Inf, Sup, Min, Max of $\{9+\frac{(-1)^n}{2^n}: n \in \Bbb{N}\}$, $\bigcup_{n=1}^{\infty} [\frac{n+1}{n},3n^2]$ I am thinking about the following characteristics of the set - Infimum, Minimum, Supremum, Maximum, (Open/closed/neither/both).
Here is what I got so far:

$S_1 :=\{9+\frac{(-1)^n}{2^n} ; n \in \Bbb{N}\}$

$\inf(S_1) = 9$, $ \sup(S_1) = 9+\frac{1}{2^2}$, $\max(S_1) = +\frac{1}{2^2}$.
Is the set open?

$S_2 := \bigcap_{n=1}^{\infty} (3-\frac{2}{n},5+\frac{1}{2n})$

I think the limiting set $S_2 =(3,5)$, $\inf(S_2) = 1$ and $\min(S_2) = 3$, $\sup(S_2) = 5.5$ and $\max(S_2)= 5$.
And since $S_2 =(3,5)$, it is open?
Lastly,

$S_3 = \bigcup_{n=1}^{\infty} [\frac{n+1}{n},3n^2]$, 

$\inf(S_3)=1, \sup(S_3)=$ does not exist neither does its maximum, as it tends to infinity? Also $S$ is neither open nor closed.
How can I think of these type of problems?
 A: The set is not open. To see this, consider $\text{Int } S_1$ (ie. The interior points of $S_1$). You'll find that $\text{Int } S_1 \neq S$ so the set is not open.
For $S_2$ you are correct that the limiting set is $(3,5)$ and this is indeed open. To show $S_2 = (3,5)$, show the subset relation both ways (ie. If $x\in (3,5)$ then $x\in\bigcap_{n=1}^\infty A_n$ and vice versa).
However since you know $S_2=(3,5)$, it does not make sense to say that it's infimum is 1 or its supremum is 5.5. You've found that $A_1 = (1,5.5)$ but this is not $S_2$. Remember $S_2$ is the intersection of the $A_n$'s. 
You are also correct with $S_3$.
A: Hints:
The infimum $c$ of a set $S$ is the greatest lower bound of a set. A number $d$ is a minimum if it is the infimum of the set $S$ and it is contained in $S$. 9 is definitely not a lower bound of the first set: the first element of $S$ is 8.5! Similarly, 1 isn't the infimum of the second set: you can find a greater lower bound.
A supremum is the least upper bound, and a maximum is a supremum which is contained in the set. You're right about the supremum and maximum for the first set. But you're not right about the maximum of the second.
A set $S$ is open if for each $x \in S$, $\exists \epsilon > 0$ such that $(x-\epsilon, x+\epsilon) \subseteq S$. Try picking any member of the first set and finding an $\epsilon$ which works. You won't be able to find one.
