Supremum Proof:

Question

Show sup{−1/n: n∈N}=0. So far I have the following: Let b be an upper bound. Suppose b<0. b<-1/n which is equivalent to -b>1/n. I don't know what to do from here and if I'm approaching the question correctly.

Answer 1

HINT: If we define a sequence $(a_n)_n = -\frac{1}{n}$, then we know that $0$ is an upper bound (since all the elements of this sequence are negative numbers). Then suppose for a contradiction that there exists a smaller upper bound than $0$, say $b$; and we know that $b < 0$. Then, we can use Archimedian Property to show that there exists a natural number $n$ such that $a_n = -\frac{1}{n} > b$ for some $n$, which is contradictory to the assumption $b$ is an upper bound.

Answer 2

You can use that $\sup(-A) = - \inf(A)$, i.e, your problem now is to compute $\inf\left \{ \frac{1}{n}:n\in\mathbb{N}\right \}$, can you proceed from here or do you want more help ?