Proving the supremum of this set

471 Views Asked by At

Suppose we have a subset of reals A and a real number s exists such that for all natural n, s + 1/n is an upper bound for A while s - 1/n is not an upper bound for A. Apparently s = supA but I can't seem to figure out why. I'm having trouble even proving the first condition of the supremum as an upper bound of the set. What would be the right direction?

1

There are 1 best solutions below

2
On

Intuition: The first condition tells you that there are upper bounds on $A$ that are just a little above $s$, suggesting that $s$ is an upper bound too. The second suggests that there's no upper bound on $A$ that's less than $s$.

Sketch: If $s$ is not an upper bound for $A$, then pick some $x$ in $A$ above $s$ and show that it violates the first condition. If $s$ is not the least upper bound, then let $t$ be a lower upper bound for $A$ and show that it violates the second condition.