[Math] Supremum Proof Question

Question

Let $a<b$ be real numbers and consider the set $T=\mathbb{Q}\cap[a,b]$. Show $\sup T=b$.

I can show that $b\geq x$ for all $x\in T$ and thus an upper bound, but am not sure how to go about showing it is the least upper bound.

Answer 1

Suppose there is a smaller upper bound for $T$, say this is some $M<b$. Then by denseness of $\mathbb{Q}$ there is some rational number $r$ such that $$M<r<b. \ \ \ (1)$$ Again by denseness of the rationals, $T$ is not empty. Take some $x \in T$. Then $a \leq x \leq M$. Combining this with the above yields $r \in T$, which gives $r \leq M$, a contradiction with $(1)$.