Supremum of $M = \{ \left \lfloor{\alpha n} \right \rfloor\frac{1}{n}: n \in \mathbb{N}_{>0}\}$

Question

Lets assume the set $M = \{ \left \lfloor{\alpha n} \right \rfloor\frac{1}{n}: n \in \mathbb{N}_{>0}\} $ with $\alpha \in \mathbb{R}, \alpha > 0$. How can I systematically find the supremum of this set? This Set apparently limited from above. I think $\left \lfloor{\alpha n} \right \rfloor \leq \left \lfloor{\alpha +1} \right \rfloor * n $, hence $ \left \lfloor{\alpha +1} \right \rfloor$ is s upper bound of M, but is this bound the smallest possible bound?

Answer 1

First of all note that for all $n\in\mathbb{N}$ we have $\lfloor\alpha n\rfloor\frac{1}{n}\leq\alpha n\frac{1}{n}=\alpha$, so $\alpha$ is an upper bound. Now I say it is the supremum. Let $\epsilon>0$. Pick a big enough $n$ such that $\frac{1}{n}<\epsilon$. So now we have:

$\alpha-\epsilon<\alpha-\frac{1}{n}=(\alpha-\frac{1}{n})\frac{n}{n}=(\alpha n-1)\frac{1}{n}\leq\lfloor\alpha n\rfloor\frac{1}{n}$

So $\alpha-\epsilon$ is already not an upper bound of your set.