A set theory problem - proof 
Prove that $X \cap (\bigcup_{i \in I} Y_i) = \bigcap_{i \in I} (Y_i \cap X) $

Is the indices finite? How do I know it isn't $1 \leq i < \infty$?
Also, isn't the RHS just $Y_1 \cap Y_2 \cap ....\cap X$? (assuming we have finite intersection).
EDIT
A quick counterexample from myself disproves this. 
$Y_1 = \{1,2 \}$ and $Y_2 = \{ 2\}$ and $X = \{1\}$
EDIT2
Yes, yes, yes; thank you all for your answers. 
 A: No, the index set $I$ could be anything, even uncountably infinite. The problem didn't say anything about $I$ being finite, so you can't assume it.
A: The indexing set needn't be finite. Maybe it should read $$X\cap \bigcup Y_i=\bigcup (X\cap Y_i)\text{ ? }$$ You can argue as follows:
First suppose $x\in X\cap\bigcup Y_i$. Then $x\in X$ and $x\in\bigcup Y_i$. Thus $x\in X$ and $x\in Y_i\;$ for at least one $i$, say it is $j$. Thus, $x\in X\cap Y_j$. This means $x\in \bigcup (X\cap Y_i)$.
Can you do the other direction?
A: Sure, the right-hand side is the intersection you indicate in the post. And the two sides are in general not equal. So proving equality will be a challenge!
For example, let $Y_1$ be the set of odd numbers and $Y_2$ the set of even numbers. Let $X=\{1,2\}$. The right-hand side is empty, the left is not, it is equal to $X$. 
As to whether $I$ is allowed to be infinite, that is context-dependent. In principle there is no restriction, but in a very finite-oriented course, there might be an implicit convention that index sets are finite. 
Remark: If we replace the outer intersection on the right by union, the modified equation becomes true, whatever the size of the index set. A straightforward proof can be given by element-chasing.
