# theorem about family of sets

• July 30th 2011, 08:23 AM
issacnewton
Here's the theorem I am trying prove

Suppose $A$ is a set, and for every family of sets $\mathcal{F}$
we have

$\left[\bigcup \mathcal{F} = A\right]\Rightarrow A \in \mathcal{F}$

so prove that $A$ has exactly one element. Now there are two parts here. One is the existence part and second is uniqueness part. I have proved the
existence part using method of contradiction. I chose $\mathcal{F}=\varnothing$ as a particular F to show the contradiction.

Now coming to the uniqueness part, the given is

Given ---->

$\forall \mathcal{F} \left[(\bigcup \mathcal{F} = A )\Rightarrow A \in \mathcal{F}\right]$

and the Goal is

$\forall y \forall z \left[(y \in A)\wedge (z \in A)\Rightarrow (y=z)\right]$

I will assume the negation of the goal and try to find the contradiction. Hence the
givens are

$\forall \mathcal{F} \left[(\bigcup \mathcal{F} = A )\Rightarrow A \in \mathcal{F}\right]$

$(y \in A)\wedge (z \in A)$

$y \neq z$

and the goal now is contradiction. Now here I am having some difficulty,
how do I choose particular $\mathcal{F}$ ? Since $y \in A$
and $z \in A$ we know that

$\{y,z\} \subseteq A$

But there could be more elements in A. Is there any way I can consider different
cases which will exhaust the possibilities ?

The method of contradiction is fine for existence, but probably not best for uniqueness. Suppose towards a contradiction that $A=\emptyset$. Then $\bigcup\emptyset=\emptyset=A$ and hence $A\in\emptyset$, a contradiction. So there is $a\in A$. Let $\mathcal{F}=\{\{a\},A\setminus\{a\}\}$. Then $\bigcup\mathcal{F}=A$ and hence $A\in\mathcal{F}$. Since $A\neq A\setminus\{a\}$ it follows that $A=\{a\}$.