The following question is motivated by the recent breakthrough result by Justin Gilmer on the union-closed sets (aka Frankl) conjecture.

Let $\mathcal{F}\subseteq\mathcal{P}(\mathbb{N})$ be a finite, union-closed family, i.e. $A,B\in\mathcal{F}\Rightarrow A\cup B\in\mathcal{F}$. Let $A_{\mathcal{F}}$ and $B_{\mathcal{F}}$ be two independent random variables, uniformly distributed over the elements of $\mathcal{F}$. Denoting by $H$ the entropy, we clearly have $$H(A_{\mathcal{F}}\cup B_{\mathcal{F}})\leq H(A_{\mathcal{F}})=\ln|\mathcal{F}|,$$ since the entropy is maximized by the uniform distribution. I am wondering whether a sharper bound of the form $H(A_{\mathcal{F}}\cup B_{\mathcal{F}})\leq \lambda H(A_{\mathcal{F}})$, for some $\lambda<1$, still continue to hold. The intuition is that the distribution $A_{\mathcal{F}}\cup B_{\mathcal{F}}$ should "deviate" enough from the uniform bound in order to get a non-trivial upper bound on its entropy. Denoting by $\mathrm{UC}$ the collection of all finite, union-closed families $\mathcal{F}\subseteq\mathcal{P}(\mathbb{N})$ (with $|\mathcal{F}|>1$), we then consider the following quantity:

$$\lambda:=\sup_{\mathcal{F}\in UC}\frac{H(A_{\mathcal{F}}\cup B_{\mathcal{F}})}{H(A_{\mathcal{F}})}=\sup_{\mathcal{F}\in UC}\frac{H(A_{\mathcal{F}}\cup B_{\mathcal{F}})}{\ln|\mathcal{F}|}.$$

We have $\lambda\leq 1$, and by considering the union-closed families $\mathcal{P}[n]\setminus\{\emptyset\}$, we get the lower bound $\lambda\geq 0.82$.

Is $\lambda=1$ or $\lambda<1$? In the latter case, it is possible to provide an explicit, non-trivial upper bound?


1 Answer 1


The best upper bound is $\lambda=1$. Here is a simple family of examples: Let $\mathcal F_n$ be $\Big\{\{1,\ldots,i\}\colon 1\le i\le n\}\Big\}$. That is $\mathcal F_n$ is the collection of all initial segments of $\{1,\ldots,n\}$, which is clearly union-closed. Clearly $|\mathcal F_n|=n$, so that $H(A)=\log n$. Also if $A:=[1,i]$ and $B:=[1,j]$ are random elements of $\mathcal F_n$, then $[1,i]\cup[1,j]=[1,\max(i,j)]$ so that $\mathcal P(A\cup B)=\{1,\ldots,k\}=(2k-1)/n^2$. We now estimate $H(A\cup B)$. This is given by $$ H(A\cup B)=-\sum_{j=1}^n \frac{2j-1}{n^2}\log \frac{2j-1}{n^2}. $$ We will approximate this by a Riemann sum to show that for large $n$, $H(A\cup B)\approx H(A)-\log 2+\frac 12$. To see this, we have $$ H(A\cup B)=\sum_{j=1}^n\frac{2j-1}{n^2}\log n-\frac 1n \sum_{j=1}^n \frac{2j-1}n\log\frac{2j-1}n. $$ Hence $$ H(A\cup B)\approx \log n - \int_{0}^1 2x\log(2x)\,dx, $$ giving $H(A\cup B)\approx\log n-\log 2+\frac 12$ as claimed.

  • $\begingroup$ Thank you! In your example there is an element ($i=1$) which appears in every member of the family. Do you think the situation changes if we restrict to the case when $Pr(i\in A_{\mathcal{F}})\leq \lambda$ for every $i$, for some $\lambda<1$? $\endgroup$ Commented Dec 26, 2022 at 23:28
  • $\begingroup$ Cheaply, we could have the sets $\{i\colon i<j\}$ where $j$ runs from 1 to $n$ so that there is no element contained in all of the sets. But this doesn't solve the question with the additional constraint (I assume that the $\lambda$ in the question (the proportional entropy drop) is supposed to be distinct from this $\lambda$: the probability that each element lies in a random element of $\mathcal F$. $\endgroup$ Commented Dec 27, 2022 at 1:18
  • $\begingroup$ Yes sorry, they are not the same $\lambda$. I suspect indeed that with the extra constraint $Pr(i\in A_{F})\leq u<1\,\forall i$ , one could get $\lambda=\lambda(u)<1$. This would lead to an improvement on the best bound for the Frankl conjecture, so either it's false or it's probably an hard task. $\endgroup$ Commented Dec 27, 2022 at 1:33
  • $\begingroup$ I'm likely wrong (not my area), but didn't Gilmer (arxiv.org/abs/2211.09055) just give a constant lower bound (0.01 I think) for Frankl's conjecture? $\endgroup$ Commented Jan 21, 2023 at 2:49
  • $\begingroup$ Oops, I now see that you referenced Gilmer's paper in your post, so obviously you didn't need me to mention it. Sorry for the poor reading comprehension! $\endgroup$ Commented Jan 21, 2023 at 3:42

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