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I've been musing over this problem over the past few days, and believe I have an answer. However, I am still a bit shaky with some of the definitions I am using, and would appreciate if anyone could either confirm or point out errors in what I have. The problem reads: Let $\delta$ be the conjunction of the following sentences:

• $(\forall x)\lnot Exx$;
• $(\forall x)(\forall y)(\forall z)\big((Exy \land Eyz) \to Exz\big)$;
• $(\forall x)(\forall y)\big(x\ne y\to(Exy \lor Eyx)\big)$;
• $(\forall x)(\exists y)(\exists z)(Exy \land Ezx)$;
• $(\forall x)(\forall y)\big(Exy \to (∃z)(Exz \land Ezy)\big)$.

Show that $\operatorname{Cn}(\delta)$ is complete.

My thought process was as follows: I believed $\delta$ to be the axiomatization of a dense linear order without endpoints. Enderton (p155) states that a set $T$ of sentences is a theory iff $T=\operatorname{Cn}(T)$. I then thought I must prove that $\operatorname{Cn}(\delta)$ is in fact the theory of dense linear order without endpoints. If this is the case, I do know how to use a back and forth argument to show the theory is $\aleph_0$ categorical, then apply the Łoś-Vaught test to get completeness.

Any guidance would be appreciated!

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Yes, Andre, that is a mistake. It should read N_0 categorical –  Nik Kumar Dec 10 '12 at 2:01

1 Answer 1

up vote 2 down vote accepted

The first three clauses of $\delta$ say that $E$ is an irreflexive, transitive, total binary relation and hence a strict linear order; the fourth says that it has no endpoints; and the last says that it’s dense. Thus, you’re quite right: $\delta$ is an axiomatization of the theory of dense linear orders without endpoints. (I’m not entirely sure whether that confirmation is all that you needed, but it not, I’m not sure what more is needed.)

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Brian, my main concern was whether not showing the theory of dense linear orders without endpoints is equivalent to showing Cn(δ) is complete. I'm not sure if I'm getting overly bogged down in definitions –  Nik Kumar Dec 10 '12 at 19:00

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