The set
is identical to :
1. (-1 , 1)
2. [-1 , 1)
3. (-1 , 1]
4. [-1 , 1]
--------------------
The set
is identical to :
1. (-1 , 1)
2. [-1 , 1)
3. (-1 , 1]
4. [-1 , 1]
For n=1,2,3,4 the interval $\displaystyle (-1-n^{-1}, 1+n^{-1})$ is equal to $\displaystyle (-2,2),\;(-\tfrac32,\tfrac32),\;(-\tfrac43,\tfrac43),\;(-\tfrac54,\tfrac54)$. You can see that as n increases, all these sets are going to contain the interval [–1,1] (including the endpoints). But as n gets very large, all points to the left of –1 or to the right of +1 are eventually going to be excluded from $\displaystyle (-1-n^{-1}, 1+n^{-1})$. So the intersection of all these intervals is exactly [–1,1].
The other part of the question appears to ask for the union of the same sequence of intervals. The union will be the largest of the intervals, which is the first one, namely (–2,2). However, I think that you probably intended to ask for the union $\displaystyle \textstyle\bigcup_{n=1}^\infty(-1+n^{-1},1-n^{-1})$. I'll leave you to work out what that is. I suggest that you use the same method as for the previous part: start by writing down what happens for n=1,2,3,4,... and then think about what happens as n gets very large.
$\displaystyle \begin{gathered}
\bigcap\limits_{n = 1}^\infty {\left( { - 1 - n^{ - 1} , 1 + n^{ - 1} } \right)} = \left[ { - 1,1} \right] \hfill \\
\bigcup\limits_{n = 1}^\infty {\left( { - 1 - n^{ - 1} , 1 + n^{ - 1} } \right)} = \left( { - 2,2} \right) \hfill \\
\end{gathered} $