[Combinatorics]
Problem: Let $n$ be an integer greater than 1 and let $X$ be an $n$-element set. A non-empty collection of subsets $A_{1}, \ldots, A_{k}$ of $X$ is tight if the union $A_{1} \cup \cdots \cup A_{k}$ is a proper subset of $X$ and no element of $X$ lies in exactly one of the $A_{i}$ s. Find the largest cardinality of a collection of proper non-empty subsets of $X$, no non-empty subcollection of which is tight.



Note. A subset $A$ of $X$ is proper if $A \neq X$. The sets in a collection are assumed to be distinct. The whole collection is assumed to be a subcollection.

Solution: The required maximum is $2 n-2$. To describe a $(2 n-2)$-element collection satisfying the required conditions, write $X=\{1,2, \ldots, n\}$ and set $B_{k}=\{1,2, \ldots, k\}$, $k=1,2, \ldots, n-1$, and $B_{k}=\{k-n+2, k-n+3, \ldots, n\}, k=n, n+1, \ldots, 2 n-2$. To show that no subcollection of the $B_{k}$ is tight, consider a subcollection $\mathcal{C}$ whose union $U$ is a proper subset of $X$, let $m$ be an element in $X \backslash U$, and notice that $\mathcal{C}$ is a subcollection of $\left\{B_{1}, \ldots, B_{m-1}, B_{m+n-1}, \ldots, B_{2 n-2}\right\}$, since the other $B$ 's are precisely those containing $m$. If $U$ contains elements less than $m$, let $k$ be the greatest such and notice that $B_{k}$ is the only member of $\mathcal{C}$ containing $k$; and if $U$ contains elements greater than $m$, let $k$ be the least such and notice that $B_{k+n-2}$ is the only member of $\mathcal{C}$ containing $k$. Consequently, $\mathcal{C}$ is not tight.



We now proceed to show by induction on $n \geq 2$ that the cardinality of a collection of proper non-empty subsets of $X$, no subcollection of which is tight, does not exceed $2 n-2$. The base case $n=2$ is clear, so let $n>2$ and suppose, if possible, that $\mathcal{B}$ is a collection of $2 n-1$ proper non-empty subsets of $X$ containing no tight subcollection.



To begin, notice that $\mathcal{B}$ has an empty intersection: if the members of $\mathcal{B}$ shared an element $x$, then $\mathcal{B}^{\prime}=\{B \backslash\{x\}: B \in \mathcal{B}, B \neq\{x\}\}$ would be a collection of at least $2 n-2$ proper non-empty subsets of $X \backslash\{x\}$ containing no tight subcollection, and the induction hypothesis would be contradicted.



Now, for every $x$ in $X$, let $\mathcal{B}_{x}$ be the (non-empty) collection of all members of $\mathcal{B}$ not containing $x$. Since no subcollection of $\mathcal{B}$ is tight, $\mathcal{B}_{x}$ is not tight, and since the union of $\mathcal{B}_{x}$ does not contain $x$, some $x^{\prime}$ in $X$ is covered by a single member of $\mathcal{B}_{x}$. In other words, there is a single set in $\mathcal{B}$ covering $x^{\prime}$ but not $x$. In this case, draw an arrow from $x$ to $x^{\prime}$. Since there is at least one arrow from each $x$ in $X$, some of these arrows form a (minimal) cycle $x_{1} \rightarrow x_{2} \rightarrow \cdots \rightarrow x_{k} \rightarrow x_{k+1}=x_{1}$ for some suitable integer $k \geq 2$. Let $A_{i}$ be the unique member of $\mathcal{B}$ containing $x_{i+1}$ but not $x_{i}$, and let $X^{\prime}=\left\{x_{1}, x_{2}, \ldots, x_{k}\right\}$.



Remove $A_{1}, A_{2}, \ldots, A_{k}$ from $\mathcal{B}$ to obtain a collection $\mathcal{B}^{\prime}$ each member of which either contains or is disjoint from $X^{\prime}$ : for if a member $B$ of $\mathcal{B}^{\prime}$ contained some but not all elements of $X^{\prime}$, then $B$ should contain $x_{i+1}$ but not $x_{i}$ for some $i$, and $B=A_{i}$, a contradiction. This rules out the case $k=n$, for otherwise $\mathcal{B}=\left\{A_{1}, A_{2}, \ldots, A_{n}\right\}$, so $|\mathcal{B}|<2 n-1$.



