@MartinSleziak I proved it by doing (not II) => (not I). If there is no model then we have vacuously T makes all sentences true. So it makes true $\varphi$ and $\lnot \varphi$.
Let $\Sigma$ be a set of formulas of $L$ and let $\varphi$ be a formula of $L$. Then $\Sigma+\{\varphi\}$ is consistent iff $\Sigma\not\vdash\not\varphi$.
If $T$ is consistent, then there is a formula $\varphi$ such that $T\not\vdash\varphi$. By version I, this implies $T\not\models\varphi$. For the later to hold, there must be at least one model of $T$.
@MartinSleziak Ah, right. $T \models \varphi \iff \varphi$ is true in every model of $T$. Hence negating both sides gives you "exists a model such that $\varphi$ is not true in it" on the right hand side.
For the other direction I did a direct proof of (I) => (II): Let $T$ be a theory such that for every sentence $\varphi$ we have $T \models \varphi$ implies $T \vdash \varphi$. Let $T$ be consistent and assume that it does not have a model.
Then $T \models \lnot \varphi$ (vacuously because it doesn't have a model) and hence by assumption we have $T \vdash \lnot \varphi$ which would be a contradiction to $T$ being consistent.
Now I'm going to do exercise 8.
Let $T$ be a theory.
Claim: $T$ has a model $\iff$ every finite subset $S$ has a model.
$\implies$ Assume $S \subset T$ does not have a model. Then by Gödel's completeness theorem version II, $S$ is not consistent. Hence $T$ is not consistent, that is, $T \vdash \varphi$ and $T \vdash \lnot \varphi$. By the soundness theorem we hence have $T \models \varphi$ and $T \models \lnot \varphi$. But this can only hold if $T$ does not have a model.
@MattN $\Longleftarrow$ Let $T$ be such that every finite subset $S$ has a model. Assume $T$ does not have a model. Then for all sentences $\varphi$ in $T$ we have $T \models \varphi$ and $T \models \lnot \varphi$. $S= \{\varphi, \lnot \varphi \}$ is a finite subset of $T$ hence by assumption has a model. But in any mode $M$, $M \models \varphi$ and $M \models \lnot \varphi$ is impossible hence $T$ has to have a model.
One direction of my proof has to be wrong since I don't seem to use that $S$ is finite anywhere.
BTW if you remember, Asaf mentioned recently in the chat that he was giving a lecture which included the proof of compactness theorem using Konig's lemma. (That's a result on infinite trees - I don't know whether you've already studied things like this.)
@MattN You've used finiteness in $\Leftarrow$. Namely, you used the fact that $S=\{\varphi,\neg\varphi\}$ is finite.
@MartinSleziak But if I drop the assumption that $S$ is finite in the statement of the theorem then $S = \{ \varphi, \lnot \varphi \}$ is still a subset of $T$ and the rest of the proof still works even though $S$ is not necessarily finite.
So "If every arbitrary subset of $T$ has a model then $T$ also has to have a model".
Since I've mentioned Konig's lemma, it's in Chapter 14 of this book. (It's 2nd volume.) They also have result about relation to compactness theorem there.
So we have $T \models \varphi$ if and only if $T \vdash \varphi$.
Hence if I want to show that $\varphi$ is consistent with $T$, which is the same as saying $T \nvdash \lnot \varphi$, then it's enough to find a model of $T$ that doesn't make $\lnot \varphi$ true.
Ah, for this to work $T$ has to be consistent.
No, it works even if $T$ is inconsistent.
Found another typo: page 79, penultimate paragraph, "nonabilian".
The whole talk about consistency has to be relative, in a philosophical sense at least, because you always at least assume the consistency of first order logic.
This form of compactness theorem: If $T$ is a theory in a first-order language $L$, then $T$ has a model iff every finite subset $S$ of $T$ has a model.
They formulated there the following weaker version of compactness theorem:
(WCT) Let $L$ be a first-order language without functional symbols, and let $(S_n)_{n\in\omega}$ be an increasing sequence of finite sets of quantifier-free sentences of $L$. If each $S_n$ has a model, then the theory $S = \bigcup_{n\in\omega} S_n$ also has a model.
And they claim that it is equivalent to Konig's lemma (in ZF) - Theorem 14.4.
Another equivalent condition from the same theorem is this:
(WTY) Every product $\prod_{n\in\omega} X_n$ of finite non empty Hausdorff spaces $X_n$ is nonempty and compact.
I would have to check where choice is used in the proof of Konig's lemma - I haven't thought about this before. I'll probably let it be at the moment.
@MattN Which is what I said: $V$ is (informally) the collection of all hereditarily well-founded sets, while the cumulative hierarchy is a way of breaking that collection into sets by rank.
I like the name a professor who taught me used for the cumulative hierarchy: Creating the universe (of sets) from nothing (=from $\emptyset$) There are sets which were created on the 1-st, 2-nd, ..., $\omega$-th, ... day
Matt, I know you don't like handwavy things but I actually added some "reading material" to the handwavy post in which I link to a non-handwavy explanation of the Levy collapse forcing.
@MattN I am not sure how much time I will have during the workdays. (The next week will probably rather busy.) So it's quite probable that I won't have much time for reading the book. But I'll try at least to check every day, whether there is something new in this chatroom.