for instance, a completely regular space $X$ is pseudocompact if and only if every countable cover of $X$ has a finite subcollection whose union is dense.

A closure system is a pair $(X,C)$ such that $X$ is a set and $C\subseteq P(X)$ is a collection of subsets of $X$ closed under arbitrary intersection. It is easy to show that $(X,C)$ is a closure system if and only if $C\subseteq P(X)$ and for each $L\subseteq X$ there is a smallest $M\in C$ with $L\subseteq M$. If $(X,C)$ is a topological closure system, then define $C^{*}:P(X)\rightarrow P(X)$ by letting $C^{*}(L)$ be the smallest element in $C$ with $L\subseteq C^{*}(L)$. We shall call $C^{*}(L)$ is the closure of $L$. A subset $A\subseteq X$ is said to be dense if $C^{*}(A)=X$. We say that a complete lattice $P$ is compact if whenever $R\subseteq P$ and $\bigvee R=1$ there is some finite subset $S\subseteq R$ with $\bigvee S=1$. The notion of compactness in lattices has been studied along with its generalizations in general topology and lattice theory (for instance, algebraic lattices are the complete lattices generated by compact elements). The following proposition has a straightforward proof.

$\mathbf{Proposition}$ Suppose that $(X,C)$ is a closure system. Then the following are equivalent.

1. The lattice $C$ is compact.

2. Whenever $A\subseteq X$ is dense, there is a finite $B\subseteq A$ such that $B$ is dense in $X$.

3. Whenever $D\subseteq C$ is a set where $\bigcup D$ is dense in $X$, then there is some finite $E\subseteq D$ where $\bigcup E$ is dense in $X$.

$\mathbf{Proof}$

$1\leftrightarrow 3$ The equivalence follows from the fact that we have $\bigvee^{C}D=X$ if and only if $\bigcup D$ is dense in $X$.

$3\rightarrow 2$ Suppose that $A$ is dense in $X$. Then let $D=\{C^{*}(\{a\})|a\in A\}$. Then $\bigcup D$ is dense in $X$, so there is some finite $E\subseteq D$ with $\bigcup E$ dense in $X$. If $E=\{C^{*}(\{a_{1}\}),...,C^{*}(\{a_{n}\})\}$, then$ C^{*}(\{a_{1},...,a_{n}\})=C^{*}(\bigcup E)=X$, so $\{a_{1},...,a_{n}\}$ is dense $X$ and $\{a_{1},...,a_{n}\}\subseteq D$.

$2\rightarrow 3$ Suppose that $D\subseteq C$ and $\bigcup D$ is dense in $X$. Therefore there is some finite subset $A\subseteq\bigcup D$ which is dense in $A$. Therefore there is some $E\subseteq D$ with $A\subseteq\bigcup E$ and where $\bigcup E$ is dense in $X$. $\mathbf{QED}$.

Furthermore, I claim that complete lattices and closure systems are pretty much the same thing. A based lattice is a pair $(L,A)$ such that $L$ is a complete lattice, $A\subseteq L$, and $L=\{\bigvee^{L}R|R\subseteq A\}$. A closure system $(X,C)$ is said to be a $T_{0}$-closure system if $C^{*}(\{a\})\neq C^{*}(\{b\})$ whenever $a\neq b$. If $(L,A)$ is a based lattice, then $(A,\{\{a\in A|a\leq x\}|x\in L\})$ is a $T_{0}$-closure system. Similarly, if $(X,C)$ is a $T_{0}$-closure system, then $(C,\{C^{*}(\{x\})|x\in X\})$ is a based-lattice. The category of $T_{0}$-closure systems is equivalent to the category of all based lattices. Therefore, closure systems are pretty much complete lattices, so compactness in lattices corresponds to a sort of anti-compactness in closure systems.

Answer 7

My reputation points (all spent on a bounty ...) do not allow me to comment where it is appropriate to do so. However I would like to add to the comment thread on Andrej's answer above.

Whilst I don't think this is an area of very active research globally, that there is an important duality between proper and open is clear from a number of sources. The most prominent reference to my mind is Chapter C3 of Johnstone's topos theory book 'Sketches of an Elephant'. Therein classes of geometric morphism (proper, open, ...) are investigated side by side, sketching at least how they are parallel. From memory Japie Vermeulen's paper 'Proper maps of locales' demonstrates the parallel between proper and open by showing how the Joyal-Tierney results on open maps also work for proper maps, essentially using the same arguments and so hinting at the parallel. Paul Taylor's work on ASD shows how to make the parallel into a formal duality by flipping the order enrichment and there is a paper by Steven Vickers ('Locales are not pointless') which effectively makes a similar observation. My website has a number of other papers on the topic including an introductory paper ('On the Parallel between the Suplattice and Preframe approaches to Locale Theory') which appeared in Annals of Pure and Applied Logic a few years ago now.