Let $L$ a countable language and $T$ a complete $L-$theory. Let $p(x)\subseteq L$ a non-isolated consistent type. Is there an homogeneous model that omits the type?
The notion of homogeneity is the following:
We say that $N$ is (elementarily) $\lambda$-homogeneous if every partial elementary map $k\colon N \to N$ of cardinality ${<}\lambda$ extends to an automorphism. We say that $N$ is homogeneous if it is $\lambda$-homogeneous and of cardinality $\lambda$.
I tried to follow the proof of the omitting types theorem with the construction of Löwenheim-Skolem and using the following Lemma:
Let $L(A)$ countable. Let $p(x) \subseteq L(A)$ and suppose that $A$ does not isolate $p(x)$. Then, if $\psi(z)\in L(A)$ is consistent, $\psi(z)$ has a solution $a$ such that $A,a$ does not isolate $p(x)$.
I added the hypothesis in the inductive construction of a homogeneous model, but I can’t understand how to construct the chain.
If this is true, is it still true dropping the hypothesis of completeness of the theory?
First, an easy observation: If $T$ has a countable atomic model (equivalently, if isolated types are dense in the type space $S_n(T)$ for all $n$), then $T$ has a countable homogeneous model omitting $p(x)$, since countable atomic models are homogeneous and omit all non-isolated types.
But in general, it is not possible to find a homogeneous model omitting a given non-isolated type. I will give a counterexample. Thanks to Primo Petri for the simplification suggested in the comments.
Let $L$ be the language consisting of countably many binary relation symbols $(R_n)_{n\in \omega}$. Let $T$ be the $L$-theory asserting that each $R_n$ is a graph relation (symmetric and anti-reflexive). Let $T^*$ be the model companion of the above theory. $T^*$ is sometimes called the theory of "kaleidoscope random graphs".
Someone more concretely, note that for any finite sublanguage $L'\subseteq L$, the class of finite models of $T|_{L'}$ is a Fraïssé class. Writing $T_{L'}$ for the theory of the Fraïssé limit, we have that $T^* = \bigcup_{L'\subseteq_{\mathrm{fin}} L} T_{L'}$. Since each $T_{L'}$ has quantifier elimination, it follows that $T^*$ has quantifier elimination.
Let $p(x,y,x',y')$ be the type $$\{x\neq y, x'\neq y', (x\neq x')\lor (y\neq y'), (x\neq y') \lor (y\neq x')\}\cup \{R_n(x,y)\leftrightarrow R_n(x',y')\mid n\in \omega\}.$$ A realization of $p$ is a $4$-tuple $(a,b,a',b')$ such that $a\neq b$, $a'\neq b'$, $\{a,b\}\neq \{a',b'\}$, and the pairs $\{a,b\}$ and $\{a',b'\}$ agree on all graph relations $R_n$.
I claim that no homogeneous model of $T^*$ omits $p$. Let $M\models T^*$ be a homogeneous model. Let $a$, $b$, and $a'$ be three distinct elements. By quantifier elimination, there is a unique $1$-type over $\varnothing$ relative to $T^*$, so by homogeneity, there is an automorphism $\sigma\in \mathrm{Aut}(M)$ such that $\sigma(a) = a'$. Let $b' = \sigma(b)$. Note that $a\neq b$ implies $a'\neq b'$. Further, since $a'\notin \{a,b\}$, $\{a,b\}\neq \{a',b'\}$. And since $\sigma$ is an automorphism, we have $R_n(a,b) \leftrightarrow R_n(a',b')$ for all $n\in \omega$. So $(a,b,a',b')$ realizes $p$.
It remains to show that $p$ is non-isolated. One way to see this is with a probabilistic construction of models of $T^*$, which may also help motivate what's going on with this example.
Take the domain of our model $M$ to be $\omega$. For each pair $\{a,b\}$ of distinct elements of $\omega$, flip countably many fair coins independently. Set $R_n(a,b)$ (and $R_n(b,a)$, for symmetry) to be true if and only if the $n^{\text{th}}$ coin came up heads.
The resulting structure $M$ is almost surely a model of $T^*$, for essentially the same reason that the infinite Erdös-Rényi random graph is almost surely isomorphic to the Rado graph.
Further, $M$ almost surely omits $p$, which establishes that $p$ is non-isolated. For any $4$-tuple $(a,b,a',b')$ such that $a\neq b$, $a'\neq b'$, and $\{a,b\}\neq \{a',b'\}$, almost surely $\{a,b\}$ and $\{a',b'\}$ differ on some coin flip (the probability that the infinite sequence of coin flips for $\{a',b'\}$ agrees exactly with the infinite sequence of coin flips for $\{a,b\}$ is $0$). So almost surely $(a,b,a',b')$ fails to realize $p$. Since there are only countably many such $4$-tuples, almost surely $p(x)$ is omitted.
Note that structures built in this probabilistic way are almost surely not homogeneous: If $a\neq a'$ in $G$, then almost surely there is no automorphism of $M$ moving $a$ to $a'$, since for any $b$ distinct from $a$ and $a'$, there is almost surely no $b'\neq a'$ such that $\{a,b\}$ and $\{a',b'\}$ agree on all their coin flips. So you can clearly see the tension between homogeneity and omitting $p$.
Caveat: The $p$ in my example is not a complete type. I think one could work a bit harder to find an example of a complete non-isolated type which cannot be omitted in any homogeneous model, but I'll leave that to you.