How can I find the intersection and union of this family of set.
For each natural number $n\ge3$ let $A_n=[\frac{1}{n},2+\frac{1}{n}]$ and the family $L=\{A_n:n\ge3\}$
$A_3=[1/3,7/3]$
$A_4=[1/4,9/4]$
$A_5=[1/5,11/5]$
It seem with this set the largest value it will reach is $\frac{7}{3}$ and the smallest will eventually be zero.
so my union over the family L would be $(0,\frac{7}{3})$
But the intersection I cannot seem to figure out.
Best Answer
You're right about the union: for any $x > 0$ there is some $n$ such that ${1 \over n} < x$, and this ensures that $x \in A_n \subset \cup_k A_k$. The set is "decreasing on the right" as you note, so indeed $\cup_k A_k = (0, {7 \over 3}]$. Note that no $x \le 0$ is in the union, as all $A_k \subset (0, \infty)$.
But a similar argument can be held for the intersection: the largest element on the left we reach is ${1 \over 3}$, in $A_3$, and points less than it cannot be in the intersection, because they're not even in $A_3$. But every point between $\frac{1}{3}$ and $2$ (endpoints included) is in all $A_n$, as for such $x$ $\frac{1}{n} \le \frac{1}{3} \le x \le 2 < \frac{1}{n} + 2$, so $x \in A_n$. But if $x > 2$, again for some large enough $x$, $2 + \frac{1}{n} < x$ and for such $n$, $x \notin A_n$, so not in $\cap_k A_k$ as well.
So $\cap_k A_k = [\frac{1}{3}, 2]$.