We know set A is countable if A is finite or in a one-to-one mapping to natural numbers.
I try to summarize my though. I think the following proposition is true. suppose $\Sigma$ is arbitrary alphabet. every one would please help me and add some hints for each one, or if I'm wrong correct me !! thanks to all.
1) Each arbitrary Language on $\Sigma$ is Countable.
2) the set of all language from $\Sigma$ is Countable.
3) for Each arbitrary Language on $\Sigma$ we have a generative formal grammar.
4) Each arbitrary Language on $\Sigma$ that generated by formal grammar, is recursive.
If $\Sigma$ is any finite alphabet (or even countably infinite), then $\Sigma^*$ is countable, hence so is any language on $\Sigma$
Let $a\in\Sigma$ be an arbitrary letter. Then for any subset $A\subseteq\mathbb N$, we have the language $\{\,a^n\mid n\in A\,\}$. As there are uncoutably many subsets of $\mathbb N$, there are uncountably many languages.
Grammars are allowed to have only finitely many rules and metavariabables, hence there are only countably many grammars - less than there are languages.
Indeed, one can transform a grammar into a Turing machine that shows recursiveness.