Want to take part in these discussions? Sign in if you have an account, or apply for one below
Vanilla 1.1.10 is a product of Lussumo. More Information: Documentation, Community Support.
As an outcome of recent discussion at Math Overflow here, Mike Shulman suggested some nLab pages where comparisons of different definitions of compactness are rigorously established. I have created one such page: compactness and stable closure. (The importance and significance of the stable closure condition should be brought out better.)
I have added to the Properties-section at compact space that in Hausdorff space every compact subset is closed.
I have tried to edit compact space just a little for readability.
For one, I tried to highlight nets over proper filters just a tad more (moving the statements for nets out of parenthesis, adding reminder to the equivalence via their eventuality filters), just so that the entry gets at least a little closer to giving the reader the expected statement about convergence of sequences.
Also I pre-fixed the text of all of the many equivalent definitions by “$X$ is compact if…” to make it read more like a text meant for public consumption, and less like a personal note.
Looks good!
I am confused about the constructive notion of “compact locale”. In the Elephant just before C3.2.8, Johnstone says
If $X$ is a locale in a Boolean topos, then the unique locale map $\gamma:X\to 1$ is proper iff $\gamma_*$ preserves directed joins… Constructively, the Frobenius reciprocity condition [$f_*(U\cup f^*(V)) = f_*(U)\cup V$]… is a nontrivial restriction even on locale maps with codomain 1; so we include it in the definition of compactness — that is, we define $X$ to be compact if $X\to 1$ is proper.
However, as far as I can tell this additional Frobenius condition is not included in the general definition of proper geometric morphism, which just says (in the stack semantics of $E$) that $f:F\to E$ preserves directed unions of subterminals. It makes sense that some extra condition is necessary when speaking externally about a general locale map $f:X\to Y$, since “$f_*$ preserves directed unions” is not internalized to the stack semantics of $Sh(Y)$. However, when talking about a map $X\to 1$, we should already be in the stack semantics of $Sh(1) = Set$. So I don’t understand why we need the Frobenius condition separately in the constructive definition of “compact locale”. Johnstone doesn’t give an example, and I don’t have Moerdijk or Vermuelen’s papers.
On the other hand, I don’t see how to prove the Frobenius condition either. I have looked through section C3.2 of the Elephant as carefully as I can, and I haven’t been able to extract an actual proof that a proper geometric morphism between localic toposes satisfies the Frobenius condition to be a proper map of locales. Prior to defining proper geometric morphisms, he says
…we may reasonably define a geometric morphism $f:F\to E$ to be proper if… $f_*(\Omega_F)$ is a compact internal frame in $E$. What does it mean to say that an internal frame… is compact in a general topos $E$? For the case $E=Set$, we saw how to answer this in topos-theoretic terms in 1.5.5: it means that the direct image functor $Sh(X) \to Set$… preserves directed colimits of subterminal objects.
and then proceeds to define $f:F\to E$ to be proper if “$f_*$ preserves directed colimits of subterminal objects” is true in the stack semantics of $E$. But C1.5.5 used “$f_*$ preserves directed unions” as a definition of “compact locale”, which the first quote above claims is not sufficient constructively, i.e. over a general base $E$. So I am confused; can anyone help?
Ok, I found Vermeulen’s paper, and I believe he has a constructive proof that if $r:X\to 1$ preserves directed joins then it satisfies the Frobenius condition. Suppose $U\in O(X)$ and $V\in O(1) = \Omega$; we must show that if $r_\ast(U \cup r_\ast(V))$ is true, then so is $r_\ast(U) \cup V$. Note that $r_\ast(W)$ is the truth value of the statement “$W=X$”, while $r^*(P) = \bigcup \{ X \mid P \}$. Suppose $r_\ast(U \cup r_\ast(V))$, i.e. that $U\cup \bigcup \{ X \mid V \} = X$. Now consider the set $\{ W\in O(X) \mid V \vee (W\le U) \}$; this is evidently directed, and our supposition in the last sentence says exactly that its union is $X$. Therefore, if $X$ is compact in the sense that its top element is inaccessible by directed joins, there exists a $W$ such that $V \vee (W\le U)$ and $X\le W$. In other words, either $V$ or $X\le U$, i.e. either $V$ or $r_\ast(U)$, which is what we wanted.
