Confusion on Proof of Marker Lemma 2.1.14, $T \vDash \phi \implies \exists \Delta \subseteq T, |\Delta| \in \mathbb{N} \land \Delta \vDash T$

Question

Here is lemma 2.1.14 from Marker's Model Theory restated exactly,

$$ \textit{If $\ T \vDash \phi$, then $\Delta \vDash T$ for some finite $\Delta \subseteq T$.} $$

To make sure I understand it properly I'll restate the lemma in my own words:

Let $T$ be an $\mathcal L$-theory, and $\phi$ and $\mathcal L$-sentence where $T \vDash \phi$. Then there exists a finite subset $\Delta$ of $T$ where $T$ is a logical consequence of $\Delta$, that is, $\Delta \vDash T$.

The provided proof of this lemma is very short, so I'll reproduce that exactly too:

Suppose not. Let $\Delta \subseteq T$ be finite. $\color{red}{\underline{\text{Because $\Delta \not\vDash \phi$}}}$, $\Delta \cup \{\neg \phi\}$ is satisfiable. Thus, $T \cup \{\neg\phi\}$ is finitely satisfiable and, by the Compactness Theorem, $T \not \vDash \phi$.

I've emphasized in red and underlined the portion that I cannot understand. By "Suppose not", we're fixing a $\phi$ and assuming that $T \vDash \phi$, but that for all finite $\Delta$, we have $\Delta \not \vDash T$. That is, for each $\Delta$, there is some $\varphi_{\Delta} \in T$, where $\Delta \not \vDash \varphi_{\Delta}$. I have no idea how we can claim that $\Delta \not \vDash \phi$, since it's fixed at the time of the "Suppose not". The way the proof is structured, when assuming the negative for contradiction, seems to imply that for all finite $\Delta \in T$ we have $\Delta \not \vDash \phi$, which I don't understand.

The other steps of the proof are fine as far as I can see. But I'm not sure if the error is in my misunderstanding or if the error is in the text, as this text has a number of errors.

If someone can clear up this specific step in the proof that would be much appreciated.

Answer 1

As indicated in the comments, this is a typo and the conclusion should be $\Delta\models \phi.$ It says if a sentence is a logical consequence of a theory, then there is a finite subtheory that it is a logical consequence of.

This is easier to understand from the deductive side. If a sentence is provable from $T,$ then since proofs are finite, it only uses a finite number of assumptions from $T,$ and thus it's provable from a finite subtheory. And the completeness theorem says that $T\models \phi$ if and only if $\phi$ is provable from $T.$