Query about Approximation Property of Suprema

Question

The statement for approximation property of Suprema states that: If E has a finite supremum and $\epsilon > 0$ is any positive number, then there is a point $a \in E$ such that $ sup E - \epsilon < a \leq sup E$.

Now, I can rewrite the inequality as follows: $ sup E - \epsilon < a \leq sup E< sup E + \epsilon$

Which implies: $ sup E - \epsilon < a < sup E + \epsilon$

Implying: $ - \epsilon < a - sup E < + \epsilon$

Implying: $ \left|a - sup E\right| < \epsilon$

and Implying: $ \left|a - sup E\right| = 0$.

This means that $a = sup E$ which is not true since there could be instances where sup E can belong outside the set E.

Could you please tell me what's wrong with my logic? Thanks!

Answer 1

Note that the point $a$ may depend on $\varepsilon$, so you cannot deduce that $|a-sup E|<\varepsilon$ implies $|a-\sup E|=0$.

For example, it may be the case that as $\varepsilon \to 0$, we define a sequence of points $a=a_\varepsilon$ so that $a_\varepsilon$ "leaves" the set $E$. For example, consider the set $[0,1)$. Given the sequence $\varepsilon(n)=1/n$, we can take $a(n) = 1-1/2n$, each of which is $\varepsilon$-close to 1. But, for any given $a(n)$, there will eventually exist a $\varepsilon(N)$ for $N$ large enough so that $a(n)$ will no longer be $\varepsilon$-close.