proof that axiom of union and axiom of pairing imply weak axiom of union

Question

The axiom of union says that to every set $\mathcal{A}$ there is a set $\{z: \exists A \; z \in A \wedge A \in \mathcal{A}\}$ or in other words: $\forall \mathcal{A} \exists B: \forall z [z \in B \leftrightarrow \exists A \in \mathcal{A} \; z \in A]$.
The weak axiom of union says that $\forall A \forall B \exists C: \forall x (x \in C \leftrightarrow x \in A \vee x \in B)$.
The axiom of pairing says: $\forall x \forall y \exists p: \forall z [z \in p \leftrightarrow z = x \vee z = y]$.

I am trying to give a (not fully formal) proof that if the axiom of union and axiom of pairing do hold, then also the weak axiom of pairing does hold.

Futhermore, I'm looking for a structure in which the axiom of union does hold, but the weak axiom of union doesn't. (So obviously this has to be a structure where the axiom of union does hold, but the axiom of pairing does not.) I'm still trying to find an apt structure, but I think that finally I'll find one.
So I'd primarily appreciate your help on giving a (nearly formal) proof, since I'm just not managing to. Thank you in advance!

Answer 1

Assume the axiom of union and the axiom of pairing hold. Then by the axiom of pairing $$\forall A\forall B\exists p:\forall S[S\in p\iff S=A\vee S=B],$$ and by the axiom of union $$\forall p\exists C:\forall z[z\in C\iff\exists S\in p:z\in S].$$ Therefore $$\forall A\forall B\exists C:\forall z[z\in C\iff\exists S:(S=A\vee S=B)\wedge z\in S].$$ I will leave the details to you.

EDIT: Full proof: Let $A,B$ be sets. By the axiom of pairing there exists a set $p$ such that $$\forall S[S\in p\iff S=A\vee S=B].$$ It follows that $\forall z:$ \begin{align} (\exists S\in p:z\in S) &\iff(\exists S:(S=A\vee S=B)\wedge z\in S) \\&\iff(\exists S:(S=A\wedge z\in S)\vee(S=B\wedge z\in S)) \\&\iff(z\in A\vee z\in B). \end{align} By the axiom of union there exists a set $C$ such that $$\forall z[z\in C\iff\exists S\in p:z\in S].$$ Hence, we find $$\forall z[z\in C\iff z\in A\vee z\in B].$$ Since $A,B$ were arbitrary, we conclude $$\forall A\forall B\exists C:\forall z[z\in C\iff z\in A\vee z\in B].$$ So if the axiom of pairing and the axiom of union hold, then the weak axiom of union holds.

Fixed counterexample: A model where the axiom of union holds but the weak axiom of union does not hold would be the model where $\emptyset$, $\{\emptyset\}$ and $\{\{\emptyset\}\}$ are the only sets. The weak axiom of union would imply that $\{\emptyset\}\cup\{\{\emptyset\}\}=\{\emptyset,\{\emptyset\}\}$ would exist, which it does not. I leave it up to you to check that the axiom of union does hold.

Answer 2

For any sets $A$ and $B$ there is a set $C = \{ A, B \}$. By Union, there must be a set $D$ that contains all and only the elements of the elements of $C$, i.e. $D$ will contain all and only elements of either $A$ or $B$. That's the set $'C'$ you are looking for in the definition of Weak Union .. not the initial $C$

The outline for the proof is so mechanical that I decided just to put it in formal proof format ... I left just the inside few steps for you if you wanted to complete this as a formal proof: