Suppose F is a family of sets that has the property that for every G ⊆ F, ⋃G ϵ F. Prove that there is a unique set A s.t. A ϵ F and ∀ B ϵ F (B ⊆ A).

Question

the solutions to this questions, 3.6.11, are in the suggested solutions in the textbook (How to Prove It, Velleman). However, I have some trouble understanding the question. Initially, I produced a simple example to get my head around the question:

G={{1,2,3,4},{2,3,4,5}}

F={{1,2,3,4},{2,3,4,5},{1,2,3,4,5},{99}}

$\bigcup$G={1,2,3,4,5}$\in$F

Indeed, G$\subseteq$F $\implies$ $\bigcup$G$\in$F, but

B={99}$\in$F but B$\not\subseteq$A

So there is no set in F such that every other element is a subset it. I am sure I am interpreting this wrong, so can someone please highlight where my error is?

Thanks!

Answer 1

We want such a set $F$ to have the following property: If $G$ is any subset of $F$, no matter which one, then $\bigcup G \in F$.

The set $F = \{\{1, 2, 3, 4\}, \{2, 3, 4, 5\}, \{1, 2, 3, 4, 5\}, \{99\}\}$ does not have this property, since $\{\{1, 2, 3, 4\}, \{99\}\} \subseteq F$ but $\{1, 2, 3, 4\} \cup \{99\} = \{1, 2, 3, 4, 99\} \notin F$.

The set $F = \{\{1, 2, 3, 4\}, \{2, 3, 4, 5\}, \{1, 2, 3, 4, 5\}\}$ also does not have this property, but for a sneaky reason: $\varnothing \subseteq F$, but $\bigcup \varnothing = \varnothing \notin F$.

On the other hand, the set $F = \{\{1, 2, 3, 4\}, \{2, 3, 4, 5\}, \{1, 2, 3, 4, 5\}, \varnothing\}$ does have this property. To see this, we need to verify it for every subset $G$ of $F$. In other words, we need to check that all 15 of the sets $$ \varnothing, \quad \{1, 2, 3, 4\}, \quad \{2, 3, 4, 5\}, \quad \{1, 2, 3, 4, 5\}, \\ \varnothing \cup \{1, 2, 3, 4\}, \quad \varnothing \cup \{2, 3, 4, 5\}, \quad \varnothing \cup \{1, 2, 3, 4, 5\}, \\ \{1, 2, 3, 4\} \cup \{2, 3, 4, 5\}, \quad \{1, 2, 3, 4\} \cup \{1, 2, 3, 4, 5\}, \quad \{2, 3, 4, 5\} \cup \{1, 2, 3, 4, 5\}, \\ \varnothing \cup \{1, 2, 3, 4\} \cup \{2, 3, 4, 5\}, \quad \varnothing \cup \{1, 2, 3, 4\} \cup \{1, 2, 3, 4, 5\}, \quad \varnothing \cup \{2, 3, 4, 5\} \cup \{1, 2, 3, 4, 5\}, \\ \{1, 2, 3, 4\} \cup \{2, 3, 4, 5\} \cup \{1, 2, 3, 4, 5\}, \quad \varnothing \cup \{1, 2, 3, 4\} \cup \{2, 3, 4, 5\} \cup \{1, 2, 3, 4, 5\} $$ are elements of $F$. And they are!