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If I'm not wrong, it is easy to prove the following statements :

1° For every natural number $k$, every union-closed family that has at least $2$ members with cardinality $\geq k$ has at least $2$ members whose intersection has cardinality $\geq k$. This is trivial : if $X$and $Y$ are two distinct members of the family with cardinality $\geq k$, we can assume that $Y$ is not contained in $X$. Then $X$ and $X \cup Y$ are two distinct members of the family whose intersection is $X$ and has thus cardinality $\geq k$.

2° For every natural number $k$, every union-closed family that has at least $k+3$ members with cardinality $\geq k$ has at least $3$ members whose intersection has cardinality $\geq k$.

3° For every natural number $k$, every union-closed family that has at least $k+4$ members with cardinality $\geq k$ has at least $4$ members whose intersection has cardinality $\geq k$.

I think I have also a proof of the following statement :

4° For every natural number $k$, every union-closed family that has at least $\frac{k^{2}+5k}{2} + 5$ members with cardinality $\geq k$ has at least $5$ members whose intersection has cardinality $\geq k$.

My question is : are there results such as : "If a union-closed family of sets has at least this number of members with cardinalty $\geq k$, then the family has at least $c$ members whose intersection has cardinality $\geq k$".

(Edit :) For every natural number $c \geq 1$ and for every natural number $k \geq 0$, let $s(c,k)$ denote the least natural number $s$ with the following property : every union-closed family that has at least $s$ members with cardinality $\geq k$ has at least $c$ members whose intersection has cardinality $\geq k$. (I presume that such an $s$ always exists, so that $s(c,k)$ always exists.) Clearly, $s(1,k) = 1$ and the above statements amount to say that $s(2,k) \leq 2$, $s(3,k) \leq k + 3$, $s(4,k) \leq k + 4$, $s(5,k) \leq \frac{k^{2}+5k}{2} + 5$. In fact, it is easy to find examples proving that $s(2,k) = 2$, $s(3,k) = k + 3$, $s(4,k) = k + 4$, $s(5,k) = \frac{k^{2}+5k}{2} + 5$.

(In order to prove that $s(5,k) \geq \frac{k^{2}+5k}{2} + 5$, look at the family of all subsets of a set with cardinality $k+2$.)

Thus, it seems reasonable to conjecture that for every natural number $c \geq 1$, there is a polynomial $f_{c}(X)$ such that for every natural number $k$, $s(c,k) = f_{c}(k)$.

The first poynomials of the sequence are $f_{1}(X) = 1$, $f_{2}(X) = 2$, $f_{3}(X) = X+3$, $f_{4}(X) = X+4$, $f_{5}(X) = \frac{X^{2}+5X}{2} + 5$.

Is there a known sequence of polynomials beginning in this manner ?

Note : for every natural number $r \geq 0$, we must have

$s(1 + 2^{r}, k) \geq {k+r \choose k} + {k+r \choose k+1} + \ldots + {k+r \choose k+r} +1$.

(Look at the family of all subsets of a set with $k+r$ elements.)

In the limits of the above results, $s(1 + 2^{r}, k)$ is equal to the found minimal value, $s(c, k)$ grows unit by unit when $c$ runs from $c = 1 + 2^{r}$ to $c=2^{r+1}$ (included) and, for $c=1+2^{r+1}$, takes again the found minimal value (with $r+1$ instead of $r$).

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    $\begingroup$ I was going to mention the current Polymath project hosted on Gowers's blog, but I see from your previous question that you already know about it. You might find a higher concentration of people able to answer your question there. $\endgroup$ – Ben Barber Mar 11 '16 at 11:03
  • $\begingroup$ Thanks for this hint, but I think that I'm not at the level of the Polymath project. And the role of this project is not to answer questions. $\endgroup$ – Panurge Mar 11 '16 at 11:59
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I do not know of a reference for your version. I provide some motivation for recasting your question into lattice terms.

Along similar lines is the following lower bound argument:

Assume there is a counterexample to Frankl's conjecture. Thus there is a collection F which contains its common nonempty union X. Since X is nonempty, there is an element z which occurs in X meaning it occurs in at least one member of F. Thus F (being a counterexample) has at least 2*1+1=3 members.

Since F has at least 3 members, two of the members are nonempty, and either one of them contains the other, or their union contains both. Either way, we find an element z in at least two members of F. So F contains at least 5 members.

Now we have at least 4 nonempty members of F to use. We assume without loss a nonempty member A of F contained in a nonempty member B of F. If third member C intersects A or A union C is distinct from B, we have z in A also in two other members of F. Otherwise B is the disjoint union of A and C and now we use D distinct from C to get D or D union A a third set containing either z or else B, C and D having a nonempty intersection. This means F has at least 7 members.

One can pursue this line with increasing complexity up to F having at least 13 members (I might remember/resurrect the argument from old notes); I am essentially looking at (semi-)lattices and trying to prove an equivalent inequality regarding the number of join-irreducible elements in the lattice. You might find some benefit in recasting your problem in lattice-theoretic terms. Poonen has shown such an equivalence in his 1990 REU paper, and by now the surveys on Frankl's conjecture should include this version and some progress on it.

I will look for my notes on derivations and post some related ideas, e.g. if union-closed finite F has k-many 2 sets then F has at least as many (-2)sets (shorthand for $|X|-2$), or F is one of finite and small number of exceptions.

Gerhard "Is Hobbling Down Memory Lane" Paseman, 2016.03.31.

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  • $\begingroup$ One can streamline/sidestep the lower bound argument above with a similar result: If F is a reduced counterexample with union of n elements, then F has at least 4n-1 members. I remember browsing the library a year or so after I proved this and found it in an article of Giovanni Lo Faro, who had a mild extension of it. Gerhard "Didn't Strike A Cold Iron" Paseman, 2016.03.31. $\endgroup$ – Gerhard Paseman Mar 31 '16 at 17:11
  • $\begingroup$ Thanks for your interest in my question. I first formulated it with a conjecture, but I now think that this conjecture has no value. I find it curious that in the first results I mentioned ($c =2$, $3$ or $4$), we have statements as "If $k$ is a natural number, if a union-closed family of sets has at least $f(c)k + c$ members with cardinality at least $k$, then this family has $c$ members whose intersection has cardinality at least $k$", with $f(c)k$ a linear function of $k$, and not something like $2^{k}$. But I didn't study the case $c=5$. $\endgroup$ – Panurge Mar 31 '16 at 17:51

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