Is this an accurate way to write the axiom of choice using transversals?

Question

Is this an accurate way to write the axiom of choice using transversals?

$\forall \mathcal{C} \exists T\, (T\subseteq \cup\,\mathcal{C} \land \forall X \exists s \forall t\, ((X\in\mathcal{C} \land t\in T\cap X) \implies t=s))$

Answer 1

No, this is not the correct way. The reason is that to write the axiom of choice using a transversal set you also need to require that $\cal C$ is pairwise disjoint.

For example, it is impossible to find a transversal to $\{\{0,1\},\{1,2\},\{0,2\}\}$.

Moreover you can omit the part where $T\subseteq\bigcup\mathcal C$, since it we can prove that if a transversal set exists, then its intersection with $\bigcup C$ is also a transversal set.

So you would like to write something like this:

$$\forall\mathcal C\Bigg(\forall x\forall y\Big(x\in\mathcal C\land y\in\mathcal C\land x\cap y\neq\varnothing\rightarrow x=y\Big)\rightarrow\exists T\forall x\Big(x\in\mathcal C\rightarrow\exists y\big(y\in x\cap T\land\forall z(z\in x\cap T\rightarrow z=y)\big)\Big)\Bigg)$$

Answer 2

Some texts like Kunen's Set Theory use the following

$\forall A\exists R(R\text{ well-orders }A)$

I like the version in Wikipedia (Axiom of Choice) but I personally like my slightly modified version:

$\forall \mathcal{X}\Big[\mathcal{X}\ne\varnothing\,\wedge\,\varnothing \notin \mathcal{X}\Rightarrow \exists f\big[f:\mathcal{X}\to\bigcup\mathcal{X}\,\wedge\, \forall A[A\in\mathcal{X}\Rightarrow f(A)\in A]\big]\Big]$

because it's easily translatable into the following: "for each nonempty collection $\mathcal{X}$ of nonempty sets, there exists $f:\mathcal{X}\to\bigcup\mathcal{X}$ such that for each $A\in\mathcal{X}$, $f(A)\in A$."

*Sorry if you insisted the "$A\cap C$ is a singleton for every $A\in \mathcal{X}$" formulation.