Uniqueness proof. Is this possible?

Question

I’m tasked with a problem from How to Prove It and I just don’t think it can be done. Any help in letting me know is greatly appreciated!

Definitions:

1.) Let $\bigcup_!F$ be defined as $\bigcup_!F = \{ x| ∃_!A ,A \in F \land x \in A\}$.

2.) Family set $F$ has the property of being pairwise disjoint meaning every pair of distinct elements of F are disjoint. This is written as $\forall A \in F\; \forall B \in F ,A \neq B \implies A \cap B = \emptyset$.

I am now asked to prove that $\bigcup_!F = \bigcup F \iff F$ is pairwise disjoint.

My dilemma is I believe $\bigcup_!F$ by definition indicates that there is exactly one and only one set in family set $F$. If I have in fact interpreted that correctly from the definition $\bigcup_!F = \{ x| ∃_!A ,A \in F \land x \in A\}$ then isn’t it impossible to prove that $F$ is pairwise disjoint? I mean, wouldn’t you need at least two sets within family set $F$ in order to prove it is pairwise disjoint? What am I not getting?

Answer 1

I suppose your UF is actually $\cup \mathcal{F}$, the union of a family of sets.

So U!F might be written as $\cup! \mathcal{F}$, though this is entirely non-standard. It means those $x$ that are in exactly one member of the family. So ie.g. if $\mathcal{F} = \{\{1,2\} ,\{2,3\}\}$, this is the set $\{1,3\}$ (check this).

Suppose $\cup! \mathcal{F} = \cup \mathcal{F}$ and suppose $A \neq B$ are two sets from $\mathcal{F}$. If $A \cap B \neq \emptyset$, then some $x \in A \cap B $ exists. For this $x$ we have $x \in \cup \mathcal{F}$ but $x \notin \cup!\mathcal{F}$ (As $x \in A$ and $x \in B$, contradiction.

The reverse is similar.

If a family has only one member, say $\mathcal{F} = \{A\}$ then both sets equal $A$ and the family is pairwise disjoint. Disprove it, if you can! (the condition is void, so true).

Answer 2

You're misunderstanding the first definition.

Writing $$ \bigcup\nolimits_! F = \{ x \mid \exists_! A: (A\in F\land x\in A) \} $$ doesn't mean that $\bigcup_! F$ is only defined when $\exists_! A$. On the contrary, $\bigcup_! F$ is defined for every family of sets.

The entire condition $\exists_! A: (A\in F\land x\in A)$ is simply something that can be either true or false, depending on what $F$ and $x$ are. Writing it down inside the set builder notation doesn't promise that it will be true -- it says that the set you're defining consists of those $x$ that make the condition true. If you find an $x$ where $\exists_! A(\cdots x\cdots)$ is false, this simply means that this $x$ is not a member of $\bigcup_! F$.

(Note that $\bigcup_!$ is not a standard concept. It is defined just for the purpose of this exercise).

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Also, even if you were right and $F$ has only one element, every family consisting of a single set only is pairwise disjoint. In that case $A\ne B$ is always false when $A,B\in F$, and putting false on the left-hand side of $\Rightarrow$ makes the entire $\Rightarrow$ true.