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.