How show that $b$ is the supremum of an interval $[a,b)$

Question

I think it is obvious that the supremum of an interval of the form $[a,b)$ is $b$ but I am trying to show it myself.

My try:

$b$ is an upper bound, if it is a supremum, then $\exists x \in [a,b)$ s.t. $x+\epsilon>b\geq x$ for some $\epsilon>0$ (epsilon characterization of the supremum)

Assume it is not, then $\forall x \in [a,b)$, $ x+\epsilon\leq b <x$ but this means that $b$ isn't an upper bound, contradiction. Thus, $b$ must be the supremum of the interval. (also, one can notice that the inequality would imply that $\epsilon<0$ which is another contradiction)

Any help in verifying my error, correcting it, or suggesting better ways to prove this is appreciated.

Answer 1

Your proof is confusingly written.

Problem 1:

if it is a supremum, then $\exists x \in [a,b)$ s.t. $x+\epsilon>b\geq x$ for some $\epsilon>0$ (epsilon characterization of the supremum)

This is not the epsilon characterization of the supremum. The characterization says that:

$s$ is a supremum of $A$ if, for every $\epsilon > 0$, there exists some $x\in A$ such that $s-\epsilon < a$.

This is completely different from what you wrote. Most importantly, the definition of supremum says something is true for every epsilon, not "for some $\epsilon > 0$ which is what you wrote.

Problem 2:

You wrote "$\forall x \in b$", which makes no sense. $b$ is not a set, so $x\in b$ is nonsensical.