If is the maximal size of a tower, is $\latex \mu$ also the maximal size of a family such that has no pseudo-intersection, and such that is injective. (where denote the set of all f-lower bound ("pseudo intersections") of a family )

Such an necessary has the fip. Note that we can obtain such a family from any fip family without pseudo intersection by considering the family of the members of that satisfy

if this could be "kind of" an indication to the way to build a tower from any with fip and npi.

——————————————————————————————-

For example by choosing $ latex C_i\in i_f(\left\{F_i,\, i<j\right\}$ for all and compose it with th operation that is defined here https://gowers.wordpress.com/2017/09/19/two-infinities-that-are-surprisingly-equal/#comment-203525

The main difference is instead of composing the $F_i$ we compose the that belong to members of a decreasing family of co-filters* with empty (usual) intersection., so that we can hope in a more realistic way that the tower we get has no pseudo -intersction (npi)

I will develop in a MSE post some ideas I've got about the way to hope to make it work or at least exhibe a reason why it's not easy by reducing the aim to special cases to discuss : one idea could be to use the preorder that is called in the paragraph "motivation" of this link on MSE : https://math.stackexchange.com/questions/2577440/decreasing-family-of-families-with-lower-bound-zero (I already wrote this in upper posts but note that $ latex A*B<_b A$ and also )

(*the set of complements of is a filter for any family $X$, so I qualify a co-filter, but I don't know if it is the right terminology)

]]>Volume 4, Issue 2, 2017, pp. 219–320DOI: 10.4171/EMSS/4-2-3

Published online: 2017-11-13

Numerical infinities and infinitesimals: Methodology, applications, and repercussions on two Hilbert problems

Yaroslav D. Sergeyev

How this could have made it through, it totally beyond me (for it has really 0 content, and the author is well-known in a bad way…).

]]>It also seems that (in the proof) we don’t need the hypothesis that if (it also comes from a previous attempt that was maybe correct but a bit more uselessly complicated, anyway it doesn’t make the proof falser or truer, it’s just useless)

Sorry for all these notations failure… I hope that everything is correct up to the notations/redondances edit…

]]>For any , and any , let's say that .

The "stronger statement" is then :

For any fip-family and any such that is a tower and such that there exists and a tower such that .

AND SUCH THAT

(I forget this that is crucial in the proof)

]]>We just have to prove a statement a bit stronger (that seems true in any lattice)

For any , and any , let's say that .

The "stronger statement" is then :

For any fip-family and any such that is a tower and such that there exists and a tower such that .

We can say that " extend according to "

To prove this we consider minimal such that it is not true, and we say without lost of generality that for any . Then if is successor, we have and then

as is minimal, we have a tower such that extend according to and we take such that if and if not. And that's a contradiction.

If type is limite

we have for any such that , a tower such that and and we built like this for any .

We have for any , so we can define such that for any , it is clear that and that so it is a contradiction. si not limite, and not successor, and it is obviously not so the "stronger statement " is true.

The result is true for all with the fip and all $T$ so it is still true if and , and that seems to be what we wanted to prove.

]]>Take if and if not, where has no f-intersection. ]]>

if there exists such that :

for all

(the decreasing of is quicker than that of , that's why the inequality is this way, that is also why I used the "greater or equal " instead of switch the places of A and B, if the bug doesn't come in this very comment, this will show that it does'nt react to "lt" the same way tan to "gt". *)

*by the way… is it " a > b" or " a \> b " that works ?

Well, I'm going to test it, so that you don't loose time by answering that kind of boring details :

test

(with the barre)

(without the barre)

There is kind of an “inverse problem” by asking who are the partial orders such that .

I say again the definition of :

for all integer, where is the usual order on and $A^*$ is the increasing injection from to $A=A^*(\mathbb{N})$

(the previous definition took $a\in \mathbb{Z}$ so there is some definition problem for the fonction if )

Let's also say again the two interesting fact that we have with

1) it is extending and compatible with (where )

2) and

So that 2) get rid of the problem we had before with that is not true… we pay this nice fact by dealing with instead of but there may be hope to get back to with new statements that are interesting independently from $p=t$.

