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 $r$).