“Let’s use this idea to prove that there do exist towers with empty f-intersections. I shall build a collection of non-empty f-sets [A_\alpha] by transfinite induction. If I have already built [A_\alpha], I let [A_{\alpha+1}] be any non-empty f-set that is strictly f-contained in [A_\alpha]. That tells me how to build my sets at successor ordinals. If \alpha is a limit ordinal, then I’ll take A_\alpha to be a non-empty f-intersection of all the [A_\beta] with \beta<\alpha.

But how am I so sure that such an f-intersection exists? I’m not, but if it doesn’t exist, then I’m very happy, as that means that the f-sets [A_\beta] with \beta<\alpha form a tower with empty f-intersection.

Since all the f-sets in this tower are distinct, the process has to terminate at some point, and that implies that a tower with empty f-intersection must exist."

Instead of building one tower let's build a couple of tower, one with $f$-sets and one with representantive element. So we have built our couple $([A_{\alpha}],A_{\alpha})$ exactly the same way, but we just ask that the normal set tower is strictly increasing (it should'nt be a problem, wether $\alpha$ is a limit ordinal or not)

But if the normal set tower is strictly decreasing, the process has to stop at most when $\alpha$ is the first uncontable ordinal. I must be doing a confusion because it seems to give the result pretty easily… I steel post it, and I hope that, if it's wrong or stupid, then it's not stupid in such a way that I would feel ridiculous.

So the problem is, once again, what to do about limit ordinals. Suppose that is a limit ordinal and that for each with I have constructed a set system with the finite intersection property and no f-intersection. I now need to construct a limiting set system somehow. This will be fine if I can ensure that no set from has been changed infinitely many times. But it’s not obvious that that will happen.

Thinking about this further, it seems to be running into the same difficulties.

]]>For example we can build for some , It seems that we are doing “nothing more” here, but maybe a similar idea would work…

]]>) ]]>

An other way to solve the first problem could be to take $G’:= B*A*B\subset A\cap B$.

Am I too optimistic?

As with jean-camille’s proof attempt, we begin by arbitrarily well-ordering . Now let us attempt to build a tower inductively. However, this time, instead of trying to use sets from , we will make the following weaker promise: that for every set in there will be a set in that is f-contained in it. This will ensure that any f-intersection of is also an f-intersection of , since is a tower.

How might this inductive construction go? Let the enumeration of the f-sets in be , where ranges over the ordinals belonging to the order type of the well-ordering of . I now want to built a tower . I’ll actually make the more specific promise that will always be f-contained in .

To start with it’s fine: we take , and so on. All these are infinite, by the finite intersection property of . I don’t actually care whether they are distinct: if they aren’t, that makes my tower smaller, and that’s only good for me.

But then we need to create . There isn’t an obvious way to do that, so what I’ll do instead is find an f-intersection Y_\omega’$ of the sets . (If I can’t find one, then I’m done, since it means I’ve got a countable tower with no f-intersection. Of course, that can’t exist, so this case won’t in fact arise until the tower is of size at least **p**.) Having found , I will let .

But it’s not clear that that works. In fact, in general there is no reason at all for it to work: I might accidentally choose an f-intersection that is disjoint from .

I think what I need in order to let the proof continue is something that feels hard to achieve: a set with the following two properties.

(i) is an f-intersection of the sets .

(ii) The set still has the finite-intersection property.

I don’t see why such a set has to exist, but it also appears to be a necessary condition for this approach to work.

]]>Also, I think limit ordinals do create problems. Suppose, for example, that you have created . How do you propose to create using the set ?

Maybe there is an idea for doing this actually. At each limit ordinal you do what I suggested and take an f-intersection. And at each successor ordinal you don’t take but rather you take *the first that you have not yet used*. In that way it will indeed be guaranteed that every set in takes part in one of the convolutions. But the first problem remains, I think.

There is two reason why I thought “I’ve got it”

1) First one is that I thought that for each we have such that, by construction, . Thus, if some is f-contained in every , it seems to give a contradiction.

Isn’t it possible to take every member of into the *-machine? I mean If the “no non empty pseudo-intersection” is not already satisfied? (please forgive the fuzzy statement here! I never did any transfinite proof, and I naîvely thought it shouldn’t be a problem.)

2) Second and (seriously wrong) reason : I translated the problem in terme of usual sets, and made the self-statement that we don’t need a (usual set) tower which is not constant by some rank, to have an empty intersection , as soon as we can remove the whole intersection from each member of the tower. .But my usual-set translation is certainly very wrong, because I could deduce from it that is …countable**!

**Actually I’m not confortable with this example of “Eratosthene” tower, that seems to have the fip and the “no non empty f-intersection” property, and is clearly countable :

, where is the -th prime number.

I cannot imagine any proper f-set that is f-contained in every anyway!

So… I’m afraid that I do a very serious confusion by now… but wait a minute, I think the set of prime numbers does the job! I feel better!

Given two sets and , with their elements written in increasing order, we are defining to be the set . Note that this is a subset of .

Now we take a collection of sets that has the finite intersection property and no pseudointersection. We well-order it somehow, and write for the well-ordering. And now we want to create a tower of cardinality at most that of and argue that it too has no pseudointersection.

For each ordinal that is smaller than the order type of , let be the corresponding set in . We shall inductively define sets as follows. (They are to form the tower.)

