I'm taking a class in set theory following the text "Elements of Set Theory" by Herbert Enderton. I had a question about the construction of the naturals and the axiom of infinity.
As I understand it, the axiom of infinity says $\exists A (A \text{ Inductive})$ (assume I had already defined what it means to be an inductive set in first order logic). Then when Enderton constructs the natural numbers (page 68), he says
Let A be an inductive set; by the infinity axiom it is possible to find such a set.
Then applies a subset axiom to construct the smallest inductive set, which we call $\omega$.
The specific line I take issue with is the line "let A be an inductive set". I had a couple questions.
1] What are we actually doing when we say "let x be an object satisfying P" in the context of set theory? What does "let" mean?
2] When, in the axiom of infinity, we give the name $A$ to the inductive set, is the letter $A$ now forever bound in the language of set theory to an inductive set which exists via the axiom of infinity? If I'm being super strict, is it the case that I should no longer use the letter "A" to refer to anything else? If not, how do I refer later to the specific set which exists via the axiom?
3] Suppose we either assume or prove that there is an object with a given property, but that object may not be the unique object with this property. For example, the property of being an inductive set. If I later want to "chose" any object - it doesn't matter which one - that has this property, how do I express this in the language of set theory?
It goes like this. Let $x=\Bbb{N}_A$ be the relation defined by a formula whose free variables are $x,A$ and says $$\text{Inductive}(A)\land\forall y(y\in x\leftrightarrow(y\in A\land(\forall z(\text{Inductive}(z)\rightarrow y\in z))))$$ what this says is that $A$ is inductive, and $x$ is the subset separated out of $A$ as all the elements of $A$ that are members of all inductive sets.
Now you prove: $\forall A\;\text{Inductive}(A)\rightarrow(\exists x\, x=\Bbb{N}_A)$
To prove this you introduce a new constant $A$, assume $A$ is inductive, then use comprehension to show $x=\Bbb{N}_A$ exists. Finally, use the logical rule of universal generalization that for all inductive $A$, $\Bbb{N}_A$ exists.
Next stage: $\forall A\forall B\forall x\forall y\;(\text{Inductive}(A)\land\text{Inductive}(B)\land x=\Bbb{N}_A\land y=\Bbb{N}_B)\rightarrow x=y$
In other words, that the $\Bbb{N}_A$ does not depend on the inductive $A$ you start with. To show this, let $z\in\Bbb{N}_A$. Then $z$ is in $A$ and $z$ is in every inductive set. Since $B$ is inductive, $z\in B$. Therefore $z$ is in $B$ and in every inductive set, so $z\in\Bbb{N}_B$. This proves $\Bbb{N}_A\subseteq\Bbb{N}_B$. By a symmetric argument, $\Bbb{N}_B\subseteq\Bbb{N}_A$. By the axiom of extensionality, $\Bbb{N}_A=\Bbb{N}_B$.
Finally define $x=\Bbb{N}$ iff $\exists A\;\text{Inductive}(A)\land x=\Bbb{N}_A$. Using existential instantiation, by the axiom of infinity, there exists an inductive $A$. Then $\Bbb{N}$ exists and is unique by the results above.