Real Analysis - Does my counter example disprove this lemma?

Question

I was reading a real analysis text that gave the following lemma after introducing the axiom of completeness and least upper bounds, it reads:

Assume $s \in \mathbb{R}$ is an upper bound for a set $A \subseteq \mathbb{R}$. Then, $s = sup A$ if and only if, for every choice of $\epsilon \gt 0$, there exists an element $a \in A$ satisfying $s - \epsilon \lt a$.

When thinking about this lemma I was testing it with some example sets, and I stumbled across one that seemed to disprove it. The set is simply the set of real numbers on the interval $[0,1)$, our supremum is $s = 1$, and I set $\epsilon = 1 - a$. Putting it all together we get

$$1 - (1 - a) \lt a$$ $$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ a \lt a$$ As far as I can tell this set fulfills all of the requirements for A, 1 is the supremum, and $1-a$ is a valid choice for $\epsilon$, as no choice of $a \in [0,1)$ will make it less than or equal to zero. But it also feels very unlikely this lemma would be disproved so easily, so I'm sure I went wrong somewhere, but I'm not sure where. If anyone has some insight into this it would be greatly appreciated, I can provide further context if necessary.

Answer 1

For every $\epsilon > 0$, there exists some $b \in A$ such that $b > \sup A - \epsilon$. In your example, you showed that for $\epsilon = 1 - a$, $b = a$ does not work. But there are other $b \in A$ that will work! For example, $b = \frac{1+a}{2}$ will work in your example.

Answer 2

for every choice of $\epsilon>0$, there exists an element $a \in A$ ...

means that we choose the number $a$ depending on the value of $\epsilon$. Since $a$ must be allowed to depend on $\epsilon$, you can't have $\epsilon$ depending on $a$.

In other words, when you say $\epsilon = 1-a$, what is this number $a$ you're using? No such variable has been introduced yet.

In formal logic, we call $a$ "bound" by the phrase "there exists an element $a \in A$ satisfying $s - \epsilon < a$" - the variable exists for the purpose of that quantified phrase, and has no meaning outside that phrase.