Subtract infinite unions, is it possible? $\cup_{n=1}^\infty F_n - \cup_{n=k}^\infty F_n$

Here, $k \ge 1$.

What is the result of this calculation? I am thinking for $k = 1$, it must be the null set, but what about $k = 2, 3 ...$? For example, for $k = 2$, is it then $F_1$?

Note that the subtraction sign means everything in the LHS set after removal of everything in the RHS set.
We have that

$$A\setminus B=A\cap B^c$$ and 

$$(\cup_{i\in I}A_i)^c=\cap_{i\in I}A_i^c.$$ So

\begin{align}\cup_{n=1}^{\infty} F_n \setminus (\cup_{n=k}^{\infty} F_n)&amp;=(\cup_{n=1}^{\infty} F_n) \cap (\cup_{n=k}^{\infty} F_n)^c\\ &amp;=(\cup_{n=1}^{\infty} F_n) \cap (\cap_{n=k}^{\infty} F_n^c.)\end{align}

Note that if $k=2$ the result is not $F_1.$ For example, $F_1=[0,2], F_2=[1,3], F_n=\emptyset, n\ge 3.$ Then

$$\cup_{n=1}^{\infty} F_n \setminus (\cup_{n=2}^{\infty} F_n)=[0,1)\ne F_1.$$
Take all $F_i$ equal a fixed set. Then you get null. However, take $F_1$ to be arbitrary, then you can get any other set. The question is not well-posed.