# theorem about family of sets

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

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

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

so prove that $\displaystyle A$ 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 $\displaystyle \mathcal{F}=\varnothing$ as a particular F to show the contradiction.

Now coming to the uniqueness part, the given is

Given ---->

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

and the Goal is

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

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

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

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

$\displaystyle y \neq z$

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

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

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 $\displaystyle A=\emptyset$. Then $\displaystyle \bigcup\emptyset=\emptyset=A$ and hence $\displaystyle A\in\emptyset$, a contradiction. So there is $\displaystyle a\in A$. Let $\displaystyle \mathcal{F}=\{\{a\},A\setminus\{a\}\}$. Then $\displaystyle \bigcup\mathcal{F}=A$ and hence $\displaystyle A\in\mathcal{F}$. Since $\displaystyle A\neq A\setminus\{a\}$ it follows that $\displaystyle A=\{a\}$.