(Hall's Theorem) Existence of two subfamilies of sets containing the same elements

Question

I came across the following claim in a textbook on combinatorics [1].

Claim (Lindström, Tverberg): Let $A_1, . . . , A_m \subseteq [n]$ be non-empty with $m > n$. There are non-empty, disjoint $I, J \subseteq [m]$, such that $$\bigcup_{i \in I} A_i = \bigcup_{j \in J} A_j.$$

The book provides a very nice proof of this claim, using characteristic vectors of the sets $A_1, . . . , A_m$ and some linear algebra basics.

It also includes a remark that says this claim can be proved using Hall's theorem. This second approach is not obvious to me but I would like to understand how you can use Hall's theorem here (it's really bugging me). If anyone could provide a solution, a hint, or a reference to a proof, I would really appreciate it.

Reference: [1] B. Sudakov, "Algebraic Methods in Combinatorics", 2023

Original Proof: Let $v_1, . . . , v_m$ be the characteristic vectors of $A_1, . . . , A_m$. Since $m>n$, they are linearly dependent over $\mathbb{R}$. So, there is a nonzero $\alpha \in \mathbb{R}^m$ with $$\sum_{i=1}^m \alpha_i v_i = 0.$$ Let $$I = \{i : \alpha_i > 0\},$$ $$J=\{j : \alpha_j < 0\}.$$ The nonzero coordinates of $\sum\limits_{i \in I} \alpha_i v_i$ correspond to the elements of $\bigcup\limits_{i \in I} A_i$, which are the same as the nonzero coordinates of $\sum\limits_{j \in J} \alpha_j v_j$ corresponding to $\bigcup\limits_{j \in J} A_j$, completing the proof.

Answer 1

As a partial response, here is a proof of the claim based on a possible proof of Hall's theorem. Maybe it will inspire further progress.

The graph we consider is the usual graph for the system-of-representatives view of the theorem: let $X = \{x_1, x_2, \dots, x_m\}$, $Y = \{y_1, y_2, \dots, y_n\}$, and put an edge $x_i y_j$ whenever $j \in A_i$. It will be important that because the $A_i$ are all nonempty, $X$ has no isolated vertices.

Pick a maximum matching $M$ in this graph. Of course, since $m>n$, this matching does not cover all of $X$. We are going to start by letting $U_0$ be the set of all vertices in $X$ not covered by $M$, and then repeat the following procedure:

1. After $U_i$ has been defined, let $V_i = N(U_i) - N(U_0 \cup \dots \cup U_{i-1})$: the set of "new" vertices in $Y$, which are adjacent to $U_i$ that we haven't found previously.
2. Let $U_{i+1}$ be the set of all vertices in $X$ matched to $V_i$ by $M$.

We stop when no new vertices get added to $U_0, U_1, U_2, \dots$ or $V_0, V_1, V_2, \dots$.

How is this related to Hall's theorem? Well, every vertex in $V_i$ has a path to it from $U_0$, a path zigzagging between the sets $$U_0 - V_0 - U_1 - V_1 \dots - U_i - V_i$$ in which every $V-U$ type edge is an edge of the matching. So if we ever find an unmatched vertex in $V_i$, this zigzagging path is an augmenting path that could be used to improve $M$. Since $M$ is assumed to be a maximum matching, no such augmenting path exists. Therefore every vertex in $V_i$ is matched by $M$; it follows that $|V_i| = |U_{i+1}|$. The procedure can only stop by having $V_i = \varnothing$ at some stage, and it must stop since our graph is finite.

At this point, we normally say that $S = U_0 \cup U_1 \cup \dots \cup U_{i+1}$ is a subset of $X$ with $N(S) = V_1 \cup V_2 \cup \dots \cup V_i$, and $|S| = |U_0| + |N(S)|$, so $S$ is a set violating Hall's condition. This proves the hard direction of Hall's theorem.

What we'll say here instead is that if we take $S_0 = U_0 \cup U_2 \cup U_4 \cup \dots$ and $S_1 = U_1 \cup U_3 \cup U_5 \cup \dots$ (even-numbered sets in $S_0$, and odd-numbered sets in $S_1$), then $N(S_0) = N(S_1) = N(S)$. That's because each vertex in $V_i$ has a neighbor in $U_i$ (that's how we got it) and a neighbor in $U_{i+1}$ (which it was matched to by $M$). The sets $S_0$ and $S_1$ are exactly what we're looking for in this question. (To relate them back to the theorem we want, we let $I = \{i \in [m] : x_i \in S_0\}$ and $J = \{j \in [m] : x_j \in S_1\}$.)

Unfortunately, I don't see how to use just the statement of Hall's theorem (and not the proof) to make this deduction. It's not clear a priori why a set $S \subseteq X$ with $|N(S)|<|S|$ must split into disjoint sets $S_0, S_1$ with $N(S_0) = N(S_1) = N(S)$, even though we see it happening here.

We still learn one useful thing from this argument: the existence of $S_0$ and $S_1$ follows not just if $m>n$, but whenever the graph has no perfect matching. And this useful fact is a consequence of Hall's theorem (as well as the theorem in the question): if we find a set $S$ with $|N(S)|<|S|$, we can apply the theorem in the question to just the sets $A_i$ with $x_i \in S$, and just the elements $j \in [n]$ with $y_j \in N(S)$.