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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).

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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.

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  • $\begingroup$ Thank you! I'd completely forgotten about the trick to take the union! For 2), we assume that $\Sigma$ is not complete. So there is $\phi$ in $\mathcal{L}$ such that it doesn't hold that either $\Sigma \vdash \Phi$ or $\Sigma \vdash \neg \phi$. We consider $\Sigma':=\Sigma \cup \phi$. We do this with all $\psi$ which cause $\Sigma$ to not be complete until we have the theory $\tilde{\Sigma}$. Then $\tilde{\Sigma}\supset \Sigma$ and $\tilde{\Sigma} \in P$ in contradiction to the maximality of $\Sigma$. But I don't see how to explain the $\tilde{\Sigma}$ is consistent? $\endgroup$ – Studentu Nov 25 '18 at 4:22
  • $\begingroup$ 3) With the Lemma of Zorn we get that there is at least one maximal element in P. Due to 2), this element is also complete. But I don't get how to explain that 3) follows from this. Intuitively, it is clear to me, but if we take the maximal element $\Sigma^{\ast}$ for example, it doesn't have to hold that $\Sigma_0 \in \Sigma^{\ast}$, because it could be another maximal element which $\Sigma_0$ is a subset of. $\endgroup$ – Studentu Nov 25 '18 at 4:23
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    $\begingroup$ For 2 you only need to show that if $\Sigma \nvdash \phi$ and $\Sigma \nvdash \lnot\phi,$ then either $\Sigma\cup\{\phi\}$ or $\Sigma\cup\{\lnot\phi\}$ are consistent (actually, both are). This shows that if $\Sigma\in P$ is incomplete, then it is not maximal in $P,$ which is what you need. For 3, yes it does have to hold that $\Sigma_0\subseteq \Sigma^*$ since all $\Sigma \in P$ have $\Sigma_0\subseteq\Sigma.$ $\endgroup$ – spaceisdarkgreen Nov 25 '18 at 6:11
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    $\begingroup$ The general theorem you can show it that $\Sigma\cup \{\phi\}$ is consistent if and only if $\Sigma \nvdash \lnot\phi .$ For the direction you need, assume $\Sigma\cup \{\phi\}$ is inconsistent then show $\Sigma \vdash \lnot \phi.$ You only need to use very general facts about the system. For instance by the deduction theorem, if $\Sigma\cup \{\phi\}\vdash \psi,$ and $\Sigma\cup \{\phi\}\vdash \lnot\psi,$ then $ \Sigma \vdash \phi\to \psi$ and $\Sigma \vdash \phi\to \lnot \psi.$ Can you show from there that $\Sigma \vdash \lnot\phi$? $\endgroup$ – spaceisdarkgreen Nov 25 '18 at 19:34
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    $\begingroup$ @Studentu Some of that is on the right track, but you need $\Sigma \vdash \lnot\phi,$ not just $\Sigma\cup\{\phi\}\vdash \lnot \phi.$ If you have the $\bot,$ symbol, you might just do $ \Sigma\cup\{\phi\}\vdash\bot,$ and then use the deduction theorem to get $\Sigma \vdash \phi\to \bot.$ How close this is to $\Sigma\vdash \lnot \phi$ depends on the system you're using (in some systems it's the same thing by definition). $\endgroup$ – spaceisdarkgreen Nov 25 '18 at 21:16

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