# Finding sup, inf, max and min for each sets

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.

• Do you not need to supply rigorous proofs? Commented Sep 14, 2016 at 22:31
• Set 2-5 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 $\{x| -3 < x < -1\}$. Commented Sep 14, 2016 at 22:50
• One 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. Commented Sep 14, 2016 at 22:53
• @fleablood yes you are right, the only reason i didn't put {} is because when i put dollar signs around them, they disappear.. Commented Sep 14, 2016 at 22:53
• Frustrating.. 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???? Commented Sep 14, 2016 at 22:57

1. $$\inf (-3, -1) =-3$$ because $$\forall x \le -3\mid x \not \in (-3,-1)$$ so $$-3$$ is a lower bound. If $$x > -3$$ then there is an $$e$$ so that $$-3 and $$e \in (-3,-1)$$ so $$-3$$ is the 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 $$e \in (-3-1)$$.

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.

1. $$\inf\{1/n\mid n\in \mathbb N\} = 0$$ 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 principle. 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.

1. $$\{1/n\mid n \in \mathbb Z; n \ne 0\}$$

For every $$1/n > 0$$ 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 $$\min = \inf = -1$$.

1. $$S= \{ x < 6\mid x \in \mathbb Q\}$$

$$\sup = 6$$ 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 $$\inf$$.

1. $$S= \{ 1/n + (-1)^n\}$$

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 in 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 $$1/n + (-1)^n = -1$$).

• can't ask for a better explanation. much thank you! Commented Sep 15, 2016 at 0:03
• 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 $\frac{3}{2}$.
• thanks! how did you get 3/2 for set 5? Commented Sep 14, 2016 at 22:53
• For $n$ even, we have $\frac{1}{n} + 1$ and the largest value for $\frac{1}{n}$, where $n$ is even is $\frac{1}{2}$. Commented Sep 14, 2016 at 23:09