# model theory - existence of a model

Let $$\Gamma$$ be a consistent set of $$L$$-sentences with infinite cardinality. If it has an infinite model, then there exists a model for $$\Gamma' =\Gamma \cup \{\lnot c_a = c_b : a \neq b \},$$ where $$c_i$$ is a constant symbol.

How would I go about proving this? I'm imagining that I can extend the model for $$\Gamma$$ to have a larger domain (to include more constant symbols) but I don't know how to show that it is actually a model for $$\Gamma'$$. Maybe with compactness?

• Yes, compactness. – Asaf Karagila Dec 8 '18 at 20:43
• Yes, with compactness. But it should be assumed that the constant symbols $c_i$ do not appear in $\Gamma$, otherwise the statement is not true. – Alex Kruckman Dec 8 '18 at 20:44
• @AlexKruckman Why is that assumption necessary? – bof Dec 8 '18 at 20:45
• For example, in the vocabulary $(c_a,c_b)$, $\Gamma$ could be the set $\{c_a = c_b\}$. – Alex Kruckman Dec 8 '18 at 20:48
• @AlexKruckman Thanks, that makes sense. Now, compactness requires that every finite subset of $\Gamma'$ has a model but I'm not sure how I would go about proving that for every subset. Any hints? – bof Dec 8 '18 at 20:53

A finite subset of $$\Gamma'$$ is a finite subset of $$\Gamma$$, together with the assertions that finitely many $$c_1,\dots,c_n$$ are pairwise distinct.
Take any infinite model of $$\Gamma$$ (which exists by assumption), interpret $$c_1,\dots,c_n$$ as distinct elements, and interpret the rest of the constant symbols arbitrarily (we are free to do this, since the constant symbols are not mentioned in $$\Gamma$$). This is a model of our finite subset of $$\Gamma'$$. That's all there is to it.
• I'm sorry but I have another question. Why can't we take any infinite model of $\Gamma$ and interpret $c_i$ as distinct elements and use it for the model of $\Gamma'$? What's special about finite subsets? – bof Dec 8 '18 at 21:19