Example of complete formula on constants.

Question

I have question considering example of complete formula, that was presented in Keisler/Chang - Model Theory:

2.3.1. EXAMPLES

(1). Let $T$ be a complete theory and let $c_0,c_1,c_2,\ldots$ be constant symbols of $\mathscr{L}$. Then any formula of $\mathscr{L}$ of the form $$x_0\equiv c_0\wedge x_1\equiv c_1\wedge\cdots\wedge x_n\equiv c_n$$ is complete in $T$. If $\mathfrak{U}$ is a model of $T$ such that every element of $A$ is a constant, then $\mathfrak{U}$ is an atomic model.

Why so? Am I right, when I think, that if theory is complete than -> every of it's formulas is complete too? So that formula with conjunction too?

Also, I pin basic definitions about it:

Consider a complete theory $T$ in $\mathcal{L}$. A formula $\varphi(x_1\ldots x_n)$ is said to be complete (in $T$) iff for every formula $\psi(x_1,\ldots x_n)$ exactly one of $$T\vDash\varphi\rightarrow\psi,\quad T\vDash\varphi\rightarrow\neg\psi$$ holds. A formula $\theta(x_1\ldots x_n)$ is said to be completable (in T) iff there is a complete formula $\varphi(x_1\ldots x_n)$ with $T\vDash\varphi\rightarrow\theta$. If $\theta(x_1\ldots x_n)$ is not completable it is said to be incompletable.

A theory $T$ is said to be atomic iff every formula of $\mathcal{L}$ which is consistent with $T$ is completable in $T$. A model $\mathfrak{U}$ if is said to be an atomic model iff every $n$-tuple $a_1,\ldots,a_n\in A$ satisfies a complete formula in $\mathrm{Th}(\mathfrak{U})$.

Answer 1

$T$ is complete iff for every sentence $\phi$ either $T \models \phi$ or $T \models \lnot \phi$. This requirement only applies to sentences, i.e., closed formulas: if $\phi = \phi(x_1, \ldots, x_n)$ has free variables then $\phi$ might hold under some interpretations of $\phi$ but not under others. E.g., consider $\phi(x, y) = x < y$ in the theory of a dense linear order without endpoints.

The formula $x_0\equiv c_0\wedge x_1\equiv c_1\wedge\cdots\wedge x_n\equiv c_n$ is complete in the sense of the definition you give, because for any $\phi$, $T \models x_0\equiv c_0\wedge x_1\equiv c_1\wedge\cdots\wedge x_n\equiv c_n \to \phi$ iff $T \models \phi(c_1, \ldots, c_n)$. Now, $\phi(c_1, \ldots, c_n)$ is a sentence, so if $T$ is complete either $T \models \phi(c_1, \ldots, c_n)$ or $T \models \lnot\phi(c_1, \ldots, c_n)$ and so either $T \models x_0\equiv c_0\wedge x_1\equiv c_1\wedge\cdots\wedge x_n\equiv c_n \to \phi$ or $T \models x_0\equiv c_0\wedge x_1\equiv c_1\wedge\cdots\wedge x_n\equiv c_n \to \lnot\phi$.

If every element of a model $\mathfrak{A}$ is a constant, then every $n$-tuple, $a_1, \ldots a_n$ satisfies the formula $x_1\equiv a_1\wedge x_1\equiv a_2\wedge\cdots\wedge x_n\equiv a_n$, which we have just shown is complete. Hence such an $\mathfrak{A}$ is an atomic model according to the definition you have quoted.

Answer 2

I changed my answer to myself:

Thus $T$ is complete, therefore it's consistent therefore it has model. So, it's sufficient to show, that $T \cup x_0\equiv c_0\wedge x_1\equiv c_1\wedge\cdots\wedge x_n\equiv c_n$ also will be consistent (and from that complete).

Suppose it isn't true.

Then, $T \models \forall x_0...\forall x_n \neg(x_0 \equiv c_0\wedge x_1\equiv c_1\wedge\cdots\wedge x_n\equiv c_n)$ or $T \models \forall x_0...\forall x_n (\neg x_0 \equiv c_0\vee \neg x_1\equiv c_1 \vee \cdots\vee \neg x_n\equiv c_n)$

Interpretation $I^{\mathfrak{A}}_{}$ of $c_{i} := t_{2i}$, as a term from signature in the model $\mathfrak{A} = <A,I^{\mathfrak{A}}_{}>$ is $I^{\mathfrak{A}}_{}[c_i] = a_{i}$.

Some labeling $\gamma_{j}$ of $x_{i} := t_{1i}$, as a term, in model $\mathfrak{A} =<A,I^{\mathfrak{A}}_{}>$ is $t^{\mathfrak{A}}_{1i}[\gamma_{j}] = \gamma_{j} x_{i} = a_{1i}$.

So, thus in the last formula with all variables bounded, it's must be true on all labelings $\gamma_{j}$ and on all of our variables.

But for all $i \in {0,...,n}$, $x_{i}$ takes all values from $A$, and eventually it takes $a_i$ in some labeling $\gamma_{j}$.

The resulting contradiction completes the proof.