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Here is a short proof that at most $2n-2$ swaps are necessary. We proceed by induction on $n$. For the base case $n=1$, it is clear that no swaps are necessary. For the inductive step, let $X_1,X_2 …

This is false in general. Let $X=\{1, \dots, n+1\}$, $Y=\{n+1, \dots, 2n+1\}$, and $Z$ be any $(n+1)$-subset of $[2n+1]$ not containing $n+1$. Let $\mathcal{F}=\mathcal{A} \setminus \{X,Y,Z\}$. The …

Here is a counterexample for $n=5$. Partition the non-empty subsets of $\{1, \dots, 5\}$ into the singleton subsets and a sixth family containing all the other non-empty subsets. So, $m=6$ and $|\ma …

This follows from Proposition 2.1 of the paper Many $T$ copies in $H$-free graphs by Alon and Shikhelman. Theorem (Alon and Shikelman) Let $T$ be a fixed graph with $t$ vertices. Then $ex(n,T,H)=\Ome …

Here are some minor remarks. Since the actual numbers do not matter, the question can be rephrased as follows. Let $\sigma: [n] \to \{-, +\}$. Say that a $k$-subset of $[n]$ is $\sigma$-positive if …

As observed by Geva Yashfe, the answer is $2^n$. This can be achieved when each of $A$ and $\overline{A}:=E\setminus A$ are bases, with $A = \{a_1,\ldots,a_n\}$, $\overline{A} = \{b_1,\ldots,b_n\}$, a …

This question initially arose out of a question in asymptotic matroid theory. The matroid question has since been answered in a different way, but the extremal set theory question remains unanswered …

The form of the Frankl-Wilson that I know is the following. Frankl-Wilson Theorem. Let $q$ be a prime power and let $\mathcal{F}$ be a family of $k$-sets of an $n$-set such that $|A \cap B| \equiv …

The same statement in fact holds for all $n$. Theorem. Let $\mathcal{F}$ be a separating, union-closed family of non-empty sets on the ground set $[n]$. Then $|\mathcal{F}| \geq n$, and if $|\math …

Edit. My previous upper bound was not correct. Thanks to Gilad for pointing that out. If $m<k$, then of course it is not possible. Otherwise, for an upper bound start with a matching $M$ saturating …

Yes, it is always possible to find regular triangle-free graphs of any degree up to half the number of vertices (as long as the number of vertices is even). To see this, by Hall's Theorem the edges o …

For a crude lower bound, one can consider the largest possible size for a set in your partition. One candidate is to take a collection of 2-subsets of $[2n]$,which are triangle-free and then add the …

The answer is $2^n -n$. Let $I$ be the $n \times n$ identity matrix and let $D$ be a diagonal whose non-zero entries are indexed by $S$. Further suppose that $D$ does not have exactly one zero entry …