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I am very confused with definition 1 in section 2.1 of this article.

Definition 1 (Access Structure [2]) Let $\left\{ P_1, P_2, ..., P_n \right\}$ be a set of parties. A collection $\mathbb{A} \subseteq 2^{\left\{ P_1, P_2, ..., P_n \right\}}$ is monotone if $\forall B, C : \text{if } B \in \mathbb{A} \text{ and } B \subseteq C \text{ then } C \in \mathbb{A}$.

  1. Following to this post, why isn't $\mathbb{A} = \left\{ \left\{ 1, 2 \right\}, \left\{ 3, 4 \right\} \right\} \subseteq 2^{\left\{ 1, 2, 3, 4 \right\}}$ monotone?
  2. Where do $B$ and $C$ come from? Are they both elements of $\mathbb{A}$? Are they any kind of sets? If yes, $B = \left\{ 1, 2 \right\}, C = \left\{ 1, 2, 3, 4, 5 \right\}$ would be true, but doesn't make sense, because $C$ is not an element of $2^{\left\{ 1, 2, 3, 4 \right\}}$, so it can't be an element of $\mathbb{A}$, too. But the definition above implies that.
Patriot
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Kluddizz
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1 Answers1

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It's for $B,C \in 2^{\{P_1, P_2, ..., P_n\}}$

So, take $B = \{1,2\} \in \mathbb{A}$ as in your example. $\forall C \in 2^{\{1,2,3,4\}}$ such that $B \subset C$, $C$ must be in $\mathbb{A}$.

Possible values of $C$ are $\{1, 2\}$, $\{1, 2, 3\}$, $\{1,2,4\}$, and $\{1,2,3,4\}$. As you can see not all of them are in $\mathbb{A}$.

The idea is that if any set has access, all sets that contain it must also have access.

Aman Grewal
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