supremum and infimum proof

Question

$$A=\left\{\frac{2^n}{3^k}: n,k \in \Bbb N,\ n \le k\right\}$$ I'd say that $\sup A=1$ (for $n=k=0$) and $\inf A=0$. I have problem with proving this. Generally I have problem with proofs that include epsilon.

Any help appreciated.

Answer 1

Hints:

• To prove that something is the supremum of a set, you need to show that it is both an upper bound and a least upper bound. To show that $1$ is an upper bound, note that $$ 2^n \le 3^n \le 3^k. $$ Then to show it is the smallest upper bound, show it is in fact an element of the set.

• Similarly, to prove the infemum is $0$, you need to show that $0$ is a lower bound and the greatest lower bound. It should be easy to see why it is a lower bound. Then you need to show that any $\epsilon > 0$ is not a lower bound; to do so, note that there exists $n$ such that $\epsilon > \frac{1}{3^n}$.

Answer 2

We know intuitively that as $k$ gets really big, the fraction goes to 0. How big does $k$ have to be to get below $\frac{1}{2}$? At $k=2$, $\frac{4}{9}<\frac{1}{2}$.

What about to get below $\frac{1}{3}$? For $k=3$, $\frac{8}{27}<\frac{1}{3}$.

Try continuing this approach. Think of $\epsilon$ as $\frac{1}{N}$ for $N$ some big number