I am trying to show that $\sup\{\frac{1}{2n}\;;\;n\in\mathbb{N}\}=\frac{1}{2}$, I managed to prove that $\frac{1}{2}$ is an upper bound. However, I am trying to show that if there exists $\alpha$ such that $\alpha<\frac{1}{2}$, then $\alpha$ would not be least upper bound. I wanted to resort to contradiction in this case. I supposed that there exists $\alpha<\frac{1}{2}\implies \frac{1}{2n}\leq\alpha<\frac{1}{2}$, I was trying to show that either such $n$ does not exist, or $\alpha>\frac{1}{2}$ this is why I am stuck. Any help would be much much appreciated.
Update : Does the fact that $\frac{1}{2}$ in the set automatically imply $\frac{1}{2}$ must be the least upper bound? or it wouldn't be valid to say that as part of a proof?
$\frac12$ is an element of the set ($n=1$ in the definition of the set). So clearly nothing smaller than $\frac12$ can be an upper bound for the set.