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A Proof of Lemma 2.5
Lemma 2.5. Consider a partitioning of the set system  $K = \{S_1, \ldots, S_k\}$  into  $m$  sets  $I_1, \ldots, I_m$ . Then,
$$
\operatorname {w i n n e r} (K) \in \left\{\operatorname {w i n n e r} \left(I _ {1}\right), \dots , \operatorname {w i n n e r} \left(I _ {m}\right) \right\}.
$$
The proof of Lemma 2.5 relies on the following observation.
Lemma A.1. Let  $X$  and  $Y$  be two subsets of  $K$ . If  $X \subset Y$ , then for every  $n$ , we always have
$$
\mathcal {R} _ {n} (Y) \cap X = \mathcal {R} _ {n} (X) \quad o r \quad \mathcal {R} _ {n} (Y) \cap X = \varnothing . \tag {9}
$$
Proof. We prove that (9) holds by induction on  $n$ . Initially, when  $n = 0$ , we have  $\mathcal{R}_0(X) = X$  and  $\mathcal{R}_0(Y) = Y$ . Therefore,  $\mathcal{R}_0(Y) \cap X = X \cap Y = X = \mathcal{R}_0(X)$ . Suppose (9) holds for  $n$ , we prove that (9) also holds for  $n' = n + 1$ . If  $\mathcal{R}_n(Y) \cap X = \emptyset$ , then  $\mathcal{R}_n(Y) \cap X$  remains empty for all  $n' \geq n$ . Therefore, (9) holds for  $n + 1$ . So, let us assume that  $\mathcal{R}_n(Y) \cap X = \mathcal{R}_n(X)$ . Consider three cases:
- If  $\omega_{n+1}$  belongs to all sets in  $\mathcal{R}_n(Y)$ , then it also belongs to all sets in  $\mathcal{R}_n(X) = \mathcal{R}_n(Y) \cap X$ . Thus, in this case, no set is eliminated in  $X$  or  $Y$ . That is,  $\mathcal{R}_{n+1}(X) = \mathcal{R}_n(X)$  and  $\mathcal{R}_{n+1}(Y) = \mathcal{R}_n(Y)$ .
- If  $\omega_{n+1}$  belongs to all sets in  $\mathcal{R}_n(X)$ , but not all sets in  $\mathcal{R}_n(Y)$ , then, at step  $n+1$ , we remove all sets that contain  $\omega_{n+1}$  and, particularly, all sets in  $\mathcal{R}_n(X)$ , from  $\mathcal{R}_n(Y)$ . Consequently,  $\mathcal{R}_{n+1}(Y) \cap X = \emptyset$ .
- If not all sets in  $\mathcal{R}_n(X)$  and not all sets in  $\mathcal{R}_n(Y)$  contain  $\omega_{n+1}$ , then we remove exactly the same sets from both  $\mathcal{R}_n(X)$  and  $\mathcal{R}_n(Y) \cap X$ . Namely, we remove sets  $S_i \in \mathcal{R}_n(Y)$  that contain  $\omega_{n+1}$ .
We conclude that (9) holds for  $n' = n + 1$ .
Proof of Lemma 2.5. Consider an arbitrary realization of the game  $\omega_{1},\omega_{2},\ldots$  . Let  $n$  be the round when all sets but the winner are eliminated from the competition i.e.,  $\mathcal{R}_n$  contains only one set, the winner. Since  $K$  is the union of  $I_{1},\dots,I_{k}$  , the winner must belong to some  $I_{j}$  . Now, by Lemma A.1 for  $X = I_{j}$  and  $Y = K$  , we have  $\mathcal{R}_n(K)\cap I_j = \mathcal{R}_n(I_j)$  or  $\mathcal{R}_n(K)\cap I_j = \emptyset$  . We know that  $\mathcal{R}_n(K) = \{\mathrm{winner}(K)\}$  and winner  $(K)\in I_{j}$  . Thus,  $\mathcal{R}_n(K)\cap I_j = \{\mathrm{winner}(K)\} \neq \emptyset$  , and
$$
\mathcal {R} _ {n} (I _ {j}) = \mathcal {R} _ {n} (K) \cap I _ {j} = \{\operatorname {w i n n e r} (K) \}.
$$
We conclude that at round  $n$ ,  $\mathcal{R}_n(I_j)$  contains only one set – the winner in  $K$ . Consequently, it is also the winner in  $I_j$  i.e., winner  $(I_j) = \mathrm{winner}(K)$ . This finishes the proof.