One of this statement could be the fact that $p_{fct}=p$ that I spoke about at the beginning of the post but there is another one (that might be very wrong, but we need a weaker version)

monotone no-intersection statement :

Suppose that is a collection of families such that , for all and suppose that no $F^k$ have a non-empty f-intersection, then have no f-intersection as well.

This statement, if true, would help us in the particular case of and that $F=F_0$ has the property that any ending segment has no-intersection ( $\left\{F^0_i,\, j<i<p\right\}$ has no f-intersection for all )

There is a construction mixing the usual argument in the main comment, the "more and more tower-like" idea in the next main comment, and the $*$ argument that failed because of not being true :

If is successor we say if , and .if not

If is limit, we take any f-intersection of and we say that if , and if we take .

We can imagine other construction, but if the statement is true, and if we can suppose that we can assume the existence** of the particular case of a family with cardinality where the ending segment have no-intersection (I think this** is not difficult) then is a tower with no-intersection.

So if either monotone no-intersection statement (and **) is true either is true then we have what we want.

]]>(1) has a non empty intersection for all .

(2) has no intersection, for all $k<c$

Let's say that a general family (necessarily with fip and npi)

that satisfy (1) and (2) is a quasi-tower, and let's call the smallest cardinality that can have such a family.

We obviously have .

What seems to be the difficult part in order to prove that

Is it or is it ? Are they both difficult?

If we have one of the two results, we can get rid of conditions that could be difficult to handle with. never the less, there is nothing to lose by having one of the two results…

]]>if there exists such that is greater integer than for all .

Sorry for these notation mistakes! (I'm going to continue these attempt and ideas on mathOverflow, with the same surname, then I will post it here, when I'll be sure that it is correct and relevant!)

I hope I'll get something nice to bring and then bring kind of a compensation for all these inaccuracies, in contrast with the smart behavior and of the high quality of other interventions !

]]>Let’s take a generalization of the problem and figure out when it is simple the cases that we can solve.

Let be a quasi-ordered relation on , such that is thinner than .

For any , we write , or if and . We also write for any , .

We say that has no o-intersection if that it has the o-intersection property (o-ip) if for any finite set , we have not equal to . We say that is o-saturated if it has the oip and no o-intersection. And we say that is a o-total if for any we have or .Let's call resp. ) the smallest ordinal such that there exists a o-saturated set ( resp. o-saturated and o-total) indexed by ordinals lower than .

We can ask our self whether …

We can say that and , and we hope to prove that they are equal.

We can also obviously notice that .

We can also state that the equality holds, for any such that , any has the x-ip. Because we can use the argument of the main comment which is precisely : https://gowers.wordpress.com/2017/09/19/two-infinities-that-are-surprisingly-equal/#comment-203713.

Let's then give an example of such a quasi order. Still calling the increasing injection from , we'll say that if there exists such that for all .

We can see that and , We can

call this (***) then make two statement :

The first one is that $t_{fct}=p_{fct}$.

(because (***) leads that any has the fct.ip. and we just said that the equality holds in that case)

the second statement is that $t_{fct}$ is not smaller than $t=t_f$. Indeed we can then use the argument here : https://gowers.wordpress.com/2017/09/19/two-infinities-that-are-surprisingly-equal/#comment-203534

(***) assure us that the "composition tower" has no fct-intersection witch is stronger then having no f-intersection. For the same reason of strength, we have the obvious opposite inclusion and we then have .

We can also build an intermediate strongly quasi-order that we can call , (for "translation") : if and only if there exists such that . We can see easily that :

(by choosing fine representative element of a tr-tower : = total family indexed by ordinals that extend )

We then have :

lower or equal to lower or equal to (+++)

This make the hope of picking a fct-tower (then a f-tower with the -argument) into a little bit more concrete than before these obervations… because we can suppose that is only when , and is a limit ordinal…

(+++) also show that

is equivelent to

Witch is a translation of the problem in a context of classes closed "up-to translation" (like fct and tr) – note that the composition is compatible with both tr-equivalence and fct-equivalence)

It might also be interesting to get other quasi order, by making an eventually non-empty f-intersection empty by some quotient, and consider a digressing family of such quasi order…

It could also be instructive to build , quasi-order where .

]]>Let be a quasi-ordered relation on , such that is thinner than .

For any , we write , or if and . We also write for any , .

