The cardinality of a language over a set of variables $V$ with $\#V= κ$

Question

In predicate logic, to define a language, we consider the following $3$ sets:

(1) $V$, the set of variables;

(2) $F$, the sets of function symbols;

(3) $P$, the set of predicate symbols with their arities.

Now I am concidering about determine the cardinality of a language over a set with arbitary cardinality, $κ$(Note that $κ$ can be either finite, countablely infinite or uncountablely infinite ).

May I please ask:

If the cardinality of $V$, $\#V = κ$, what conditions on $V$ , $F$ and $P$ will give us a language over $V$, $F$ and $P$ with cardinality $κ$?

And may I please ask if it is the fact that if $\kappa$ is finite, then to get a language with cardinality $\kappa$ we need $(V+F+P)^{<\omega}=(\kappa+F+P)^{<\omega}=\kappa$, so we can never have a language of cardinality $\kappa$. I think it is something weird here. Am I wrong with something?

I had googled it but I have get nothing from websites, now I am confused. Could someone please help? Thanks so much!

EDIT: Latter I am told that in my course, the cardinality of a language refers to the set of expression instead of this of the set of strings. So now I need to consider further about when a string makes sense.

EDIT':Now I am considering that if we can find out a general way to determine the cardinality of expression, which may related to combinatorics. See this question: In general, how to determine the cardinality of the set of expression over a language? Any help would be appreciate.

Answer 1

If $\kappa=0$, then the language is empty if there are no nullary function symbols (= constant symbols), and otherwise has nonzero (hence $\not=\kappa$) size. If $\kappa\ge 1$ but $\kappa$ is finite, then the language is infinite (no nonempty language in the sense of the question is finite) and hence it is not of size $\kappa$.

So all that's left is $\kappa$ infinite, and it turns out that the issue of (un)countability is irrelevant here. Each formula in the language is a finite string built out of symbols from $V, F, P$, and the logical symbols (of which there are finitely many). Since $V$ is infinite, the cardinality of the set of formulas is thus $(V+F+P)^{<\omega}$.

Now clearly if $F$ or $P$ has size $>\kappa$, we can't have the language have size $\kappa$. Meanwhile, for infinite $\kappa $, $(3\kappa)^{<\omega}=\kappa $. So the language has size $\kappa$ exactly when F and P have size at most $k $.

Answer 2

If $\Sigma$ is an alphabet of infinite cardinality, then the language $\Sigma^\star$ of finite strings over $\Sigma$ has the same cardinality as $\Sigma$.