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Hey friends of maths,

I am trying to find an example that fulfills the requirements mentioned above, more precisely:

I'd appreciate your help with finding two (or better more than 200) non-isomorphic finite structures where the axiom $\forall x \exists P \forall z (z \in P \leftrightarrow z = x)$ does hold, and also, if possible, where the axioms $\forall A \forall B ([\forall x: x \in A \Leftrightarrow x \in B] \Rightarrow A = B)$ and $\exists N:(\forall x: x \in N \Leftrightarrow x \neq x)$ - do hold.

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This will help: it is impossible to satisfy both the axiom of the empty set and the singleton axiom using a finite sized structure:

First, by axiom of the empty set, you need an object, call it $d_0$, and for whatever other objects you have in the domain, those objects cannot stand in the relation as denoted by $\in$ (let's call this relarion $R$) with it.

Second, by the axiom of the singleton, there needs to be an object such that $d_0$ is the one and only object standing in the relation $R$ with it. For the reason just stated earlier, this cannot be $d_0$ itself, and so we need a new object $d_1$. So, we have $(d_0,d_1)\in R$ ... and there cannot be any other objects standing in relation $R$ to $d_1$.

OK, but then by the axiom of the singleton again, there needs to be some objects such that $d_1$ is the one and only object standing in relation $R$ with it ... and this cannot be $d_0$ (since nothing stands in relarion $R$ to $d_0$), not can it be $d_1$, since there cannot be any object other than $d_0$ standing in relation $R$ to $d_1$. So, we need to have anew object, $d_2$, and we need that $(d_1,d_2)\in R$, and we need to make sure that nothing but $d_1$ stands in the relation $R$ to $d_2$.

... ok, I think you see where this is going ... you indefinitely need to keep adding one more object $d_{i+1}$ to the already existing objects $d_0$ through $d_i$, because in order to satisfy the singleton axiom for $d_i$ we need some object that $d_i$ stands in the relation $R$ to, and by the axiom of the empty set that object cannot be $d_0$ and since we have $(d_j,d_{j+1})\in R$ for all $0\le j \le i-1$, all objects $d_1$ through $d_i$ already have some object other than $d_i$ standing in the relation $R$ to them and are not allowed to have any other object stand in relation $R$ to them.

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    $\begingroup$ This answers the title question with room to spare, since extensionality wasn't needed. But the text of the question demands only singletons, with the empty set thrown in "if possible". Since it's impossible in a finite universe, it might be worth noting that the singleton axiom and extensionality have lots of finite models: Arrange any finite number of entities in a cyclic order and let the "membership" relation hold just between each element and the next in the cyclic ordering. $\endgroup$ – Andreas Blass Dec 26 '18 at 2:22
  • $\begingroup$ @AndreasBlass Ah, I missed the 'if possible' in the body of the question. Thanks! $\endgroup$ – Bram28 Dec 26 '18 at 2:30
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    $\begingroup$ @Studentu I meant it just the other way around: $(d_0,d_1)$ would stand for $d_0 \in d_1$ $\endgroup$ – Bram28 Dec 26 '18 at 21:55
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    $\begingroup$ @Studentu Also, to explain what AndreasBlass was saying: suppose you have a domain with exactly one object: $d_1$. Also assume that $(d_1,d_1)\in R$, i.e. we basically have $d_1 \in d_1$. Then notice that for all objects (which is just $d_1$ of course) there is something $again $d_1$!) such that that second something has exactly one 'element' ... so, the singleton axiom is satisfied. And, the axiom of extensionality is satisfied ... there is only one object that has the same 'elements' 'inside' it. OK, so this is one model $\endgroup$ – Bram28 Dec 26 '18 at 21:59
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    $\begingroup$ @Studentu OK, now consider the following: we have two obejcts, $d_1$ and $d_2$. we have $d_1 \in d_2$, and $d_2 \in d_1$ ... a '2-cycle' of the kind Andres was talking about. Do you see how this satisfies the axioms as well? And then of course you can do 3 objects, and have $d_1 \in d_2$, $d_2 \in d_3$, and $d_3 \in d_1$ .... $\endgroup$ – Bram28 Dec 26 '18 at 22:01

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