Why are maximal consistent sets essential to Henkin-proofs of Completeness? As well-known, the Completeness theorem states that
$$\Gamma \vDash \varphi \Rightarrow \Gamma \vdash \varphi$$
The proof we find in didactic textbooks are usually called "Henkin-proofs". Let $\mathcal{L}$ be our referring language. A Henkin-proof (for propositional logic) goes more or less along the lines of


*

*Let $\Gamma$ be consistent.

*Extend $\Gamma$ to a maximal consistent set $\Delta$

*Show that $\Delta$ preserves consistency and that $\Gamma \subseteq \Delta$

*Define a valuation $v$ for $\Delta$ such that $v(\psi)=1$ iff $\psi \in \Delta$ for all atomic $\psi \in \mathcal{L}$

*Define $v$'s unique extension $\bar v$ as usual.

*Then $\bar v \vDash \Delta$ and, since $\Gamma \subseteq \Delta$,

*$\bar v \vDash \Gamma$.
Now my question is:

Why can't we just define $v$ using $\Gamma$ directly?

I'm obviously missing something, but why can't we simply forget the maximal consistent part? That is, for every atomic $\psi \in \mathcal{L}$, define:
$v(\psi) = \begin{cases}
1 & \text{if $\psi \in \Gamma$} \\
0 & \text{if $\psi \notin \Gamma$}
\end{cases}$
What is the problem with this definition? Is it that we don't know the elements of $\Gamma$? But if it is so, why wouldn't this apply to $\Delta$? In other words, what is the essential point of extending $\Gamma$ to a maximal consistent set?
Thanks in advance.
 A: You give a method for producing a valuation, $\nu$, corresponding to a theory, $\Gamma$. This is well-defined, but $\nu$ might not make $\Gamma$ true! For instance: what happens if $\Gamma=\{\psi_0\iff\neg\psi_1\}$ for some atomic $\psi_0, \psi_1$? Then your valuation would make both $\psi_0$ and $\psi_1$ false, so $\Gamma$ would not hold!
Essentially, in the example above, we need to make a choice between making $\psi_0$ true and making $\psi_1$ true. The point of extending to a complete theory is exactly to make these sorts of choices.
In fact, if you prefer, we can phrase the argument in terms of making choices, instead of forming a completion. Given $\Gamma$, we list the atomic propositions as $\psi_0, \psi_1, . . .$; at stage $\alpha$, we have built a partial valuation $\nu_\alpha$ of the first $(\alpha-1)$-many atomic propositions. We then extent $\nu_\alpha$ to $\psi_\alpha$ as follows: $\nu_{\alpha+1}(\psi_\alpha)=1$ if for every finite subset $\Gamma_0$ of $\Gamma$, there is a valuation $\mu$ which makes $\Gamma_0$ true, extends $\nu_\alpha$ and such that $\mu(\psi_{\alpha})=1$; and we make $\nu_{\alpha+1}(\psi_\alpha)=0$ otherwise. If you unpack this, of course, this is really just forming a completion of $\Gamma$, but it might seem more intuitive.
A: I usually think about this theorem in terms of the contrapositive. The completeness is equivalent to:
$$ \Gamma \not\vdash \varphi \Longrightarrow \Gamma \not\vDash \varphi$$
So $\Gamma$ doesn't need to be consistent, but rather non-trivial with respect to $\varphi$ - that is, you can just assume it not to derive $\varphi$. This is a standard argument in non-classical logics dealing with contradictory yet non-trivial systems.
Then you strive to prove that such $\Gamma$ can be maximally extended to some $\Gamma'$ such that also $\Gamma' \not \vdash \varphi$ and expect to be able to obtain a model of $\Gamma'$, which will be a model of $\Gamma$ also. By the maximality of $\Gamma'$, $\varphi$ cannot be satisfied by this model.
You can think of the maximality of $\Gamma'$ as both a way to set the truth of all sentences (there cannot be room for ambiguities in any model) and to enforce the model obtained not to satisfy $\varphi$, as this would imply that $\Gamma' \vdash \varphi$.
I'm being vague on purpose, as this argument can be used also for non-classical logics and the exactly nature of model and satisfaction would vary.
