I have a set $A = \bigl\{\frac{n}{2^n} : n \in \Bbb N\bigr\}$. Now I want to prove that $0$ is the $\inf A$ and $\frac{1}{2}$ the $\sup A$.
I'm doing it this way:
I use the definition:
$\sup(A)$ exists only if: 1)$\forall a\in A,~a\leq s$
2) $\epsilon>0$ their exists $n$ such that $s-\epsilon<\frac{n}{2^n}$.
So I'm showing first condition: $n \geq \frac{1}{2}$
And second: $n > 0 -\epsilon = - \epsilon$
The conditions are true, but how I have to prove this? Maybe I should do it in another way?
On
Clearly $0$ is a lower bound of the set. Suppose $\exists$ $\alpha>0$ such that $\alpha$ is the infimum of the set $A$. Then by for $\alpha >0$ and $1$, by Archimedian property $\exists~n \in \mathbb{N}$ such that $n\alpha >1$ or $\alpha >\frac{1}{n}$. Now it can be seen that for $n>5$ we have $\frac{1}{n}>\frac{n}{2^n}$. Therefore $\alpha>0$ can not become a lower bound. It is clear the $<\frac{n}{2^n}>$ is monotonically decreasing and. As $\frac{1}{2}$ is an upper bound of $A$ and is achived at $n=1$. Hence $0$ is infimum and $\frac{1}{2}$ is supremum of $A$.
The supremum is easy $\frac 12$ is in $A$ and the sequence $a_n = \frac{n}{2^n}$ is monotonically decreasing.
The infimum. $0< a_n$ for all $n$
And for any $\epsilon > 0$ there exists an $n$ such that $a_n < \epsilon.$