So my current conclusion is that the first Elephant quote above is wrong about the Frobenius condition being an additional restriction constructively. This did seem like the most likely resolution, since otherwise the definition of proper geometric morphism would probably be wrong, which seems unlikely.
I have added this proof to covert space, with pointers to it from compact space and proper map.
There seems to be something wrong right at the beginning of compact space:
After def. 2.1, the usual definition about existence of finite open subcovers, there is def. 2.2 which is the immediate reformulation in terms of closed subsets: a collection of closed subsets whose intersection is empty has a finite subcollection whose intersection is still empty.
But this def. 2.2 is introduced with the remark that it needs excluded middle to be equivalent to 2.1, which is not true.
Probably what that remark about excluded middle was meant to refer to is instead the further formulation in terms of closed subsets, the one which says that a collection of closed subsets with the finite intersection property has non-empty intersection.
[edit: I have added what seems to be missing at compact space to finite intersection property]
How are closed sets being defined here?
If we are defining closed sets to be precisely the complements of open sets (in symbols, $\neg U$), then I can see that the usual open set formulation implies the closed set formulation you just mentioned: if $\bigcup_{i \in I} U_i = X$, then surely $\bigcap_{i \in I} \neg U_i = \neg \left(\bigcup_{i \in } U_i \right) = \neg X = \emptyset$ since $\neg$ takes unions to intersections.
But then how would we turn this implication around (to assert equivalence of the two formulations)? Without excluded middle, I don’t see how every open set $U$ would be the complement of a closed set.
Oh, okay. I’ll make all that explicit in the entry now.
Okay, I have edited statement and proof at compact+space#fip. I made explicit the use of excluded middle for identifying opens with complements of closed subsets, and a second use of excluded middle for getting what is usually labeled “fip”, which is the contrapositive of what was formerly stated here.
FWIW, constructively there are at least two distinct definitions of closed set.
I have added some elementary details at compact space – Examples – General, such as the proof that closed intervals are compact.
I have added the statement about unions and intersections of compact subspaces, here
In accordance with a recent discussion, I moved the detailed elementary proof of the example of closed intervals to its own page. Also, I added the example of cofinite topology to compact space.
Re: #16, $[0,3] \setminus (1,3)$ is $[0,1] \cup \{3\}$, which is compact.
Re: #17, I think you’re right that 2.4(4) and 2.10 are redundant. I don’t have an opinion on which of them should be kept.
I have now spelled out a detailed proof of prop. 3.2 here.
Regarding the duplication of the statement of the closed-projection characterization of compactness:
I have removed item 4 from prop. 2.4, but I also moved the former “Definition” 2.10 up to what is now prop. 2.5, so that it is still the next statement after item 3 of prop. 2.4.
(Todd should please have a look.)
My main request about this entry is: Somebody should turn the long list of “Definitions” 2.7 to 2.13 of compactness into a list of propositions that state that certain statements are equivalent.
Thanks, Urs. It looks good to me. Does the mention of excluded middle now make sense to tphyahoo? To circumvent excluded middle (as in closed-projection characterization of compactness), one has to change the statement to say the dual image operator $\forall_\pi: P(X \times Y) \to P(Y)$ takes open sets to open sets; the statement as it is, that the direct image along projection takes closed sets to closed sets, is equivalent to the other statement by De Morgan duality, but that’s where excluded middle comes in.
I agree, with #24, that’s what I was referring to at the end of #21.
This needs to go to the attention of Toby and Todd, I think. The request would be: Turn these terse remarks into something a little more self-contained and inviting.
Okay, I’ll begin having a look. Much of this was written (I think) by Toby quite a while ago, and his presence here has lately become more sporadic.
Please be patient though. I am finding some of the tone harsh (e.g., “now seems more obviously incoherent” – that I find excessive), and it’s a bit of a barrage of comments now to process.
Okay, I’ve gone through and reorganized section 2 of compact space according to comments/suggestions above and my own personal knowledge. Roughly I classified the various “definitions” (now called propositions) under three headings: elementary reformulations, via convergence, and via stability properties. There is still plenty left to do: plenty of proofs which can be filled in or farmed out to other parts of the nLab, as appropriate, and still links left to be made, among other things.
