Prove the intersection of transitive relations is transitive.
Let $\left\{\mathcal\ R_{i}:i\ \in\ \mathcal I\right\}$ be the set of transitive relations on $A$ indexed by $\mathcal I$,it should be shown that:
$$ (\forall i :i\in \mathcal I \implies\mathcal R_i \text{ is transitive}) \implies \bigcap_{i \in \mathcal I}\mathcal R_i \;\text{ is transitive}$$
This can be proved by induction on the number of elements in $\mathcal I$.
Let $\left|\mathcal I\right|=0$,then:
$$ (\forall i :i\in \varnothing \implies\mathcal R_i \text{ is transitive}) \implies \bigcap_{i \in \mathcal I}\mathcal R_i =A \times A $$
Which by the definition of Cartesian product implies $A \times A $ is transitive.
Assume the theorem is true for $\left|\mathcal I\right|=n$ and consider $\left|\mathcal I\right|=n+1$:
Assume $ (\forall i :i\in \mathcal I \implies\mathcal R_i \text{ is transitive}) $ $$\bigcap_{i \in \mathcal I}\mathcal R_i =\bigcap_{i \in \mathcal I \setminus \left\{n+1\right\} }\mathcal R_i \; \cap \mathcal R_{n+1}$$
By the induction hypothesis and the assumption that $\mathcal R_{n+1}$ is transitive it implies that:
$$\bigcap_{i \in \mathcal I \setminus \left\{n+1\right\} }\mathcal R_i \; \cap \mathcal R_{n+1}$$
Is transitive.
Is what have I done true?Also can the index set be an infinite one? And is it necessary for the transitive relations to be over the same set?if yes then why?
Unfortunately, your proof doesn't work. Your inductive proof only shows that if any intersection of $n$ transitive relations is transitive, then any intersection of $n+1$ transitive relations can be represented as an intersection of $2$ transitive relations. However, your argument fails to show that an intersection of $2$ transitive relations is transitive, hence the result doesn't follow.
Also, standard induction can't work for infinite families.
The simplest proof would involve going back to the definition of transitivity: if $(x,y), (y,z) \in R$ then $(x,z) \in \mathcal R$.
Assume that $(x,y),(y,z) \in \bigcap_{i \in \mathcal I} \mathcal R_i$. By the definition of intersection, this means that, for each $i \in \mathcal I$, $(x,y),(y,z) \in \mathcal R_i$. Since each $\mathcal R_i$ is transitive, that means that $(x,z) \in \mathcal R_i$ for each $i \in \mathcal I$. Again, by the definition of intersection,this means that $(x,z) \in \bigcap_{i \in \mathcal I} \mathcal R_i$.
So $(x,y),(y,z) \in \bigcap_{i \in \mathcal I} \mathcal R_i$ implies $(x,z) \in \bigcap_{i \in \mathcal I}$. Therefore $\bigcap_\mathcal{i \in \mathcal I} \mathcal R_i$ is transitive.
Note that this proof can be generalized way further. Let $\{S_i : i \in I\}\subseteq 2^X$. Let $A \subseteq 2^X$. Let $\phi: A \rightarrow 2^X$. The argument above shows that if any family $S_i$ satisfies the condition:
For any $S \subseteq S_i: S \in A$, $\phi(S) \subseteq S_i$.
Then $\bigcap_{i \in I} S_i$ satisfies it as well.