I am trying to understand what the author means by $\mathcal{F}^N$ on the bottom of page 3 of the paper "Entropy approach for a generalization of Frankl's conjecture" by Veronica Phan, and why this sentence is true: "Another strategy is note that the union-closed conjecture is true for $\mathcal{F}$ if it is true for $\mathcal{F}^N$ for some $N$". If it is the cartesian product $\mathcal{F} \times \cdots \times \mathcal{F}$ then I don't understand what is the union-closed sets conjecture for that cartesian product. Maybe it is quite simple but I cannot get it. Someone can help?
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$\begingroup$ What goes wrong with $\mathcal{F}^N = \mathcal{F} \times \cdots \times \mathcal{F}$? $\endgroup$Sam Hopkins– Sam Hopkins ♦2025-03-31 09:23:30 +00:00Commented Mar 31 at 9:23
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$\begingroup$ @SamHopkins so I suppose that in that case the operation between $(A_1, \ldots, A_N)$ and $(B_1, \ldots, B_N)$ under which $\mathcal{F}^N$ is closed is $(A_1 \cup B_1, \ldots, A_N \cup B_N)$, right? $\endgroup$Fabius Wiesner– Fabius Wiesner2025-03-31 09:43:12 +00:00Commented Mar 31 at 9:43
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$\begingroup$ And the union-closed sets conjecture requirement for $\mathcal{F}^N$ is that there is an element belonging to at least half of the elements of the tuples in $\mathcal{F}^N$ with a fixed index $i$, $1 \le i \le N$? $\endgroup$Fabius Wiesner– Fabius Wiesner2025-03-31 09:54:14 +00:00Commented Mar 31 at 9:54
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1$\begingroup$ I would view $\mathcal F^n$ as a collection of subsets of the disjoint union of $n$ copies of the largest set in $\mathcal F$. This agrees with what you say. $\endgroup$Will Sawin– Will Sawin2025-03-31 12:36:46 +00:00Commented Mar 31 at 12:36
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$\begingroup$ @WillSawin could you give an example e.g. for $\mathcal{F}^3$ or $\mathcal{F}^2$ when $\mathcal{F} =\{\emptyset, \{1\}, \{1,2\}\}$? $\endgroup$Fabius Wiesner– Fabius Wiesner2025-03-31 13:09:37 +00:00Commented Mar 31 at 13:09
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1 Answer
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For $S$ a set and $\mathcal P(S)$ the power set, we have a natural bijection $\mathcal P(S)^N = \mathcal P(S \times \{1,\dots,N\})$ sending a tuple $A_1,\dots, A_N$ of sets to the set $A_1 \times \{1\} \cup A_2 \times \{2\} +\cup \dots \cup A_N \times \{N\}$.
For $\mathcal F$ a subset of $\mathcal P(S)$, $\mathcal F^N$ is naturally a subset of $\mathcal P(S)^N$ and thus, under this bijection, a subset of $\mathcal P(S \times \{1,\dots,N\})$.
For example for $\mathcal F = \{\emptyset, \{x\} , \{x,y\}\}$ we have
$$\mathcal F^2 = \{( \emptyset,\emptyset), (\{x\},\emptyset), (\{x,y\},\emptyset),( \emptyset,\{x\}), (\{x\},\{x\}), (\{x,y\},\{x\}),( \emptyset,\{x,y\}), (\{x\},\{x,y\}), (\{x,y\},\{x,y\})\} $$ sent to
$$\{ \emptyset, \{ (x,1)\}, \{ (x,1), (y,1)\}, \{(x,2)\}, \{(x,1),(x,2)\}, \{(x,1), (y,1), (x,2)\} , \{ (x,2), (y,2)\}, \{(x,1),(x,2),(y,2)\}, \{(x,1), (y,1),(x,2),(y,2)\} \}$$
It is straightforward to check from this that the union operation is given by the formula you guess and it is straightforward to check from that that if $\mathcal F$ is union-closed that $\mathcal F^N$ is union-closed.
It is equally straightforward to check that one element of $S \times \{1,\dots,N\}$ is in at least half the elements of $\mathcal F^N$ if and only if the criterion you guessed is satisfied and easy to check that this implies one element of $S$ is in at least half the elements of $\mathcal F$.
So indeed the union-closed conjecture for $\mathcal F^N$ implies the union-closed conjecture for $\mathcal F$.
