Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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?

share|improve this question
This is not really a natural definition of dual compactness ... –  Martin Brandenburg May 4 at 9:27
Similar question at MSE: Terminologies related to “compact?”. Some authors call spaces such that every closed cover has a finite subcover strongly S-closed. Some equivalent conditions are listed in my question here. (Sorry for the self-promotion, but link seemed better solution than posting the same list again here in a comment or an answer.) –  Martin Sleziak May 4 at 10:05
anti-compact as a property name is already taken: $X$ anti-compact means that the only compact subsets of $X$ are the finite ones. E.g. a co-countable space is an example. –  Henno Brandsma May 4 at 11:59
At the risk of adding one more distracting terminological comment, let me point out that a space $X$ where every subset is 'anti-compact' in your sense is quite reasonably the dual idea of a Noetherian space. Though this would follow a nearly systematic naming tradition, I do not recommend calling such spaces 'Artinian spaces'! –  François G. Dorais May 4 at 13:28
Let's define the anti-compactness number of a space $X$ to be the smallest cardinal $\lambda$ such that every closed cover of $X$ has a subcover of cardinality less than $\lambda$. Then the anti-compactness number of a Hausdorff space is the least cardinal above its cardinality. However, the anti-compactness number of a space can be generalized to point-free topology. Therefore, the anti-compactness number seems to be a point-free generalization of the notion of the cardinality of a topological space. –  Joseph Van Name May 5 at 0:54

5 Answers 5

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):


  • 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.

share|improve this answer
'...this is an example of a monoculture being unable to make progress for a long time'. Progress on what, exactly? –  HJRW May 5 at 18:31
Progress on discovering that there is a concept dual to compactness. –  Andrej Bauer May 5 at 19:26
Was this a topic of active research? It seems harsh to criticize topologists for failing to discover a vacuous (to them) concept. And has the discovery of overtness led to any progress in topology? (As opposed to in point-free, computable, or constructive topology, where it seems to have been discovered when it was needed.) –  HJRW May 5 at 20:14
I am not criticizing anyone, just observing how plurality of views on what topology is about was necessary for some concepts to emerge. And yes, there are clear benefits to topology as a whole, as well as to classical topology. –  Andrej Bauer May 5 at 20:46
As you can guess, I was going for the drama. –  Andrej Bauer Jul 27 at 18:37

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 therfore $\{f(x_1),\dots,f(x_n)\} \subseteq f(X) \subseteq \overline{\{f(x_1),\dots,f(x_n)\}}$.

share|improve this answer

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.

share|improve this answer

Are you perhaps looking for overtness?

share|improve this answer
I am not sure that the definition given in nLab actually defines anything non-trivial. Whereas other topological concepts mean much the same thing in different logical formulations of topology, overtness is very sensitive to the ambient logic. At the classical extreme, all (sub)spaces are overt, whilst in ASD (which is essentially computable) the concept is more useful. Using a basis makes things less prone to obliteration by overpowerful logic. –  Paul Taylor Jul 28 at 14:37

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.

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.