It is good that you (tphyahoo) brought your concerns to our attention, so thanks for that. I do think the article has a better shape now. The “obviously incoherent” former 2.12 is, I hope you will now see, coherent after all. I also changed the link from logic to quantification which is more precise I think.
Thanks, Todd!! That’s great.
Something seems wrong with Proposition 3.3, that the category of compact spaces has limits. (Of course compact Hausdorff spaces have limits.) The problem is that the equalizer of two maps need not be a closed subspace; that kind of thing is true if we are working with Hausdorff spaces, but not for more general spaces.
An explicit example is where we take two maps $[0, 1] \to \{0, 1\}$ where the codomain is given the indiscrete topology, where one of the maps $f$ has $f^{-1}(1) = [0, 1/2)$, and the other is the constant map at $1$. If an equalizer in $Comp$ existed, then $\hom(1, -): Comp \to Set$ would have to preserve it, so set-theoretically it would have to be $[0, 1/2)$. The topology on $[0, 1/2)$ would have to be the same as or finer than the subspace topology in order for the equalizer map to be continuous. But if the subspace topology isn’t compact, then no finer topology would make it compact either. (Here I’m taking the contrapositive of the proposition that if $(X, \tau)$ is compact and $\tau' \subseteq \tau$ is a coarser topology, then $(X, \tau')$ is also compact.)
I guess I could go in and change it to a true statement, but I’d want to know first about the situation for compact locales. Again, for compact regular locales, there’s no problem.
I feel like if compact locales had limits we would know about it. Could you do something similar to your counterexample with the Sierpinski space instead of the indiscrete 2-point space?
Yes, it seems the same counterexample works with Sierpinski space, since $1$ is usually taken to be the open point. (Maybe this is discussed somewhere in Stone Spaces? I don’t have my copy within easy view.) So yeah, if I think about it a little longer, maybe it will become obvious that compact frames lack coequalizers, or maybe you already see that’s true.
That “Idea” section should be read at a very intuitive level; the language is somewhat fuzzy (I suppose by design). The author of those words should be granted some poetic license.
Any space, compact or not, is closed in itself, so under strict interpretations some of the language of the Idea section won’t make a lot of sense. The real sense of that passage is concentrated in the words “every net has an accumulation point” (that certainly has a precise meaning), and that should be the main takeaway from what the author is trying to convey. The rest of it seems to be a simple appeal to the archetypal image we all carry around in our heads of compact sets: sets in $\mathbb{R}^n$ which are closed and bounded, and the author is trying to draw a connection between that intuitive conception of “closed and bounded”, based on that (finite-dimensional) picture, and the more precise mathematical conception “every net must have an accumulation point”.
In infinite-dimensional Hilbert space (for example), “closed and bounded” do not imply compact. So that intuition comes with a (largish) grain of salt! In fact the word “bounded” isn’t quite a topological concept; it makes sense for metric spaces but it doesn’t have a meaning for general topological spaces.
Turning to Proposition 3.1: now we’re doing mathematics, not waving intuitive wands. As it happens, compact subspaces of Hausdorff spaces are closed (and Hausdorff spaces make up the majority of spaces one encounters when first learning topology), but in general compact subsets need not be closed in the ambient space. Thus the closure hypothesis has to be inserted by hand for the proposition to work in the generality given there.
Anyway, you may be right that the Idea section is (for some readers anyway) more confusing than enlightening. It’s hard to say, but perhaps the opening should be reconsidered.
No, the “it” here is the net (nets are generalized sequences).
Re #46: no, the definitions/characterizations section is general. Heine-Borel is for a specialized set of circumstances and belongs to Examples.
Okay, I tweaked the opening paragraph just a little. I don’t agree with joining “closed” to “bounded” as in #45, but I did mention Heine-Borel. Hopefully it’s now clear that it’s more about nets and convergence than it is about being closed and bounded.
But there was no mention of compactness in the 2nd paragraph of your edit; attention was momentarily deflected away to talk about “closed and bounded”. So I had to reword again.
Please do not remove the parentheses. They signal that a side remark is being made, which the reader can pursue if she likes, but the main focus needs to be kept on what the property is about.
