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 ?

please comment.