# Proof of alternative definition of supremum. Is my proof understandable?

I'm trying to improve my proof writing due to exams. Is my reasoning even correct in the first place and understandable? If not what could be improved?

Edit: $$s$$ is an upper bound; s=sup(A) $$\iff$$ $$\forall \epsilon>0 \exists x \in A : s-\epsilon< x \leq s$$

We're going proof the two directions separat.

$$\forall \epsilon>0 \exists x \in A : s-\epsilon< x \leq s$$ $$\implies$$ $$s=sup(A)$$. Assume this were not the case. Since $$s$$ is an upper bound due to $$x\leq s$$ and because $$s\neq sup(A)$$ we find $$sup(A)< s$$. We can also rephrase this slightly. There is an $$\epsilon>0$$ such that $$sup(A)+\epsilon=s$$. But then $$sup(A)=s-\epsilon$$ and substituting this into $$s-\epsilon yields sup(A)<x. A contradiction, because $$sup(A)$$ is an upper bound.

$$s=sup(A) \implies$$ $$\forall \epsilon>0 \exists x \in A : s-\epsilon< x \leq s$$. The first inequality $$x \leq s$$ is true virtually by definition. $$s$$ is an upper bound. The second inequality is true because $$s$$ is the least upper bound but $$s-\epsilon < sup(A)$$ so $$s-\epsilon$$ cannot be an upper bound thus there exists an $$x$$ $$s-\epsilon. qed.

## 1 Answer

The proposition is not true the way it is written.

$$\forall \epsilon>0 \exists x \in A : s-\epsilon< x \leq s \implies s = \sup(A)$$

Let $$A$$ be the interval $$[0,1]$$ Let $$s = \frac 12$$

$$s$$ is clearly not an upper bound on A and hence not the supremum.

However, for any epsilon greater than $$0,$$ there exists an x in the inerval ($$x = \frac 12$$) such that $$\frac 12-\epsilon< \frac 12 \le \frac 12$$

We need an additional qualifer that will assert that $$\forall x\in A, s\ge x$$

• This makes total sense I'm blind Commented Jan 28, 2022 at 6:55