Is mathematics done in an arbitrary model of ZFC? Following up a previous thread I posted, I have tried to refine my questions. I would be happy with answers simply confirming that I have understood matters correctly, but of course I would also be happy to read longer and more elaborate answers!
Suppose I want ZFC (as formulated in first-order logic) to be my foundations of mathematics, i.e. I want it to be possible to formalise all of my reasoning about for example number theory as consequences in first-order logic of the ZFC axioms.
These are my questions:


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*When I consider a "set", for example $\mathbb{N}$, when doing mathematics, is it reasonable to view $\mathbb{N}$ as living in a fixed model of ZFC? So if I wanted to be very pedantic, I would refer to $\mathbb{N}$ as $\mathbb{N}^\mathcal{A}$ where $\mathcal{A}$ is my fixed model of ZFC?

*Suppose I am doing mathematics and discussing properties of $\mathbb{N}$ with a friend of mine. Can any "problems" arise from me (implicitly) thinking of $\mathbb{N}$ as being $\mathbb{N}^\mathcal{A}$ living in my ZFC-model $\mathcal{A}$, while my friend thinks of $\mathbb{N}$ as being  $\mathbb{N}^\mathcal{B}$ where $\mathcal{B}$ is his ZFC-model? I suppose since all assumptions we have made about $\mathcal{A}$ and $\mathcal{B}$ is that they are ZFC-models, no "substantial" problems can arise, so my friend and I can therefore safely agree to ease notation by referring to $\mathbb{N}^\mathcal{A}$ and $\mathbb{N}^\mathcal{B}$ as simply $\mathbb{N}$?

*Given that my understanding of the above questions is correct, it seems reasonable to view mathematics in a "semantic" way as "being done in an arbitrary model of ZFC"? Alternatively, if I insist on first-order logic ZFC to be my foundations, I could view doing mathematics as a sort of informal natural deduction, i.e. I could have a "syntactic" view on mathematics. The soundness and completeness theorems ensure that the "semantic" and "syntactic" views in some sense are the same, but perhaps one of the approaches has some philosophical advantages? I obviously find the semantic view more pleasing, since the semantic view is "consistent" with my view of a sets as living in models of ZFC...
Thanks in advance!
Edit:
I realise that perhaps "the standard model of ZFC" might be relevant to my questions. That is, I should perhaps view $\mathbb{N}$ as always referring specifically to $\mathbb{N}^\mathcal{S}$ where $\mathcal{S}$ denotes the standard model of ZFC? Similarly, I should perhaps also view mathematics as being done specifically in the standard model of ZFC. If we disregard problems with even defining which model of ZFC is the standard one, this view also seems to give us problems with translating our informal mathematical proofs to natural deduction: If mathematics is done in a very specific model of ZFC, we can no longer rely on completeness of first-order logic to guarantee existence of a derivation in natural deduction.
To summarise my edit: If the standard model of ZFC is relevant to my questions, please illuminate how. I do not think the standard model is relevant for my questions in any other way than being an "intuitive model" when thinking of ZFC.
 A: Modern maths avoids hazards like Russel's paradox by making the assumption

There exists a model of ZFC

Given this one model we can construct other models with different properties. For example we can construct a model $\mathcal A$ where the Continuum Hypothesis holds and another $\mathcal B$ where it doesn't hold. So if are doing maths over $\mathcal A$ and your friend is doing maths over $\mathcal B$ it is true for you that $\aleph_1= 2^{\aleph_0}$ but not for your friend.
The solution is, rather than imagine you are working over an arbitrarily chosen model, to imagine you are working over ALL models of ZFC at once. When you say  $1+1=2$ you do not mean $1^\mathcal A+1^\mathcal A=2^\mathcal A$ for your favourite model $\mathcal A$. You mean $1^\mathcal A+1^\mathcal A=2^\mathcal A$ for each and every model $\mathcal A$. 
An alternate approach might be to try and define a standard model as one which obeys all the axioms of ZFC and does not have any additional sentences that could be added as axioms, and do your maths there. This avoids the continuum problem above. However it is impossible because we can take any non-axiom sentence $P$. Either $P$ or $\neg P$ will hold in the model. Therefore it is also a model of either $ZFC + P$ or $ZFC + \neg P$.
Another alternative would be to try and find a minimal model by somehow taking the intersection of all possible models. This cannot be done as our ultimate assumption does not assume there is some larger space where all the models coexist.