Thank you for your input, but now that some clarity has been reached, I suggest that we not spend a lot of time on what Feynman called “wordsmithing”. I’ll add that I made a change suggested by comment #37.
I took a look and ended up making some last changes myself to the Idea section:
made explicit that “everything” in the first sentence referred to sequences and nets
linked the line about not needing and ambient space for the definition with the line saying that nevertheless one does often consider compact subspaces;
after the claim that one likes to consider compact Hausdorff spaces, I added one reason
grouped the two lines about compact locales together, now at the end of the Idea-section.
Those are useful improvements, Urs – thanks.
Urs, in the idea section, did you really mean to say “paracompact” and not “locally compact”?
Argh. :-)
Thanks. I have fixed it, and also added pointer to the proof.
It is so stated. See closed projection characterization, point 5. under Variant Proofs. It’s mentioned that this approach is in Escardo’s 2009 paper. See lemma 4.3 there.
Yes, that’s what’s meant. But this page is written in classical mathematics as the default, so I don’t think there’s a need to say so explicitly here, especially since the subsequent paragraph clarifies the situation in constructive mathematics.
sorry, wrong thread
An open cover of a subset is just an open cover of that set regarded as a space with the induced topology. But we could add that as a remark afterwards.
Typos like that don’t have to be mentioned.
I don’t think much of these are really improvements. In some cases you have been replacing other people’s words, which were fine and clear enough, with words that you would have chosen instead, such as “unioning”, a word I find grating and cacophonous outside of very informal speech.
But let’s look at this stuff about directed open covers, where you have the paragraph with the word “Firstly”.
Firstly, note that unions of finite opens give a direction on any open cover. This gives us the notion of a directed open cover, which is useful for locales.
As samples of mathematical writing, both sentences of that paragraph are flawed. What one should be saying – and what I read proofs in former revisions to be saying – is that given an open cover, one can form a new cover that is directed. Not that the old cover was “given a direction”. (By the way, “give a direction” sounds clunky to my ears. I realize that the link for “directed” goes to a page titled “direction”, but I think that’s partly because of a rule we have about using noun phrases as titles. But anyway, nobody ever “gives a direction” to an open covering or to a preorder: either an open covering is directed or it isn’t. What one does is replace an open covering that may not be directed with another that is, by adding in more open sets.)
The second sentence is flawed because “this”, whatever the antecedent is, is not “giving us a notion of” anything. “Giving a notion”, to my mind, means performing an act of conceptual analysis, as in abstracting a new concept that captures or expresses a variety of observed phenomena. “This gives us a directed open cover” is closer to what one should say, but only after fixing the first sentence.
Really, I think it might be better to roll back to an earlier revision, and then fix whatever tiny typos needed fixing. I’m sorry to say, but I think some of the newer revisions are worsening the article. I’m not sure why you’re doing this.
(By the way, I agree that “diffs” are often hard to read. It might sometimes be easier reading different revisions in different tabs.)
It’s really very simple: a poset is directed if it is nonempty and any two elements have a common upper bound. The upper bound doesn’t have to be the union.
I sort of wish you’d get away from this idea of “defining a direction” here. In practice, one defines a partially ordered set, and then that partially ordered set is either directed or it isn’t: there is no extra step of “defining a direction” that needs to be done. (Even “defines a partial order” sounds too elaborate, because in this context the partial order is invariably subset inclusion.) As far as any defining goes: at some point in the proof, one introduces (or defines) an open cover $\mathcal{U}'$ and then verifies that it is directed.
I don’t know what is confusing you in this article, but let’s talk about it here. We have the traditional notion of compact space: every open cover has a finite subcover. The proposition in question states that compactness of a space $X$ is equivalent to another condition: that any directed open cover of $X$ has $X$ among its elements. I’ll call that condition D.
Let’s just take a moment to be very explicit what all this means. First, an open cover of $X$ is a collection $\mathcal{U}$ of open subsets $U \subseteq X$ whose union is $X$. An open cover $\mathcal{U}$ is directed if, whenever one is given finitely many elements $U_1, \ldots, U_n \in \mathcal{U}$, there is an element $U \in \mathcal{U}$ such that $U_1 \subseteq U, \ldots, U_n \subseteq U$, in other words there exists an element $U \in \mathcal{U}$ that is an upper bound of the $U_1, \ldots, U_n$ with respect to subset inclusion. This $U$ doesn’t always have to be the actual union (i.e., the least upper bound), although it can be.
