I am stuck at this problem.
Let $\Sigma$ be a (finite/ infinite) set of logical expression (I.e. strings of the form $(P\land Q)$ or $\lnot(P\lor \lnot (Q\land R))$ etc.).
That satisfies the following two properties:
(1) for all $\alpha,\beta\in\Sigma$, $\alpha \Rightarrow \beta$ or $\beta \Rightarrow\alpha$.
(2) for each $\alpha\in \Sigma$ there exists a model $M$ such that $M \vDash \alpha$.
(The symbol $\Rightarrow$ denotes Logical consequence (or implication), That is, if $\alpha$ and $\beta$ are logical expressions such that $\alpha \Rightarrow \beta$, then it must be the case that for each row in the truth table where $\alpha$ is $T$, $\beta$ must also be $T$. And if $\alpha \nRightarrow \beta$, then it must be the case that there exist a row in the truth table where $\alpha$ is $T$ and $\beta$ is $F$).
Prove that there exists a model $M$ such that for all $\alpha\in\Sigma$ we have $M\vDash\alpha$.
I tried to prove it by contradiction but I got stuck since the set $\Sigma$ can be uncountable. Maybe the compactness theorem can help.
Here's what I've done so far:
Suppose by contradiction that:
(*) for each model $M$ there exists $\alpha\in\Sigma$ such that $M\nvDash\alpha$
Now we will try to show that there exists a model $M$ such that for all $\alpha\in\Sigma$ we have $M\vDash\alpha$ and so we will get a contradiction
if $\Sigma\neq\emptyset$ is a finite set (similiar proof if $\Sigma$ is countable).
Let $\alpha\in\Sigma$ we get by (2) that there exists a model $M_1$ such that $M_1\vDash \alpha$
Now by (*) we get that there exists $\beta\in\Sigma$ such that $M_1\nvDash \beta$.
Now we will show that $\alpha\nRightarrow\beta$
if $\alpha\Rightarrow\beta$ then we will get that for each model $M$ such that $M\vDash\alpha$ we get $M\vDash\beta$, and so in particular we will get that for the model $M_1$ we have that $M_1\vDash\beta$ since $M_1\vDash\alpha$ which contradicts the fact that $M_1\nvDash\beta$ and so we must get that $\alpha\nRightarrow\beta$.
Now by (1) we get that it must be the case that $\beta\Rightarrow\alpha$ and so we get that for each model $M$, if $M\vDash\beta$ then $M\vDash\alpha$.
No since $\beta\in\Sigma$ we get by (2) that there exists a model $M_2$ such that $M_2\vDash\beta$ and so we get from the fact $\beta\Rightarrow\alpha$, that it must be the case that $M_2\vDash\alpha$.
By contiueing in this process we can add more and more elements from the set $\Sigma$ such that we will get at the end that we have found a model $M$ that satisfies the property:
for all $\sigma\in\Sigma$ we have $M\vDash\sigma$
which contradicts the supposition (*).
And so we must have that:
there exists a model $M$ such that for all $\alpha\in\Sigma$ we have $M\vDash\alpha$.
As was to be shown.
Q.E.D.
Thanks for any hint/help.