Let $a < b $ be real numbers and consider set T = $\mathbb Q\cap [a,b].$ Show $\sup T =b$
I needed help checking if my proof is correct. If it isn't correct can you please provide the correct proof or a hint in the correct direction.
Proof. $T$ is the set of all rational numbers x such that $a \le x \le b.$ Let $\sup T = \alpha.$
Assume $\alpha < b.$ Since the rationals are dense in $\mathbb R$ there exists a rational $r$ such that $a < \alpha <r< b .$ This means $r \in [a,b] $ and $r< \alpha$ which is a contradiction.
Assume $b< \alpha.$ Since the rationals are dense in $\mathbb R$ there exists a rational r such that $b <r < \alpha$ but this implies that $r$ is an upper bound of $[a,b]$. However since $r < \alpha$ this means that $\alpha $ cannot be a supremum of $[a,b]$, which is also a contradiction.
Hence $b = \alpha $ and is the supremum of $T.$
Thanks in advance.