Proving that supremum does not exist

Question

Given the set $A = \{x\in \mathbb R : x > \frac{3}{2}\} $

I need to prove that the supremum of $A$, $sup A$ does not exist.

My first idea was to prove by contradiction:

Suppose that $sup A$ exists, then $(\forall a \in A)[a \le supA]$

Intuitively, this looks like a contradiction because $\mathbb R$ is not bounded from above, but I dont really know how to explain it. Any help or advice is appreciated here!

Answer 1

You have pointed out the keypoint, $A$ is not bounded from above, hence there is no upper bound and hence no least upper bound.

Answer 2

Following your first idea.

The assumption that $\sup A$ exists in $\mathbb R$ (i.e. that $\sup A\in\mathbb R$) combined with e.g. the fact that $2\in A$ leads to: $$2\leq\sup A<\sup A+1\in A$$which is absurd.

This allows the conclusion that the assumption is wrong.