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 $s-\epsilon<x$ 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 $x$ $s-\epsilon<x$. qed.


1 Answer 1


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 $\forall x\in A, s\ge x$

  • $\begingroup$ This makes total sense I'm blind $\endgroup$
    – Iwan5050
    Commented Jan 28, 2022 at 6:55

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