We define $ \Gamma = \{~ \forall x_1 \neg (F(x_1) = c ), ~ \forall x_1 \forall x_2 ((F(x_1) = F(x_2)) \to (x_1 = x_2 )) ~ \} $.
Note: All of this is in what's denoted $ L $, or first-order logic, with a constant symbol denoted $ c $, with $ F $ being a 1-ary function defined on a respective domain.
Firstly, I am required to give an infinite model of $ \Gamma $.
My answer: $ ~ $ We let $ D_{I} = \mathbb{R}^+ \cup \{0\}, $ and define our constant $ c = 0 $, and let our function $ F(x_n) = x +1 $. This is correct.
Now, my question: Is there a finite model of $ \Gamma $?
If there were to be a finite model of $ \Gamma $, it seems that would imply that the cardinality of the domain would have to be greater than that of the range, which would be impossible granted that your function is one-to-one, as it indeed is defined as such.
Thank you so much in advance!
$\Gamma$ has no finite model. This is because $\forall x_1. \forall x_2. F(x_1) = F(x_2) \rightarrow x_1 = x_2$ states that $F$ is an injective function. An injective function from a finite set to itself is in fact a bijection, so $\forall y. \exists x. F(x) = y$, and in particular $\exists x. F(x) = c$.
Are you sure you defined your infinite model correctly? If you define $F(x) = x^2$, then $F(0) = 0^2 = 0 = c$, so $\exists x. F(x) = c$ again.