We start by setting to be . If we have already defined , then . (I think you meant in your comment above, but that’s a small detail.)

What I don’t see in your comment is a definition of when is a limit ordinal. But here’s an idea. Suppose that is a limit ordinal and the collection of sets has a pseudointersection. Then let be such a pseudointersection. If the collection of sets does not have a pseudointersection, then we are done.

The final step in this argument would need to be to prove that the tower created does not have a pseudointersection. I’m not sure I see why this has to be the case.

]]>Let’s define for any , the only increasing function from to . And let’s define the convolution of and as . We now build a transfinite sequence of convolutions between repesentative elements in (we take well-ordered by , and we build whose elements are indexed by , in this way : , where is the smaller element of the set of – majorant of , and then should be our tower… ) ]]>

I think the general idea of my first post probably doesn’t work. Here’s why, according to me :

Let’s have with . Then any sequence would be such as is a finite set. If it was infinite we would be able to chose for all such as with infinite, we would then have also infinite, and would be a countable tower, and this is not possible, as you said it in the article.

So the general first idea may go nowhere, as soon as this leads to the fact that the that I built in the first post is a finite set. But maybe the general following next idea can help, somehow… : it’s not an proper advance, but just a remark : I think we can build a topological space, with same -cardinality than

(let’s say , and have the same f-cardinality if and have the same cardinality).

Indeed, if is a set of representative elements of , where is closed under finite intersection, then , the general union closure of is a topological space. Of course we cannot say (yet?) that and have the same cardinality, but I think , that we can deduce from what I just said previously in this post, that and have the same f-cardinality.( I hope this is not a irrelevant triviallity, or even some wrong statement!)

]]>I’m in the latter camp, though I don’t take it to extremes. For example, I completely understand that in algebraic topology, if you develop the basics of homology theory and use those to prove results like the Brouwer fixed-point theorem and the Borsuk-Ulam theorem, you are saving yourself from having to repeat a lot of steps that you need to provide if you give direct proofs of all the things you can do using homology. But quite a lot of my research has been aimed at removing machinery — for example, I have been involved in finding more direct (and quantitative) proofs of results that were obtained using ergodic theory, and I think that has led to genuine gains in understanding.

Incidentally, it’s a slight oversimplification to say that Malliaris and Shelah improved Szeméredi’s regularity lemma. What they did was to improve the bounds significantly if you add a rather strong hypothesis about the graph. My instinct for this result too is to try to find a more elementary proof. The reason is that I find myself asking, “What is this model theory actually doing that makes the proof come out at the end?” And for me, the ideal answer would be something like, “Here’s a direct proof. Now the model theory wraps up these steps into a convenient form and allows one to prove similar results without laboriously writing out similar steps for each one.” Though I suppose I’d be even happier with a direct proof that was clearer and more transparent than the model-theoretic one rather than being essentially the same argument written out at greater length.

]]>***I’m not triying to say that we shoud get rid of $f$-set formulation, but just that it would be nice if we study statement that are not “so much more easy” to say with $f$-set than without… the $f$-statement would then be an elegant translation of an equivalent usual-set statement. If we need $f$-formulation for more than elenace and nice notations, that we might give the “indication” that we are deviating from the elementary proof goal.

]]>If I understand notation $\mathcal F= C(\mathit F)$ for some choice fonction $C$ where $\mathit F$ is the Daniel Soukup equivalent special case. The “worse thing” that could happen in order to use the “general idea “(of the upper post) is that, for all $(F,G)\in \mathcal F^^2$, $F\cup G\in [\mathbb N]$. And the “best case” we can have, is that there is infinitly many distinct $[F\cup G]$ . That give me an idee, to considere not only $\mathcal F’$, the finite intersection closure of $\mathcal F$, but also $\mathcal F”$ the closure of $\mathcal F’$ under general union! We have $\bigcap \mathcal F”\in [\emptyset]$, and also $\mathcal F”$ has the intersection property (any two member have an infinite intersection). The only thing that we have to be sure, is that $[\mathcal F”]=\mathit F”:=\left\{[X], \, X\in \mathcal F”\right\}$ and $\mathit F$ have the same cardinality… But if it has not, we are in the “best case” ! So we can assume that $\mathcal F”$ is the set of the open sets of a topological space. We could then try to study the compacity of $\mathbb N\in \mathcal F”$. ]]>

http://www.ams.org/journals/tran/2014-366-03/S0002-9947-2013-05820-5/home.html

]]>We start with a set that has the finite-intersection property and no non-empty f-intersection. This has a partial order, which we extend to a total order . Then for each element we let denote the set of all such that . (This isn’t precisely your notation, but I think it should be OK. For me consists of f-sets — that is sets up to finite symmetric differences.)

If I understand you, chooses for each f-set some representative set. I was going to suggest that this was unnecessary, but I now see that you are going to take infinite unions, so maybe it makes a difference after all. It means that is equal to the union of all such that . That is, you take all f-sets that precede in the order and take the union of their representatives.

It’s clear that the sets form a tower. It’s not clear to me that if then and are distinct. In fact, it isn’t obvious to me that the sets couldn’t all be equal to .

But as you suggest, the general idea might have a chance, since we have a lot of freedom to choose the total order . A natural special case to look at is Daniel Soukup’s first example, where no set in is f-contained in any other set in . Then all total orderings on extend the initial trivial ordering, so basically we’d be looking for some ordering, and some choice of representatives, that worked.

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