To show that a compact space $X$ satisfies condition D, let $\mathcal{U}$ be any directed open cover. (We are not obliged to assume that $\mathcal{U}$ is closed under finite unions.) By compactness, $\mathcal{U}$ has a finite subcover, meaning there are finitely many elements $U_1, U_2, \ldots, U_n$ of $\mathcal{U}$ whose union (i.e. least upper bound) is $X$. By the assumption that $\mathcal{U}$ is directed, there exists an upper bound $U$ of $U_1, \ldots, U_n$ belonging to $\mathcal{U}$. I claim $U = X$. First, $U \subseteq X$ because by definition of open cover of $X$, an element $U \in \mathcal{U}$ is automatically a subset of $X$. But also $X \subseteq U$, because $U$ is an upper bound of the $U_1, \ldots, U_n$, and the union $X$ which is the least upper bound is contained in the upper bound. So $U = X$ belongs to $\mathcal{U}$, and thus we have verified that condition D holds.
Now suppose that condition D holds for $X$. We want to show that $X$ is compact. So, let $\mathcal{U}$ be any open cover of $X$; we want to show that under condition D, there exists finitely many $U_1, \ldots, U_n \in \mathcal{U}$ whose union is $X$. Of course we can’t apply condition D directly to $\mathcal{U}$ because $\mathcal{U}$, regarded as being partially ordered by subset inclusion, might not be directed. But if we introduce a new open cover $\mathcal{U}'$ whose elements are precisely the possible finite unions of elements of $\mathcal{U}$, then $\mathcal{U}'$ is directed. (Of course this should be proven. So, suppose given elements $V_1, \ldots, V_n$ of $\mathcal{U}'$. According to how $\mathcal{U}'$ was defined, each $V_i$ is a finite union $U_{i, 1} \cup \ldots \cup U_{i, k_i}$ of elements of $\mathcal{U}$. But then $V = V_1 \cup \ldots \cup V_n = \bigcup_{i=1}^n \bigcup_{j=1}^{k_i} U_{i, j}$, being a union of finitely many elements of $\mathcal{U}$, belongs to $\mathcal{U}'$. This $V$ is an upper bound of $V_1, \ldots, V_n$.) Since $\mathcal{U}'$ is directed and since condition D holds, we have that $X \in \mathcal{U}'$. But then by definition of $\mathcal{U}'$, the set $X$ is a union of finitely many elements $U_1, \ldots, U_n$ of $\mathcal{U}$. This completes the proof.
By the way, I don’t mean to sound overly harsh in my last comment. In view of the way the article direction is written, it’s not necessarily wrong to speak of “defining a direction” (on a set) – it’s just that in this context, it sounds really weird and overly elaborate to write that way. It would be more appropriate to write that way if one started with a set $S$ and was contemplating, among the infinitely many ways in which $S$ could carry a structure of directed partial order, which one of them one wishes to single out as the topic of discussion. But in the present discussion, there’s really no choice in the matter: the only relevant partial ordering is given by subset inclusion (or the given partial order if one is starting with a frame or subsets of a frame), and it’s only a question of whether the poset under discussion is directed or not.
So much so that if a reader encounters the phrase “define a direction by” in the article compact space, it would be natural to wonder if the author was confused or didn’t quite understand what they were talking about.
I don’t know if this is what tphyahoo has in mind, but in constructive mathematics one might want the “directedness” of a directed poset to be given by a function assigning a particular upper bound to any two elements, since in the absence of choice having such a thing is a stronger assumption than the mere existence of upper bounds. However, in a join-semilattice there is always a function that selects the least upper bound, since those are uniquely defined.
Right, that would be a more structural (actually, algebraic over posets) notion of direction where the phrase “define a direction” would make sense. It seems that direction (written largely by Toby, I think) was not written with this possibility in mind – and offhand, it strikes me as a superfluous consideration for the compact space article.
Yes, the distinction isn’t relevant for compactness.
1 to 86 of 86