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This is Velleman's exercise 3.6.5.b:

Prove that for any family of sets $\mathcal F$, $∪!\mathcal F = ∪\mathcal F$ iff $\mathcal F$ is pairwise disjoint.

First let's do some translations:

$\bullet$ $∪!\mathcal F = \{x | ∃!A(A ∈ \mathcal F ∧ x ∈ A)\}$.

$\bullet$ $∃!A$ means that "there exists a unique $A$ such that..." thus "$x ∈ ∪!\mathcal F$" which we will use later is equivalent to $∃A(A ∈ \mathcal F ∧ x ∈ A) ∧ ∀B((A ∈ \mathcal F ∧ x ∈ A) \Rightarrow B = A)$.

$\bullet$ A family of sets $\mathcal F$ is said to be pairwise disjoint if $∀A ∈ \mathcal F∀B ∈ \mathcal F(A \neq B → A ∩ B =∅)$.

And now here's my proof of it:

Proof.

($\rightarrow$) Suppose $∪!\mathcal F = ∪\mathcal F$ i.e. $x ∈ ∪!\mathcal F \iff x ∈ ∪\mathcal F$ i.e. $x ∈ ∪!\mathcal F \Rightarrow x ∈ ∪\mathcal F$ and $x ∈ ∪!\mathcal F \Leftarrow x ∈ ∪\mathcal F$. Let $A$ and $B$ be arbitrary elements of $\mathcal F$. Now we prove the contrapositive and suppose $A ∩ B \neq ∅$ which means $∃x(x ∈ A ∧ x ∈ B)$. We can now choose some $x_0$ such that $x_0 ∈ A$ and $x_0 ∈ B$. Since $A$ was an arbitrary element of $\mathcal F$ then from $x ∈ ∪\mathcal F$, we have $x ∈ ∪!\mathcal F$. Since $B$ was an arbitrary element of $\mathcal F$ and $x_0 ∈ B$, then from $x ∈ ∪!\mathcal F$: particularly from $∀B((A ∈ \mathcal F ∧ x ∈ A) \Rightarrow B = A)$, we get $B = A$. Since $A$ and $B$ were arbitary then $\mathcal F$ is pairwise disjoint. Therefore if $∪!\mathcal F = ∪\mathcal F$ then $\mathcal F$ is pairwise disjoint.

Here are my questions:

  1. Is my proof of the forward direction valid?

  2. What about the backward direction? Isn't the statement $∪!\mathcal F = ∪\mathcal F$ kind of self-explanatory (i.e. each side kind of prove the other side)? In other words I cannot see a way to use $∀A ∈ \mathcal F∀B ∈ \mathcal F(A \neq B → A ∩ B =∅)$ to prove those conditional-existential statements (they seem to sufficiently rely on each other for their existence)!

Thanks.

Heptapod
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1 Answers1

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Your proof seems good, but it can be simplified.

Suppose $A,B\in\mathcal{F}$, with $A\ne B$. If $A\cap B\ne\emptyset$ and $x\in A\cap B$, then $x\notin\mathop{{\bigcup}{!}}\mathcal{F}$, but $x\in\bigcup\mathcal{F}$. Hence $\mathop{{\bigcup}{!}}\mathcal{F}\ne\bigcup\mathcal{F}$.

The other direction is essentially obvious: if the sets in $\mathcal{F}$ are pairwise disjoint, any element in $\bigcup\mathcal{F}$ belongs to exactly one element of $\mathcal{F}$, hence it belongs to $\mathop{{\bigcup}{!}}\mathcal{F}$. Alternatively, assuming $\mathop{{\bigcup}{!}}\mathcal{F}\ne\bigcup\mathcal{F}$, there exists $x\in\mathcal{F}$ with $x\notin\mathop{{\bigcup}{!}}\mathcal{F}$. Thus, by definition, $x$ belongs to at least two distinct elements of $\mathcal{F}$, so its members are not pairwise disjoint.

egreg
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