There is the theorem:
If $E$ has a finite supremum and $\epsilon >0$ is any positive number, then there is a point $a\in E$ such that $$\sup E-\epsilon <a\leq sup E.$$
And there is the proof:
I wanted to proof the theorem which must be false:
If $E$ has a finite supremum and $\epsilon >0$ is any positive number, then there is a point $a\in E$ such that $$\sup E-\epsilon <a< sup E.$$ But when I I do my proof in the simmilar manner I do not see any problem. Where is my error?
My proof:
Step 1: Suppose that the theorem is false. Then $\exists \epsilon_0 >0$ such that for all $a\in E$ we have $s_0=\sup E-\epsilon_0 \geq a$ or $a\geq \sup E$.
Step 2: Now we see that we have two ways. The first way: We can say that $s_0$ is an upper bound which will lead us to contradiction ($\epsilon_0\leq 0$). The second way: $a=\sup E$. And I dont understand what to do with this? $a$ can be equal to $\sup E$. And it is true. But in the statement of theorem we say $\sup E-\epsilon <a< sup E.$ So I conclude that the second way also is the contradiction.
What and where do I miss?
EDIT: Maybe my mistake is at the start when I write $\sup E-\epsilon <a< sup E$. Because by definition the second inequality is wrong. But then why in the proof I get not any error?
Best Answer
Your step 1 does not exclude elements of $E$ from being equal to $\sup E$. So then your step 2 is false.
That should appear more clearly if you use mathematical signs instead of words: "no element of $E$ lies between etc." in step 1 is ambiguous, because "between" is ambiguous.
Similarly, the reasoning for step 2 could be more detailed, explicitely showing the use of step 1. That would show that the case "an element of $E$ can be equal to $\sup E$" is forgotten.