0
$\begingroup$

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.

$\endgroup$
7
  • $\begingroup$ Do you not need to supply rigorous proofs? $\endgroup$
    – K.Power
    Commented Sep 14, 2016 at 22:31
  • $\begingroup$ 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 $\{x| -3 < x < -1\}$. $\endgroup$
    – fleablood
    Commented Sep 14, 2016 at 22:50
  • $\begingroup$ 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. $\endgroup$
    – fleablood
    Commented Sep 14, 2016 at 22:53
  • $\begingroup$ @fleablood yes you are right, the only reason i didn't put {} is because when i put dollar signs around them, they disappear.. $\endgroup$
    – Allie
    Commented Sep 14, 2016 at 22:53
  • 1
    $\begingroup$ 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???? $\endgroup$
    – fleablood
    Commented Sep 14, 2016 at 22:57

2 Answers 2

2
$\begingroup$
  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 $-3<e <x$ 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 $1/n + (-1)^n = -1$).

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

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .