Let $A=\{x:\frac{[b\cdot n]}{n}\}$ when $n\in \mathbb{N}$ find the supremum
we know that $b\cdot n-1<[b\cdot n]<b \cdot n$ therefore $b- \frac{1}{n}<\frac{[b\cdot n]}{n}<b$ so b is a upper bound.
Let Assume that there is $M<b$ so that M is the supremum. So $\frac{[b\cdot n]}{n}<M$ but we know that $\frac{[b\cdot n]}{n}<\frac{b \cdot n}{n}=b$ so we have $\frac{[b\cdot n]}{n}<\frac{b \cdot n}{n}=b<M$ contradiction.
Is the proof valid?