Existance of an (in)finite theory having infinite model Please help me to study the following simple cases:
Let $P$ be a binary predicate symbol. I am trying to find out, if there exists a satisfiable $T$ having infinite models only, for the following cases:
1) T is under {P} language with "="
2) T is under {P} language with "=" and T is finite
3) T is under {P} language without "="
4) T is under {P} language without "=" and T is finite
Where I am:
1)
$\varphi_i \equiv \forall x_1 \forall x_2 ... \forall x_i \exists y \land ^i_{j=1} \lnot( x_j=y)$
$T = \{\varphi_i | i>0 \}$ so there exists a T having only infinite models
2) 4) can probably have use of Löwenheim-Skolem theorem, but I don't see how to apply it properly.
3) No adequate idea so far

To make Mike's answer just a little more explicit:
When $P$ is the inequality symbol of a partial order, it interprets equality in the sense that the definable set $\{m \in M \operatorname{ | } P(m,m)\}$ is exactly the diagonal relation $\{m \in M \operatorname{ | } m = m\}$.
In particular, we know that such a $P$ must satisfy: $$\forall a \forall b, P(a,b) \land P(b,a) \rightarrow P(a,a),$$
$$\forall a \forall b \forall c, P(a,b) \land P(b,c) \rightarrow P(a,c),$$
$$\forall a, P(a,a).$$
After writing these out, can you think of just one more axiom we can add to $T$ such that any chain in our partial order must be infinite? (Hint: given an element $a$ in a partial order $M$, the set of all elements $b$ less than $a$ is $a$-definable.)