To rule out the case $k<n$, consider an extra element $x^{*}$ outside $X$ and let



$$

\mathcal{B}^{*}=\left\{B: B \in \mathcal{B}^{\prime}, B \cap X^{\prime}=\varnothing\right\} \cup\left\{\left(B \backslash X^{\prime}\right) \cup\left\{x^{*}\right\}: B \in \mathcal{B}^{\prime}, X^{\prime} \subseteq B\right\}

$$



thus, in each member of $\mathcal{B}^{\prime}$ containing $X^{\prime}$, the latter is collapsed to $\operatorname{singleton} x^{*}$. Notice that $\mathcal{B}^{*}$ is a collection of proper non-empty subsets of $X^{*}=\left(X \backslash X^{\prime}\right) \cup\left\{x^{*}\right\}$, no subcollection of which is tight. By the induction hypothesis, $\left|\mathcal{B}^{\prime}\right|=\left|\mathcal{B}^{*}\right| \leq 2\left|X^{*}\right|-2=2(n-k)$, so $|\mathcal{B}| \leq 2(n-k)+k=$ $2 n-k<2 n-1$, a final contradiction.
Proceed again by induction on $n$ to show that the cardinality of a collection of proper non-empty subsets of $X$, no subcollection of which is tight, does not exceed $2 n-2$.



Consider any collection $\mathcal{B}$ of proper non-empty subsets of $X$ with no tight subcollection (we call such collection good). Assume that there exist $M, N \in \mathcal{B}$ such that $M \cup N$ is distinct from $M, N$, and $X$. In this case, we will show how to modify $\mathcal{B}$ so that it remains good, contains the same number of sets, but the total number of elements in the sets of $\mathcal{B}$ increases.







Consider a maximal (relative to set-theoretic inclusion) subcollection $\mathcal{C} \subseteq \mathcal{B}$ such that the set $C=\bigcup_{A \in \mathcal{C}} A$ is distinct from $X$ and from all members of $\mathcal{C}$. Notice here that the union of any subcollection $\mathcal{D} \subset \mathcal{B}$ cannot coincide with any $K \in \mathcal{B} \backslash \mathcal{D}$, otherwise $\{K\} \cup \mathcal{D}$ would be tight. Surely, $\mathcal{C}$ exists (since $\{M, N\}$ is an example of a collection satisfying the requirements on $\mathcal{C}$, except for maximality); moreover, $C \notin \mathcal{B}$ by the above remark.



Since $C \neq X$, there exists an $L \in \mathcal{C}$ and $x \in L$ such that $L$ is the unique set in $\mathcal{C}$ containing $x$. Now replace in $\mathcal{B}$ the set $L$ by $C$ in order to obtain a new collection $\mathcal{B}^{\prime}$ (then $\left|\mathcal{B}^{\prime}\right|=|\mathcal{B}|$ ). We claim that $\mathcal{B}^{\prime}$ is good.



Assume, to the contrary, that $\mathcal{B}^{\prime}$ contained a tight subcollection $\mathcal{T}$; clearly, $C \in \mathcal{T}$, otherwise $\mathcal{B}$ is not good. If $\mathcal{T} \subseteq \mathcal{C} \cup\{C\}$, then $C$ is the unique set in $\mathcal{T}$ containing $x$ which is impossible. Therefore, there exists $P \in \mathcal{T} \backslash(\mathcal{C} \cup\{C\})$. By maximality of $\mathcal{C}$, the collection $\mathcal{C} \cup\{P\}$ does not satisfy the requirements imposed on $\mathcal{C}$; since $P \cup C \neq X$, this may happen only if $C \cup P=P$, i.e., if $C \subset P$. But then $\mathcal{G}=(\mathcal{T} \backslash\{C\}) \cup \mathcal{C}$ is a tight subcollection in $\mathcal{B}$ : all elements of $C$ are covered by $\mathcal{G}$ at least twice (by $P$ and an element of $\mathcal{C}$ ), and all the rest elements are covered by $\mathcal{G}$ the same number of times as by $\mathcal{T}$. A contradiction. Thus $\mathcal{B}^{\prime}$ is good.



Such modifications may be performed finitely many times, since the total number of elements of sets in $\mathcal{B}$ increases. Thus, at some moment we arrive at a good collection $\mathcal{B}$ for which the procedure no longer applies. This means that for every $M, N \in \mathcal{B}$, either $M \cup N=X$ or one of them is contained in the other.



