# Approximation property for supremum

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

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 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 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 . Because by definition the second inequality is wrong. But then why in the proof I get not any error?

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.
• Step 1: $\exists \epsilon_0$ such that no element $a\in E$ satisfies $sup E−\epsilon_0\geq a<supE$ Is it correct? Commented Jun 10, 2022 at 7:31
• In step 1, the case "an element of $E$ can be equal to $\sup E$" is still not managed. You should add to the last sentence "or $a$ is equal to $\sup E$". Commented Jun 10, 2022 at 8:01
• The contrary of "$\sup E - \epsilon < a < \sup E$" is "$a \le \sup E - \epsilon$ or $a \ge \sup E$". The second part of the "or" is missing. Commented Jun 10, 2022 at 8:11
• Yes. Then the case "$a > \sup E$" is discarded as impossible. But in your new theorem, "$a \ge \sup E$" cannot be discarded because $a = \sup E$ is possible. Commented Jun 10, 2022 at 8:29