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If it helps anything, please assume that everything below is finite.

Let $\mathcal A$ be a family of subsets of a set $X$. I want to consider the following independence condition (C) on $\mathcal A$.

(C). The function $\bigcup: 2^\mathcal A\to2^X$ is an injection.

In other words, (C) says that each subfamily of $\mathcal A$ has a different union.

Every pairwise disjoint family $\mathcal A$ satisfies this condition. If I put together pairwise disjoint pieces, I can recognize them in the union.

But there are other examples. For $|\mathcal A|=1,$ (C) simply says that $\mathcal A\neq \{\varnothing\}$ because

  • if $\mathcal A=\{\varnothing\},$ then $\varnothing,\{\varnothing\}\subseteq\mathcal A,$ and $\bigcup\varnothing=\bigcup\{\varnothing\}=\varnothing,$ but $\varnothing\neq\{\varnothing\},$ so $\bigcup$ is not injective;
  • if $\mathcal A = \{Y\},$ $Y\neq\varnothing,$ then the only subfamilies of $\mathcal A$ are $\varnothing$ and $\{Y\}$, and they have different unions.

For $|\mathcal A|=2$, $\mathcal A=\{Y,Z\},$ (C) says that $Y\setminus Z\neq\varnothing\neq Z\setminus Y.$ (In other words, neither of the sets is a subset of the other.) Indeed,

  • if $Y\setminus Z=\varnothing,$ then $\bigcup\{Y,Z\}=\bigcup\{Z\}$ (and the same for $Z\setminus Y=\varnothing$);
  • otherwise, the unions of $\varnothing,\{Y\},\{Z\},\{Y,Z\}$ are all different.

For $|\mathcal A|=3,$ I already don't see a simple condition equivalent to (C). Is there one in the general case? (C) is of course a simple condition in terms of formulation, but it is quite a complex one in terms of the "little conditions" it comprises. I'm not certain that a condition I'm asking for exists, but perhaps it does, and someone here can help me see it?

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"Every pairwise disjoint family $\mathcal A$ satisfies this condition" - No, we must not have $\emptyset$ itself in the family. –  Hagen von Eitzen May 21 '13 at 15:59

1 Answer 1

up vote 2 down vote accepted

The condition that $$\tag1 A\not\subseteq\bigcup(\mathcal A\setminus\{A\})\qquad\text{for all }A\in \mathcal A$$ is necessary to distinguish $\mathcal A$ from $\mathcal A\setminus\{A\}$ and sufficient as it implies that $A\in \mathcal B\subseteq \mathcal A$ iff $A\subseteq \bigcup\mathcal B$.

For each $A\in \mathcal A$, $(1)$ allows us to select an element $a(A)\in A\setminus \bigcup(\mathcal A\setminus\{A\})$ that works as a "sensor" for $A$. That is, we can restrict everything from $X$ to $\{a(A)\mid A\in\mathcal A\}$ and by that end up in the simple case of disjoint singleton sets. In other words, an equivalent condition is:

There exists a subset $X'\subseteq X$ such that $A\cap X'$ is a singleton for every $A\in\mathcal A$ and $A_1\ne A_2$ implies $A_1\cap X'\ne A_2\cap X'$.

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