Now let $M$ be a minimal (with respect to inclusion) set in $\mathcal{B}$. Then each set in $\mathcal{B}$ either contains $M$ or forms $X$ in union with $M$ (i.e., contains $X \backslash M$ ). Now one may easily see that the two collections



$$

\mathcal{B}_{+}=\{A \backslash M: A \in \mathcal{B}, M \subset A, A \neq M\}, \quad \mathcal{B}_{-}=\{A \cap M: A \in \mathcal{B}, X \backslash M \subset A, A \neq X \backslash M\}

$$



are good as collections of subsets of $X \backslash M$ and $M$, respectively; thus, by the induction hypothesis, we have $\left|\mathcal{B}_{+}\right|+\left|\mathcal{B}_{-}\right| \leq 2 n-4$.



Finally, each set $A \in \mathcal{B}$ either produces a set in one of the two new collections, or coincides with $M$ or $X \backslash M$. Thus $|\mathcal{B}| \leq\left|\mathcal{B}_{+}\right|+\left|\mathcal{B}_{-}\right|+2 \leq 2 n-2$, as required.
We provide yet another proof of the estimate $|\mathcal{B}| \leq 2 n-2$. Consider any collection $\mathcal{B}$ of proper non-empty subsets of $X$ with no tight subcollection (we call such collection good). Arguing indirectly, we assume that there exists a good collection $\mathcal{B}$ with $|\mathcal{B}| \geq 2 n-1$, and choose one such for the minimal possible value of $n$. Clearly, $n>2$.



Firstly, we perform a different modification of $\mathcal{B}$. Choose any $x \in X$, and consider the subcollection $\mathcal{B}_{x}=\{B: B \in \mathcal{B}, x \notin B\}$. By our assumption, $\mathcal{B}_{x}$ is not tight. As the union of sets in $\mathcal{B}_{x}$ is distinct from $X$, either this collection is empty, or there exists an element $y \in X$ contained in a unique member $A_{x}$ of $\mathcal{B}_{x}$. In the former case, we add the set $B_{x}=X \backslash\{x\}$ to $\mathcal{B}$, and in the latter we replace $A_{x}$ by $B_{x}$, to form a new collection $\mathcal{B}^{\prime}$. (Notice that if $B_{x} \in \mathcal{B}$, then $B_{x} \in \mathcal{B}_{x}$ and $y \in B_{x}$, so $B_{x}=A_{x}$.)



We claim that the collection $\mathcal{B}^{\prime}$ is also good. Indeed, if $\mathcal{B}^{\prime}$ has a tight subcollection $\mathcal{T}$, then $B_{x}$ should lie in $\mathcal{T}$. Then, as the union of the sets in $\mathcal{T}$ is distinct from $X$, we should have $\mathcal{T} \subseteq \mathcal{B}_{x} \cup\left\{B_{x}\right\}$. But in this case an element $y$ is contained in a unique member of $\mathcal{T}$, namely $B_{x}$, so $\mathcal{T}$ is not tight - a contradiction.



Perform this procedure for every $x \in X$, to get a good collection $\mathcal{B}$ containing the sets $B_{x}=X \backslash\{x\}$ for all $x \in X$. Consider now an element $x \in X$ such that $\left|\mathcal{B}_{x}\right|$ is maximal. As we have mentioned before, there exists an element $y \in X$ belonging to a unique member (namely, $B_{x}$ ) of $\mathcal{B}_{x}$. Thus, $\mathcal{B}_{x} \backslash\left\{B_{x}\right\} \subset \mathcal{B}_{y}$; also, $B_{y} \in \mathcal{B}_{y} \backslash \mathcal{B}_{x}$. Thus we get $\left|\mathcal{B}_{y}\right| \geq\left|\mathcal{B}_{x}\right|$, which by the maximality assumption yields the equality, which in turn means that $\mathcal{B}_{y}=\left(\mathcal{B}_{x} \backslash\left\{B_{x}\right\}\right) \cup\left\{B_{y}\right\}$.



Therefore, each set in $\mathcal{B} \backslash\left\{B_{x}, B_{y}\right\}$ contains either both $x$ and $y$, or none of them. Collapsing $\{x, y\}$ to singleton $x^{*}$, we get a new collection of $|\mathcal{B}|-2$ subsets of $(X \backslash\{x, y\}) \cup\left\{x^{*}\right\}$ containing no tight subcollection. This contradicts minimality of $n$.

Final answer: $2n-2$