We say that has no -intersection if that it has the $o$-intersection property (oip) if for any finite set , we have not equal to . We say that is o-saturated if it has the oip and no o-intersection. And we say that is a o-total if for any we have or .Let's call resp. ) the smallest ordinal such that there exists a o-saturated set ( resp. o-saturated and o-total) indexed by ordinals lower than .

We can ask our self whether …

We can say that and , and we hope to prove that they are equal. We can state that . And we can also state that the equality holds, for any such that , any has the xip. Because we can use successfully the argument of the comment of the main comment.

Let's now give an example of such a quasi order. Still calling the increasing injection from $\mathbb{N}\to A$, we'll say that if there exists such that for all .

We can see that and , We can

call this (***) then make two statement :

The first one is that $t_{fct}=p_{fct}$. (because (***) leads that any has the fct.ip. and we just said that the equality holds in that case)

the second statement is that $t_{fct}$ is not smaller than $t=t_f$. Indeed we can then use the argument here : https://gowers.wordpress.com/2017/09/19/two-infinities-that-are-surprisingly-equal/#comment-203684

(***) assure us that the "composition tower" has no fct-intersection witch is stronger then having no f-intersection. For the same reason of strenght, we have the obvious opposite inclusion and we then have .

We can also build an intermediate strongly quasi-order that we can call , for translation : if and only if there exists such that . We can see easily that (by choosing fine representative element of a tr-tower : = total family indexed by ordinals that extend )

We then have :

lower or equal to lower or equal to $p_{fct}=t_{fct}=t_{tr}=t$

So, the problem have a "translations 's translation" (joke with both meaning of "translation")

We have

Note that this a nice thing about this equivalent condition, is that the composition is compatible with "translations" (the tr-equivalence) and also the fct-equivalence.

It might also be interesting to get other quasi order, by making an eventually non-empty f-intersection empty by some quotient, and consider a digressing family of such quasi order…

It could also be instructive to build quasi-order where does not hold.

]]>(where such that is the only increasing injection from to $A$)

I thought this tower has no intersection because , but it is not so clear… anyway I post this comment not only to correct the previous one but to give a possible idea quickly before the correct (basic) things that might be in the previous post get spoiled by its wrong end.

]]>Indeed if it is not the case, we consider instead of .

It seems te me (but I don't feel so sure about it) that we can also assume that , and then we have tower (with no intersection), because is clearly in every with … ]]>

I call down-f-set is a usual set of f-sets such that if is one of its element, and if then is also one of its element.Let’s call the set of all down-f-sets

let be such that (condition 1)

Let’s now consider , the smallest topology such that elements of are closed set. As usual we extend the inclusion order in to a total order $<<$. Now we consider for all the set and : the adherence of the union of elements in . Let's make the supposition that if is a finite set, then is not the set of all down -f-sets.(condition 2). Condition 1) tells us that is an infinite inclusion ordered set such that the . And we also have lower or equal to .

Now we are going to choose in each a f-set $choice(\mathcal{H}$ that does not belong to any such that .

But as soon as and are down-f-set, we will have , and is a tower!

Let's get back to: with no f-intersection and the f.i.p.

We now remark that if we take , where is the set of f-sets f- included in the f-complement of , then we can conclude.

I'm going to turn everything into complement point of view, it will be more direct…

]]>Actually, Dordal shows that for a ‘fairly closed’ set of regular cardinals and some regular , one can force that there is a tower of order type iff , and the continuum is . E.g. can be any finite set of regular cardinals and the continuum anything regular above .

This is in quite the contrast with other families we study (e.g. almost disjoint, independent) where maximal witnesses of size continuum always exist.

]]>Let me try to develop the idea of the extended total order in this context :let be a set of downs-f-set that have the usual-set intersection property. Let be the support of , meaning the set of all the down f sets that are include in some . It is an ordered set for usual inclusion. We then extend the usual inclusion in a total order $<<$. What is nice here is that is a down-f-set for all . The set is now a chain of usual set of f-sets, and the set is a chain of down-f-sets that has an empty (usual) intersection. We can pick a tower with no pseudo intersection inside.We take for our original with no pseudo-intersection and the fip and it seems done…

I don't see the mistake right now, but I'm sure I will see it just after posting!

line 16 (not

]]>