Difference of two sets and distributivity

If $A,B,C$ are sets, then we all know that $A\setminus (B\cap C)= (A\setminus B)\cup (A\setminus C)$. So by induction $$A\setminus\bigcap_{i=1}^nB_i=\bigcup_{i=1}^n (A\setminus B_i)$$ for all $n\in\mathbb N$.

Now if $I$ is an uncountable set and $\{B_i\}_{i\in I}$ is a family of sets, is it true that: $$A\setminus\bigcap_{i\in I}B_i=\bigcup_{i\in I} (A\setminus B_i)\,\,\,?$$

If the answer to the above question will be "NO", what can we say if $I$ is countable?

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Yes. If $x$ is in the left hand side then it's in $A$ but not in $\bigcap_{i\in I} B_i$, so it's missing from one of the $B_i$'s. Therefore it's in $A \setminus B_i$ for some $B_i$, and it's in the right hand side. The other direction is similar.
De Morgan's laws are most fundamental and hold for all indexed families, no matter the cardinalities involved. So, $$A-\bigcap _{i\in I}A_i=\bigcup _{i\in I}(A-A_i)$$ and dually $$A-\bigcup_{i\in I}A_i=\bigcap _{i\in I}(A-A_i).$$ The proof is a very good exercise.
Note that $$A\setminus \bigcap_{i\in I}B_i=A\cap \left(\bigcap_{i\in I}B_i\right)^c,$$ where $^c$ denotes the complement. Using De Morgan's laws (which holds for a general $I$) we get $$A\cap \left(\bigcap_{i\in I}B_i\right)^c =A\cap \left(\bigcup_{i\in I}B_i^c\right)=\bigcup_{i \in I}\left(A\cap B_i^c\right).$$ Now note that $A\cap B_i^c =A\setminus B_i$ and we are done.