$A = \{x+ 1/n : x \in (0,1), n \in \mathbb Z^+ \}.$ Determine the supremum and infimum of $A$, if they exist.

Question

Let $A =\{x+ 1/n : x \in (0,1), n \in\mathbb{Z}\}$. Determine the supremum and infimum of $A$, if they exist. Rigorously justify your answers

Answer 1

If $a \in A$ then $ \exists x \in (0,1)$ and $\exists n \in \mathbb{Z}^+$ such that $a = x+1/n$. As $x<1$ and $n \geq 1$, $a < 2$ so 2 is an upper bound.

To show 2 is the least upper bound, then suppose $\exists m$ such that $0 < m < 2$ and $m$ is an upper bound. $0 < m/2 < 1$. Therefore, $a = m/2+1/1 \in A$, but $m/2+1/1 = m/2+1 > m/2 + m/2 = m$ so $a>m$ and $m$ cannot be an upper bound.

To show 0 is the greatest lower bound, suppose $\exists m$ such that $0 < m < 1$ and $m$ is a lower bound. Now simply set $x=m/2$ and $n$ to be an integer greater than $2/m$. Then $a = x+1/n < m/2+1/(2/m) = m/2 + m/2 = m$. So $a<m$ but $a \in A$ so m cannot be a lower bound of A.