Upper bound on number of models of a theory of a given cardinality

94 Views Asked by At

Let $T$ be any theory and $\lambda$ any uncountable cardinal. Then $I(\lambda , T)$ denotes the number of non-isomorphic models of $T$ of cardinality $\lambda$.

I've seen a number of sources assert that $I(\lambda , T) \leq 2^\lambda $ in general. Why is this true?

1

There are 1 best solutions below

0
On BEST ANSWER

Assuming $T$ is a theory over a language with $\kappa$ symbols, a model of $T$ with an underlying set $S$ of size $\lambda$ consists of $\kappa$ subsets of $S^n$ for certain integers $n$. There are $2^\lambda$ choices for each of these subsets, so there are at most $(2^\lambda)^\kappa=2^{\lambda\kappa}$ such models with underlying set $S$. If the language is countable so $\kappa\leq\aleph_0$ (or more generally if $\kappa\leq \lambda$), $2^{\lambda\kappa}=2^\lambda$ so there are at most $2^\lambda$ such models.

(The bound on $\kappa$ is necessary. For instance, if you have $\kappa$ unary relation symbols there are at least $2^\kappa$ non-isomorphic models of any nonzero cardinality since each relation can be either satisfied by every element or not satisfied by every element.)