Finding sup, inf, max and min for each sets
Solution 1
1) $\inf (3, 1) =3$ because $\forall x \le 3 x \not \in (3,1)$ so $3$ is lower bound. If $x > 3$ then there is an $e$ so that $3<e <x$ and $e \in (3,1)$ so $3$ is greatest lower bound.
There is not $\min$ of $(3,1)$ so for and $x \in (3,1)$ we can find an $e$ so that $3 < e < x < 1$ so $e \in (31)$.
Same reasoning: we can determine $\sup (3,1) = 1$ and there is no max.
Or we could say "there is not square bracket". I guess. That is how the square bracket was meant to be defined.
2) $\inf\{1/n n\in \mathbb N\} = 0$ because: $0 < 1/n \forall n \in \mathbb N$ so $0$ is a lower bound. If $\epsilon > 0$ we can find an $n \in \mathbb N$ so that $0 < 1/n < \epsilon$. (Why? Archimedean principal. But you can take it as a given usually, I think.) So $0$ is the greatest lower bound.
There is no minimum as for all $1/n \in $ the set, $1/(n+1)$ is in the set and smaller.
As $1 > 1/2 > 1/3 > ...... > 1/n > 1/(n+ 1).....$, $\max$ of set is $1$. $\sup = 1$ because, if maximum exists then all members of the set are smaller than the maximum and anything smaller than the maximum will have the maximum larger than it. So $\sup$ must equal $\max$ if $\max$ exists.
3) $\{1/n n \in \mathbb Z; n \ne 0\}$
For ever $1/n > 0$ in the set $1/n$ is also in that set (and, of course, 1/n < 0 < 1/n). So $\max = \sup = 1$ by the same argument above and $\min = \inf = 1$.
4) $S= \{ x < 6x \in \mathbb Q\}$
$\sup = 6$ because $6$ is an upper bound (all $x \in S$ are such $x < 6$ by definition) and for any $x < 6$ there is a rational number $q$ such that $x < q < 6$. So $6$ is least upper bound.
There is no maximum as $6 \not \in S$.
$S$ is not bounded below as for all real $n$ we can find a rational $q$ such that $q < n$. So there is no minimum nor is there any $\inf$.
5) $S= \{ 1/n + (1)^n\}$
It might be worthwhile listing a few of these. For even $n$ we have 1 1/2, 1 1/4, 1 1/6, etc. and for odd we have 0, 2/3, 4/5 etc. Okay, got it.
Okay $3/2$ is both the max and the sup. because 1)for $n = 1$, $1/n + (1)^n = 0$. For $n \ge 2$ we have $1/n + (1)^n \le 1/n + 1 \le 3/2$. So $3/2$ is an upper bound and as $3/2 \in S$ it is the least upper bound and the maximum value.
$1 = \inf S$ because. $1 < 1 + 1/n \le 1/n + (1)^n$ so $1$ is lower bound. For any $x > 1$ we can find $0 < 1/(n+1) + 1/n < \epsilon = x  (1)$. Thus $1 < 1 + 1/(n+ 1) < 1 + 1/n < x$. One of $n$ or $n+1$ must be odd and so one of $n$ or $n+1$ must be is S. So $x$ is not a lower bound. So $1$ is the greatest lower bound.
There is no minimum as $1 \not \in S$. (There is no $1/n = 0$ so there is no $1/n + (1)^n = 1$).
Solution 2
 Set $2$: Supremum and maximum are $1$, infimum is $0$ and minimum does not exist.
 Set $3$: Maxmimum and supremum are $1$, infimum and minimum are $1$.
 Set $5$: Infimum is $1$, minimum does not exist, supremum and maximum are $\frac{3}{2}$.
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Comments

Allie about 3 years
Set $1$ : $(3,1)$
Set $2$ : $\{{\frac{1}{n} : n \in \mathbb{N}}\}$
Set $3$ : $\{{\frac{1}{n} : n \in \mathbb{Z}} , n \not= 0\}$
Set $4$ : $\{{r < 6 : r \in \mathbb{Q}}\}$
Set $5$ : $\{{\frac{1}{n}+(1)^n: n \in \mathbb{N}}\}$
TRIED: I have these $5$ sets as listed. I have figured that for Set $1$, sup = $1$, inf = $3$, and max/min doesn't exist because it is not a squared bracket. As far as set $4$ is concerned, I figured that sup is $6$ where inf is negative infinity with max/min doesn't exist.
I am struggling with sets $2,3,5$ I feel like sup and inf for set $2$ is $1,0$ respectively but not quite sure if that is also its max and min. Also if I could get some help with sets $3,5$ that'd be wonderful. thank you.

K.Power about 7 yearsDo you not need to supply rigorous proofs?

fleablood about 7 yearsSet 25 are using "()" rather than "{}" for set notation. Should we be concerned? If we are to be consistent than set 1 should be {3,1} rather than $\{x 3 < x < 1\}$.

fleablood about 7 yearsOne thing to note. You can only have one of the following three cases: Neither sup nor max exist; sup exists and max doesn't; max = sup. Same for inf and min.

Allie about 7 years@fleablood yes you are right, the only reason i didn't put {} is because when i put dollar signs around them, they disappear..

fleablood about 7 yearsFrustrating.. Isn't that? You need to "escape" by putting a back slash before them. \$\{ \}\$ will be $\{\}$. Escaping with backslash is a standard coding convention. The only one I can't figure out is ~. They disappear if you use them but if you escape them they get marked as invalid code????

Allie about 7 yearsahha! thats looks more like sets. thanks for editing. and thanks for your general answer. Do you know why sup and max are 3/2 for set 5?

fleablood about 7 yearsBecause 3/2 is in set 5 and all n>2, $\pm$1 + 1/n $\le$ 1 + 1/2.


Allie about 7 yearsthanks! how did you get 3/2 for set 5?

TheGeekGreek about 7 yearsFor $n$ even, we have $\frac{1}{n} + 1$ and the largest value for $\frac{1}{n}$, where $n$ is even is $\frac{1}{2}$.

Allie about 7 yearscan't ask for a better explanation. much thank you!