Let B be a set of real numbers and suppose that u is an upper bound for B. Show that u is the least upper bound if and only if for every n ∈ Z+ there exists b ∈ B such that b > u $-$ $\frac{1}{2^n}$.
How do I go about answering this? I see that $0$ $<$ $\frac{1}{2^n}$ $\leqslant$ $\frac 12$, but am unable to make progress.
We assume in the following $B\ne\emptyset$ and bounded (so the LUB exists).
$\Rightarrow$
If $u$ is the LUB of $B$ and $n\in\mathbb Z_+$then $u-\frac{1}{2^n}$ cannot be UB, so there is $b\in B$ with $b>u-\frac{1}{2^n}$
$\Leftarrow$
We will use $0<\frac{1}{2^n}<\frac{1}{n}$
If $u$ is not LUB, then the LUB is $u'<u$. Then for $n>\frac{1}{u-u'}$ we have $u-u'>\frac{1}{n}$, so $u-\frac{1}{n}>u'$. But $2^n>n$ (induction), so $u-\frac{1}{2^n}>u-\frac{1}{n}>u'$. By the hypothesis, there is $b\in B$ with $b>u-\frac{1}{2^n}>u'$. So $u'$ is not LUB. Contradiction.