Trying to solve a problem, I fell on the following statement :


If $k$ and $r$ are natural numbers such that $r \leq k$, if a union closed family of sets ("union closed" means that the union of two sets from the family is always a member of the family) has at least ${k \choose r} + 1$ members with cardinality $r$, then this family has at least two members with cardinality $\geq k$.


I think I have a proof. Could anybody disprove this statement or indicate a mention of it in the literature ? Thanks in advance. 


**Edit:** Perhaps I should say why I'm interested in this question.
In a comment about a blogpost of Timothy Gowers,


<https://gowers.wordpress.com/2016/01/29/func1-strengthenings-variants-potential-counterexamples/>


Thomas Bloom noted that if F is a union-closed family with $m$ members, then Frankl's conjecture "trivially implies that, for every $k\leq \log\_2 m$, there exists some $k$-set which has abundancy at least $1/2^k$." (In other words, there is a $k$-set contained in at least $m/2^k$ members of F.)


He added: "Note that the other extremal case, when $k=\log\_2m$, is trivially true."


Now, if $k$ denotes the largest integer such that $2^{k} \leq m$ and if $k$ is not exactly $\log\_2m$, then the "other extremal case" is not so trivial (for me, in any case), but it results (without use of Frankl's conjecture) from the statement in my question.


**Edit:** Perhaps it is better that I prove that the above statement implies Thomas Brown's first step towards Frankl's conjecture.


Le $m > 0$ be a natural number, let $k$ denote the largest integer such that $2^{k} \leq m$. The case where $m = 2^{k}$ being trivial, assume that $m > 2^{k}$. Let $F$ be a union-closed family of sets with $m$ members. Thomas Brown's first step is the following statement :


There are two members of $F$ whose intersection has at least $k$ elements. (Since $F$ is union-closed, it amounts to say that $F$ has at least two members with cardinality $\geq k$. Indeed, if $X$ and $Y$ are two members with cardinality $\geq k$, we can assume that $Y$ is not a subset of $X$. Then $X$ and $X \cup Y$ are two distinct members of $F$ whose intersection has cardinality $\geq k$.)


Assume it is false. (Denying hypothesis.) Thus $F$ has at least $m-1$ members with cardinality $< k$. Since $m > 2^{k}$, there are at least $2^{k}$ members of $F$ with cardinality $\leq k-1$. For each $r \leq k-1$, let $n\_{r}$ denote the number of members of $F$ with cardinality $r$. Our last result expresses that $\sum\_{=0}^{k-1} n\_{r} \geq 2^{k}$. If for each $r$, we had $n\_{r} \leq {k \choose r}$, then we would have $\sum\_{r=0}^{k-1} {k \choose r} \geq 2^{k}$, i.e. $2^{k}-1 \geq 2^{k}$, which is false. Thus there is at least an $r \leq k-1$ such that the number $n\_{r}$ of members of $F$ with cardinality $r$ is $> {k \choose r}$. By Petrov's theorem (i.e. the first statement in this post, which Fedor Petrov proved in an elegant manner on this thread), this implies that $F$ has a least two members with cardinality $\geq k$. So the denying hypothesis contradicts itself.



Literature about a property of union closed families?
This is similar to your proof but without induction.


We prove that there are at least 3 such sets. For $r=k$ this is clear, so assume that $k>r$. Consider our $\binom{k}{r}+1$ $r$-sets. Call an element $v\in V$ appropriate if $v$ belongs to at most $\binom{k-1}{r-1}$ our sets. Then there exist at least $\binom{k}{r}+1-\binom{k-1}{r-1}=\binom{k-1}{r}+1$ our sets not containing $v$. Their union contains at least $k$ elements, and does not contain $v$. Now I claim that between any $k$ elements $x\_1,\dots,x\_k$ there exists an appropriate element $v$. Indeed, if not, then total number of pairs (our $r$-set $A$, $x\_i\in A$) is at least $k(\binom{k-1}{r-1}+1)>r (\binom{k}{r}+1)$, a contradiction. So, we may find appropriate element $v$, the union $U$ of our sets not containing $v$ has cardinality at least $k$. Thus there exists appropriate $u\in U$ and the union of our sets which do not contain $u$ is a third set after $V,U$.


I wonder whether bound 3 may be further improved (for some values of $k,r$, of course for $k=r$ it can not.)


