I read the following theorem here today (see Theorem 1). I'm not trying to prove it right now; I'm trying to understand it. For some reason, I am missing something because I don't think it holds true. The supremum would be equivalent to the least upper bound as it is bounded above which is what I personally use for a definition (it's the same definition as the site).
Here is an example... Consider $S=\{x\in \mathbb{R}|1<x^2<\sqrt{2}\}$ in regards to the poset $\mathbb{R}$ here. Clearly, $u=\sqrt{2}$ is a least upper bound for the poset $\mathbb{R}$. Here $\sqrt{2}$ should be the supremum as well without using Theorem 1 here. Now, I want to consider using Theorem 1 here to make sense of it. This means we should be able to find some $x_e\in S$ such that $\underbrace{u}_{=\sqrt{2}}-\epsilon<x_\epsilon$ holds true when $\epsilon >0$. I'm going to find a value say $1.3\in S$ for $x_e$ that is just less than $\sqrt{2}$ here to see what happens. So, $\sqrt{2}-\epsilon <1.3$ doesn't work which is interesting. I should be able to find a value that makes it work. Now, let's try a better guess like $1.4\in S$ which is still less than $\sqrt{2}$ to get $\sqrt{2}-\epsilon <1.4$ which looks better as $\epsilon$ could be closer to zero than before but still doesn't work when $\epsilon=0.01$. At this point, I want to get closer and closer to $\sqrt{2}$ from the left but I give up! I mean how is it possible for me to find this $x_e\in S$ such that $u-\epsilon<x_\epsilon$ to validate $u=\sqrt{2}$ here. Shouldn't it work? I feel like I am missing something. What am I doing wrong here?
First, you pick your $\epsilon$ first, and then we can decide our $x_\epsilon$.
For example, if you pick $\epsilon$ to be $0.01$, then I can pick $x_\epsilon=\sqrt2 - \frac{0.01}2 \in S$.
For an interval $S=(a,b)$, given $\epsilon>0$, if $\epsilon \ge b-a$, we can pick $x_\epsilon = \frac{a+b}2$. If $0<\epsilon < b-a$, we can pick $x_\epsilon = b-\frac{\epsilon}2$.