Based on $ \epsilon$ have a new definition of supremum:

Let there be a nonempty set $ X$ with supremum $ s$ , then $ X\cap(s – \epsilon, s]\ne \emptyset, \,\, \forall \epsilon\gt 0$ .

The conventional definition is given by:

Let $ X$ be a nonempty set of real numbers. The number $ s$ is called the supremum of $ X$ if $ s$ is an upper bound of $ X$ and $ s \le y$ for every upper bound of $ X$ .

Let, the conventional definition be denoted by ‘Def. 1’, while the new definition by ‘Def. 2’.

*Have two questions below. I need help in attempting them.*

Q. 1 : Need show that the two definitions are equivalent by proving the following two conditional statements:

(i) If $ s = sup(X)$ , as given by Defn. 1, then $ s$ is the supremum, as given by Defn. 2. Here, *assume that Defn. 1 holds*, and use this assumption *to prove that Defn. 2 holds*.

Let $ s’$ is supremum as per Defn. 2. Also, the relation between the magnitudes of $ s,s’$ is unknown, & need be established.

$ s$ will have set $ X$ elements in the range $ (s-\epsilon, s]$ if $ s-s’ \lt \epsilon$ , by the below proof:

Let us assume that $ s-s’ \ne 0 $ , let $ s-s’=k.\epsilon, k\lt 1$ , then $ s = s’+k.\epsilon \implies s -\epsilon = s’+(k-1).\epsilon \implies s -\epsilon \lt s’$ .

$ s-\epsilon\lt s’\implies \exists x \in X: X\cap (s – \epsilon, s]\ne \emptyset$ .

But, Def. 2 can take any $ \epsilon\gt 0$ to ensure $ \exists x \in X: X\cap (s’ – \epsilon, s’]\ne \emptyset$ .

So, if Def. 1 is to have ability to take any $ \epsilon\gt 0$ , need the lower bound of $ (s – \epsilon, s]$ to equal at least to $ s’ – \epsilon$ .

But, $ s – \epsilon= s’+(k-1)\epsilon \ge s- \epsilon, \forall k, 0\lt k\lt 1$ .

So, the only possible value is $ k=0$ to have the lower bound of $ (s – \epsilon, s]$ equal to $ s’ – \epsilon$ .

But, by this cannot impose any restriction on the upper bound $ s$ (of Def. 1) to equal $ s’$ (of Def. 2).

(ii) If $ s = sup(X)$ , as given by Defn. 2, then $ s$ is the supremum, as given by Defn. 1. Here, *assume that Defn. 2 holds*, and use this assumption *to prove that Defn. 1 holds*.

Let us modify for consistency with part (i) sake, $ s$ replaced by $ s’$ .

If Defn. 2 holds, then the upper bound of the interval is bounded by $ s’$ , which is also the last element that can possibly be (if, $ s’\in X$ ) in $ X$ . For Defn. 1 to hold, the upper bound must *then* be the same as the upper bound of Defn. 2, i.e. $ s’$ .

Q. 2: What is the practical significance of showing that these two definitions are logically equivalent?

The step (i) of showing that if Defn. 1 holds, then Defn. 2 holds, leads to having the lower bound of $ (s – \epsilon, s]=s’ – \epsilon$ .

The step (ii) of showing that if Defn. 2 holds, then Defn. 1 holds, leads to having the upper bound of $ (s – \epsilon, s]=s’$ per bound implies the max. value is $ s=1$ at $ n= \infty$ . All possible values of $ n$ are covered in the interval $ (s-\epsilon,s]$ with $ n=\infty$ at the upper bound.