# Let $A_i= \left \{...,-2,-1,0,1,...,i \right\}$. Find $\bigcup_{i=1}^{n} A_i$ and $\bigcap_{i=1}^{n} A_i$

I have the following assignment:

Let $A_i= \left \{...,-2,-1,0,1,...,i \right\}$.

Find

a) $\displaystyle \bigcup_{i=1}^{n} A_i$

b) $\displaystyle \bigcap_{i=1}^{n} A_i$

I think the first one is:

$\displaystyle \bigcup_{i=1}^{n} \left \{...,-2,-1,0,1,...,i \right\} = \left \{...,-2,-1,0,1,2...,n \right\}$

But what about the second one?

• The first one is correct. For the second try to see what happens with $n=3$, for instance. Commented Apr 9, 2015 at 21:34

HINT: $A_i\subseteq A_j$ whenever $i\le j$. You can use this to answer both questions. (Your answer to the first question is correct.)
Part (a) looks good. For the second one, remember what intersection means. An intersection will only give you elements that all sets in the intersection have in common. For example, $\{1,2,3\} \cap \{3,4,5\} = \{3\}$, since the only element in common between sets is $3$. Notice that $2 \notin A_1$, and $3 \notin A_1$ and $\ldots$ and $n \notin A_1$, but $A_1 \subseteq A_2 \subseteq \ldots \subseteq A_n$.
These sets are increasing; that is, $A_1\subseteq A_2\subseteq A_3\subseteq\cdots$ (as graydad notes at the end of his answer). Thus, the union of any collection of these sets is just the one with the largest subscript, and the intersection is just the one with the smallest subscript. Thus, we have $$\bigcup_{i=1}^n A_i = A_n = \{\ldots,-2,-1,0,1,\ldots,n\},$$ and $$\bigcap_{i=1}^n A_i = A_1 = \{\ldots,-2,-1,0,1\}.$$