Suppose $\emptyset \neq A \subset \mathbb{R} $. Let $A = [\,0,2).\,\,$ Prove that $\sup A = 2$
This is my attempt:
$A$ is the half open interval $[\,0,2)$ and so all the $x_i \in A$ look like $0 \leq x_i < 2$ so clearly $2$ is an upper bound.
To show it is the ${\it least}$ upper bound, suppose that $2 \neq \sup A$, that is there exists a number $M < 2$ for some real $M$ qualifying as $\sup A$. Certainly this $M \in [0,2) $ so $ M > 0 \Rightarrow 2 -M > 0$.
By the Archimedean Principle, for all real numbers $r > 0\,\, \exists\,\, n \in \mathbb{N}$ such that $0 < \frac{1}{n} < r $. By the Approximation Property of Suprema, there exists $a \in [0,2)$ such that $\sup A - \epsilon < a \leq \sup A$, where $\epsilon > 0$.
Suppose $\sup A = M < 2$. Then the above gives $M - \epsilon < a < 2\,\,\,\,\forall \epsilon > 0$. Also, by Archimedean, we have $0 < \frac{1}{n} < 2-M$, so choose $\epsilon = 2-M$. Then $M - (2-M) < a < 2 \,\Rightarrow 2(M-1) < a < 2$
We can assume $M - 1>0$ and so $2(M-1) > 2$ This results in a contradiction in the previous inequality. Hence $M < 2$ cannot be the supremum.
I realise there is probably a simpler way, but is what I have written all good?