First order logic,Step in Derivation of Non standard real numbers This is from first order logic, specifically a section detailing the construction of the non standard real numbers, after Los' theorem
And we have that:\
\
Let $L = \{+, ×, <, 0, 1\}$ be the language we usually use for this structure.
Let $M = R$. Let I be the usual interpretation of L in R. Let $M = (M, I)$.
For each $r \in R$, let $c_r$ be a new constant symbol. Do this in such a way that, for all $r, r' \in R, c_r \neq c_r' $ and $c_r \notin \{ 0, 1\}$. 
Let $L_R = L \cup \{c_r : r \in R\}.$
Let $I_R$ be the natural interpretation of $L_R$ in R (so $I_R$ agrees with I on L and, for each r $\in$ R, $I_R(c_r) = r$). Let $M_R = (R, I_R)$.
Finally, let $b \notin \{c_r : r \in R\} \cup {0, 1}$ be a new constant symbol and let
$L_b = L_R \cup \{b\}$. For each $a \in R$, let $I_a$ be the interpretation of $L_b$ in R
which agrees with $I_R$ on $L_R$ and is such that $I_a(b) = a$. For each $a \in R$, let $M_a = (R, Ia)$.
Let $\Sigma$ be the set of all $L_b$-sentences $\sigma$ such that
\begin{eqnarray}
(i) & \sigma \in L_R \wedge M_R \models \sigma \quad or \nonumber \\
(ii)& \sigma \in \{(< (0, b)) \wedge (< (b, c_r)) : r \in \{ \frac{1}{n}: n \in N\}\}.
\end{eqnarray}
Let $\Sigma ' \subseteq \Sigma$ be finite. Let $a(\Sigma ')$ be r where r is the smallest positive
real number such that $c_r$ appears in at least one of the sentences in $\Sigma '$, if such exists. Otherwise, let $a(\Sigma '
) = 2$ (or any other real number you like). 
Show that $M_{a(\Sigma ')} \models \Sigma '$
My approach was to assume that $M_{a\Sigma '} (\neg\models) \Sigma '$
and then try to show that there does exist a satisfiable $\sigma \in \Sigma '$ and then that $\sigma$ is satisfiable and hence $M_{a(\Sigma ')}\models \Sigma '$ But I'm uncertain on whether this is the best approach and what forces the $\sigma$ to be satisfiable.``
 A: Your approach is overcomplicating the task at hand. Instead, just show that $M_{a(\Sigma')}\vDash \sigma'$ for each $\sigma' \in\Sigma'$ separately.
Such a $\sigma'$ will either be something that is true in $M_R$, and therefore also in every $M_a$, or it will assert that $0<b<c_r$ for some $c_r$ where the interpretation of $c_r$ is by construction larger than $a(\Sigma')$.

Actually there's something fishy about the construction which I think looks like typos that are easily fixed but it's not quite clear to me which direction they're supposed to be fixed in.
Your $\Sigma$ adds new axioms $0<b<r$ for some set of $r$s -- which looks like $b$ is supposed to become an infinitesimal. But what you have written is actually only that $b$ must be squeezed between $0$ and $r$ for every $r\in \mathbb N$, which is easily achievable without any nonstandard elements -- as written, $M_{1/2}$ will satisfy all of $\Sigma$.
What must be meant is either to aim for $b$ being infinitesimal and add axioms
$$ 0<b \land b<c_r \qquad\text{for every }r\in(0,\infty) $$
or to aim for $b$ being infinite and add axioms
$$ c_r <b \qquad\text{for every }r\in\mathbb N $$
The definition of $a(\Sigma')$ that works in each of those cases will be either (say) half of the smallest positive $r$ such that $\Sigma'$ mentions $c_r$, or (say) one plus the largest $r$ such that $\Sigma'$ mentions $c_r$.
