Why Can we work with $M$ model countable transitive model of some finite fragment of $\mathrm{ZFC}$ and why is it exist.? When we say that let $M$ be a countable transitive model of some finite
fragment of $\mathrm{ZFC}$. 
Why Can we work with $M$ model countable transitive model of some finite
fragment of $\mathrm{ZFC}$ and why is it exist.?
someone can explain these questions. Where can I find information about these questions.
thanks
 A: If $T$ is a finite fragment of $\mathsf{ZFC}$, then by the reflection theorem there are infinitely many ordinal $\alpha$ so that $V_\alpha \models T$. Then by the Downward Loweinheim Skolem theorem, there is a countable elementary substructure $N \prec V_\alpha$ so $N \models T$. Let $M$ be the Mostowski collapse and $\pi : N \rightarrow M$ be the Mostowski isomorphism. $M$ is then a countable transitive model of $T$. 
The question "Why Can you work with $M$ model countable transitive model ... " depends on exactly what you are doing?
I presume you are using countable transitive models for some type of consistency result in conjunction with forcing arguments. 
For example, if you are trying to show $\mathsf{ZFC + CH}$ is consistent, you could start off by assuming it as not consistent. Then there is some finite $T \subseteq \mathsf{ZFC}$ so that $T \cup \{\mathsf{CH}\}$ proves a contradiction (since proofs are finite). (In order to get forcing to work, you may need to extend $T$ to a bigger finite theory $T'$) Then by the above argument, let $M$ be a countable transitive model of $T'$. Then Cohen technique of forcing shows that if $M \models T'$, then the forcing extension $M[G] \models T \cup \mathsf{CH}$. But this is a contradiction since you produced a model of $T \cup \mathsf{CH}$ even though you assumed $T \cup \mathsf{CH}$ was inconsistent. 
