If $a = \mathrm {sup}\ B$, how to show that the following holds? Let $B \subseteq \mathbb R$ be a nonempty set. If $a = \mathrm {sup}\ B$, then it will be the case that for all $n \in \mathbb N$ that an upper bound of $B$ is $$a +\frac {1}{n}$$ while $$a - \frac {1}{n}$$ won't be an upper bound of $B$.
I attempted to prove this using mathematical induction. It is easy to show that the base case for $n = 1$ is true (by definition of supremum). Then, assuming that the property holds for some $k \in \mathbb N$ such that $k \ge 1$, we proceed to show that it holds for $k + 1$. How can we show that $$a - \frac {1}{k + 1} \le a$$ and that  $$a \le a +\frac {1}{k + 1}$$

If $a + \frac 1 n$ is not  an upper bound of $B$, then there is an element $a'$ in $B$ such that $a' \gt a + \frac 1 n \gt a$ leading to a contradiction since $a$ is an upper bound of $B$. 
Similarly, if $a - \frac 1 n$ is an upper bound of $B$ then $a - \frac 1 n \lt a$ contradicting the fact that $a$ is the least upper bound of $B$. 
As T. Bongers has suggested induction is not at all required..
