Let $\mathcal{L}$ be any language of predicate logic, $\Sigma_0$ a consistent theory in $\mathcal{L}$. Let P be the set of all consistent theories $\Sigma \supseteq \Sigma_0$ in $\mathcal{L}$.
With the relation $\subseteq$ P becomes partial ordered.
I'm trying to show that
1) every chain in P (i.e., every through $\subseteq$ total ordered subset K$\subseteq$ P) is limited. (I.e. for every chain $K \subseteq P$ there is $\Sigma^{\ast} \in P$ such that for all $\Sigma \in K$ $\Sigma \subseteq \Sigma^{\ast}$ holds.
The suggestion ist to check, if the constructed set $\Sigma^{\ast}$ really is in P and to not forget that not every chain is orderisomorphic (I hope that's the correct English word for it) to $(\mathbb{N}, <)$;
2) if $\Sigma \in P_{\mathcal{L}}$ is maximal (i.e. there's no $\Sigma' \supseteq \Sigma$ in P), then $\Sigma$ is complete;
3) conclude with the Lemma of Zorn that for every consistent theory there is a complete consistent theory, which is superset of it.
My problem lies within point 1). Neither do I see intuitively how this holds nor do I have an idea of how to proof it.
3) Will be easy once 1) and 2) are shown, I think.
So I'm looking forward to your ideas for 1) and 2).
Best Answer
For part one, you want to take $\Sigma^*$ to be the union of the chain $K$. You need to show it is a consistent theory extending $\Sigma_0.$ That it extends $\Sigma_0$ is obvious. If it were inconsistent, the inconsistency would come from some finite number of sentences $\varphi_1,\ldots \varphi_n\in \Sigma^*$. For each $\varphi_i,$ choose a $\Sigma_i\in K $ such that $\varphi_i\in \Sigma_i.$ Then take $\Sigma$ to be the maximum of $\Sigma_1,\ldots, \Sigma_n.$ Then $\Sigma\in K,$ and is inconsistent, contradicting the fact that $K$ only contains consistent sets of statements.
I’m not sure what to make of the suggestion that $K$ is not necessarily order isomorphic to $\mathbb N.$ I guess maybe people in the past have written the above proof making a tacit assumption that it is and they want to warn you against that.
For 2, show that if $\Sigma$ is not complete, then there is a sentence $\varphi\notin \Sigma$ such that $\Sigma\cup\{\varphi\}$ is consistent.