First order symbolic logic question regarding model/domain

Question

Specify a model in which the sentence is true and another model in which it is false. The domain of the model must be {1,2,3}.

$ \exists y \forall x ((F(x) \iff x = y) $

I want to confirm my understanding for this problem to see if I have it right. So this is an existential closure where x and y are bound. I believe it reads "there is a y such that for all x, the function of x [output] if and only if x equals y".

So in my model $ M=<D,I> $ such that D is the domain and I is the interpetation such that for every $ a \in D $.

Would letting y be a natural number, x be an integer and F(x) be positive work as a sentence being true? I believe it would read:

"There exists a natural number y such that for all integers x, F(x) is positive if and only if x is y (the integer x is natural)".

Or do am I not allowed to have large sets of numbers due to the domain being restricted?

Answer 1

Would letting y be a natural number, x be an integer and F(x) be positive work as a sentence being true? I believe it would read:

No, not in any way.   You can not declare types.   $x,y$ are unrestricted in the statement, so they belong to the same implicit domain, of which is you were instructed to use $\{1,2,3\}$.

Also all members of that domain are positive, so that won't work at all.

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The sentence reads: "There exists some $y$ in the domain, for all $x$ in the domain, where $F(x)$ is equivalent to $x=y$."

That is to say: "There is some value in the domain where $F(x)$ will be true if and only if the argument, $x$, equals that value."

Hmm...

Answer 2

That model would actually make the sentence false, since there are multiple natural numbers that are positive, so there is not one that is equal to all of them.

To make a model in which the sentence is true, just consider a domain with exactly one object, and where that one object has property $F$, e.g. you could take as the domain $D = \{ 1 \}$, and still use $F(x)$: '$x$ is a positive number'