How to find the supremum and infimum of this set?

Question

$\{3-\frac{17}{n}:n\in\Bbb{N}\} \subseteq A \subseteq (-\infty,3].$ Prove that A has an upper bound and find $\sup(A) $.

Answer 1

First is the statement $\{3-17/n\} \subseteq A \subset (-\infty,3] $ possible.

If $x = 3-17/k \in \{3-17/n\}$ then $x <3$ so $x \in (-\infty,3] $ so, yes, the statement is possible.

If $x \in A$ then $x \in (-\infty,3]$ so $x \le 3$ so A is bounded above and $\sup A \le 3$ exists.

If $y < 3$ we can find a natural $n $ where $n > 17/ (3-y)$. So $0 <1/n < (3-y)/17$ So $0 <17/n <3-y $ so $y <3-17/n \in \{3-17/n\} \subseteq A$.

So $y < 3$ means $y$ is not an upper bound.

So $\sup A = 3$.

Now $(-\infty,3] $ is not bound below so $A $ needn't be bound below either. ($A $ could be $(-\infty,3] $ for all we know, or it could be any other subset). So $\inf A $ need not exist and is indeterminable if it did.

Well, almost, $-14=3-17/1 \in \{3-17/n\}\in A $ so if $\inf A $ existed (which it might not) it would be less than or equal to $-14$.

Answer 2

$A$ is bounded above by $3$, hence $\sup(A)$ exits, and it equals to $3$. To see this, if $\epsilon > 0$, choose $ n > \dfrac{17}{\epsilon}$, then $3- \dfrac{17}{n} > 3-\epsilon$

Answer 3

We know that $(3-\frac{17}{n})\in A$ for every $n\in\Bbb N$. Further, we know that every $a\in A$ satisfies $a\leq 3$ since $A\subseteq (-\infty,3]$.

These imply two things, if a supremum exists, it must be at least as large as $3$ (since $3-\frac{17}{n}$ gets arbitrarily close to $3$ and a supremum must be at least as large as each element) and further that it cannot be any larger than $3$ (since no elements larger than $3$ exist in $A$). Since it is a bounded set from above, a supremum must exist, and the observations here imply that it must be exactly three.

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About the infimum however, we do not have enough information to conclude anything. It suffices to show two examples of sets $A$ where the hypotheses are satisfied, but they have different infimums.

Both $A=[-100,3)$ and $A'=[-1000,3]$ have $\{3-\frac{17}{n}~:~n\in\Bbb N\}$ as a subset and $(-\infty,3]$ as a superset, however $A$ has $-100$ as an infimum and $A'$ has $-1000$ as an infimum.

It is possible that there is no infimum as well, as is the case with $A''=(-\infty,3]$

Answer 4

The definition of the reals $R$ implies that there is no $x\in R$ such that $0<x<q$ for all $q\in Q^+.$ From this, it is immediate that for any $r\in R,$ there is no $x\in R$ such that $r-1/n<x<r$ for all $n\in N.$ And therefore $\sup \{r-1/n:n\in N\}=r.$ In your Q, $r=3.$

This property of the reals is not provable without a def'n of $R.$ It is possible to extend $Q,$ or $R $, to an arithmetic system $S$ with "infinitesimals" : Positive members of $S$ that are less than every positive rational. (And their reciprocals are greater than any natural number.)