Logic: $\text{Mod } \Sigma$, the class of all models of $\Sigma$ and $\text{Th Mod }\Sigma$, how do these relate? While reviewing a question I had asked earlier here:
Proving and understanding the Fixed point lemma (Diagonal Lemma) in Logic - used in proof of Godel's incompleteness theorem
I have the following:
"$\text{Mod } \Sigma$ is the class of all models of $ \Sigma$. $\text{Th Mod } \Sigma$ is the set of all sentences which are true in all models of $\Sigma$.  This however is just the set of all sentences logically implied by $\Sigma$.  We call this set the set of consequences of $\Sigma$ or $\text{Cn }\Sigma$.  Thus we have that $\text{Cn }\Sigma = \{\sigma \mid \Sigma \models \sigma \} = \text{Th Mod } \Sigma$."
Letting $\Sigma$ denote a set of axioms. Since $\text{Th Mod } \Sigma$ is the set of all sentences which are true in all models of $\Sigma$ then this seems to imply that there are sentances which may not be true in all $\text{Th Mod } \Sigma$.  So in other words $\text{Th Mod } \Sigma$ is derivable from $\Sigma$ i.e. we can build some sentence in $\text{Th Mod } \Sigma$ from axioms in $\Sigma$.  I however am trying to understand better: $\text{Th Mod } \Sigma$ is implied from $\Sigma$.  This would mean that  $\Sigma \Rightarrow \text{Th Mod } \Sigma$.  Now this is always true (is a Tautology) as the only case which would be false would be that something derived from a set of axioms (or just a set of axioms?) in $\Sigma$ implying $\text{Th Mod } \Sigma$ would be false as all statements in $\text{Th Mod } \Sigma$ are true by definition.  I am thinking this explains their relationship, but could use some confirmation and maybe some more insight.  
Thanks,
Brian
 A: Suppose that $\Sigma$ is a consistent set of axioms, and that $\sigma$ is a sentence that cannot be derived from $\Sigma$. Then $\Sigma\cup\{\neg\sigma\}$ is consistent, so it has a model $\mathfrak{M}$. Clearly $\mathfrak{M}\vDash\Sigma$, so $\mathfrak{M}\in\operatorname{Mod}\Sigma$, but also $\mathfrak{M}\vDash\neg\sigma$. Thus, $\sigma$ is not true in $\mathfrak{M}$, and therefore $\sigma\notin\operatorname{Th}\operatorname{Mod}\Sigma$.
As a simple example, consider a language with one binary relation symbol $E$, and let $\Sigma$ be the set consisting of the following axioms:
$$\begin{align*}
&\forall x R(x,x)\\
&\forall x,y\big(R(x,y)\leftrightarrow R(y,x)\big)\\
&\forall x,y,z\big(R(x,y)\land R(y,z)\to R(x,z)\big)
\end{align*}$$
These simply say that $R$ is an equivalence relation. Let $\sigma$ be the sentence $$\exists x,y,z\Big(x\ne y\land x\ne z\land y\ne z\land \forall u(u=x\lor u=y\lor u=z)\Big)$$ saying that there are exactly three objects. Let $A=\{0,1,2\}$ and $B=\{0,1,2,3\}$, and let $R_A$ and $R_B$ be equivalence relations on $A$ and $B$, respectively. Let $\mathfrak{A}=\langle A;R_A\rangle$ and $\mathfrak{B}=\langle B;R_B\rangle$. Then $\mathfrak{A}\vDash\Sigma\cup\{\sigma\}$ and $\mathfrak{B}\vDash\Sigma\cup\{\neg\sigma\}$, so $\mathfrak{A},\mathfrak{B}\in\operatorname{Mod}\Sigma$, and neither $\sigma$ nor $\neg\sigma$ is in $\operatorname{Th}\operatorname{Mod}\Sigma$. This should be no surprise, because nothing in the axioms for an equivalence relation implies either that the underlying set does have cardinality $3$ or that it does not.
