When reading about set theory, I came across a simple question about the "uniqueness" of the empty set:
Let $\mathfrak{M} \models ZF$, let $M$ be the domain of $\mathfrak{M}$. Let's create an inner model for $ZF$, namely $\mathfrak{S}$ with domain $S$ in $\mathfrak{M}$ the following way:
For some $k_1, k_2 \in M$, construct $S$ the following way: It should contain $\{k_1\}, \{k_2\}$ and from there on every pair and union and power set and so forth for all existing sets (maybe more, I don't think it matters, also I'm omitting the axiom of infinity here), but $S$ shall not contain $k_1$ or $k_2$ (that's just the way $V_\omega$ is constructed). Take $\in_\mathfrak{S}$ just as $\in_\mathfrak{M}$.
This structure $\mathfrak{S}$ is a model for most of $ZF$ (in my opinion), but I'm not so sure about extensionality:
$$\forall a, b, c: [(c \in a \leftrightarrow c \in b) \rightarrow a = b]$$
Consider now the relation between $\{k_1\}$ and $\{k_2\}$. Since neither $k_1$ nor $k_2$ are elements of $S$, the left part is true in $\mathfrak{S}$.
Say we define "$=$" semantically via: $\mathfrak{A} \models a = b$ iff $\mathfrak{I}(a) = \mathfrak{I}(b)$. (As btw. my logic course in university did, just as most of the texts I've read.) This is clearly not the case, since both sets actually differ, so this axiom is not satisfied in the inner model, so $\mathfrak{S} \not \models ZF$. But by this argument, there is actually no model of $ZF$ since there can always be some lurking element somewhere.
Say we define "$=$" syntactically via: $\forall x: x = x$ and for all $\varphi(x)$ without usage of the variables $y, z$: $y = z \rightarrow \varphi(y/x) \leftrightarrow \varphi(z/x)$. This would work, since I cannot possible distinguish the two elements with $FO(\{\in\})$. But then I would get two different empty sets - not different from within $\mathfrak{S}$, but for the outside world. Also, I would fall victim to the compactness theorem. I do anyway for $ZF$ thanks to replacement and separation, but any theory with equality would. This also weakens equality down to a "usual" equivalence relation - there cannot by any stronger one within the model by design, but it's just not right.
So, after this bunch of explanation, my question is:
What's up with all of this? Do we have multiple empty sets when looking from outside a model, is my equality just not equal, is there a flaw in my argumentation or something completely differnt?
You're missing the fact that $\bigcup\{x\}=x$, since everything is a set. Therefore $k_1$ and $k_2$ are not only elements of $S$, but also their elements and so on.
Moreover, since you're taking the power sets to be the "real ones" (as computed in $\frak M$), the empty set is there, and $S$ is closed to unions and power sets from $\frak M$. Therefore you can prove by induction on rank that $S=M$.
One can construct models of ZF without extensinoality (namely ZF with Atoms (or Urelements)) using similar techniques. But one has to be more careful and not just take the whole power set.