Supremum proof based on $\epsilon$.

Question

Need help in vetting my answers for questions in section 2.3, 2,4 (on page # 7,8) in chap. 2 (page 7) in CRM series book by MAA: Exploratory Examples for Real Analysis, By Joanne E. Snow, Kirk E. Weller. here. The exercises refer to my first post & second post for sec. 2.2.

Q. 2.3:
Based on the data you gathered in Sec. 2.2, answer the following questions:
1. If $s$ is the lub of a nonempty set $X$ of real numbers, and if $ \epsilon \gt 0$, can we always find elements of $X$ in the half-open interval $(s - \epsilon, s]$. Why, or why not?

Yes, for both cases of $s$ being positive or negative, as shown below:

1. if $s$ is positive:
   (a) $\epsilon\le s$ : Let us denote the affected lower bound of the interval by $s'=s - \epsilon$. So $s'\ge 0$, leading to $(s',s]$ being a valid interval.
   (b) $\epsilon\gt s$ : So $s'\lt 0$, and $(s',s]$ still a valid interval.

2. if $s$ is negative:
   $s'\lt 0$, and $(s',s]$ still a valid interval.

It only depends on $X$ if it has elements in the given interval. This is particularly relevant for non-continuous sets.

2. If $u$ is an arbitrarily chosen upper bound of $X$, i.e. not equal to the supremum, and if $\epsilon \gt 0$, can we always find elements of $X$ in the half-open interval $(u - \epsilon, u]$. Why, or why not?
   

(i) Consider the lower bound : Only if $u-s \lt \epsilon$.
Let $u-s = k.\epsilon, k\lt 1$, then
$u-s= k.\epsilon \implies u = s+k.\epsilon \implies u -\epsilon = s+(k-1).\epsilon \implies u -\epsilon \lt s$.

(ii) Consider the upper bound : The upper bound of the interval $(u - \epsilon, u]$ is $u \gt s$.

This ensures that $\exists x \in X$ for $(u - \epsilon, u]$.

3. If your answers to questions 1 & 2 differ, can you explain why? In what way, does the supremum differ from any other bound.
   

supremum is the least upper bound that may lie in the set. No $\epsilon \gt 0$ can exclude elements of range from the interval $(s - \epsilon, s]$.
But, any other upper bound (let, $u\gt s$) must have $(u - \epsilon\lt s]$ to have any element of the set in the interval $(u - \epsilon, u]$.

4. Writing the definition: Based on your answers to the prior three questions, try to write "new" definition. The statement of this definition will involve a nonempty set $X$, a positive real number $\epsilon$, & the half-open interval $(s - \epsilon, s]$, where $s$ denotes the supremum.

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

Q. 2.4:
Two statements $p,q$ are said to be equivalent if the biconditional statement $p \iff q$ ($p$ if and only if $q$) is true. The biconditional $p \iff q$ is shorthand for the conjunction: $\cdots$

Definition 1: 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 $X$.

Definition 2: This is the "new" definition you derived in Sec. 2.3.

In order to show that the two definitions are equivalent, we must prove 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.

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

Let $s-s' = k.\epsilon, k\lt 1$, then $s-s'= k.\epsilon \implies s = s'+k.\epsilon \implies s -\epsilon = s'+(k-1).\epsilon \implies s -\epsilon \lt s'$. This keeps lower bound $\lt s'$, ensuring that $\exists x \in X: X\cap (s - \epsilon, s]\ne \emptyset$.
But, Def. 2 can take any $\epsilon\gt 0$.
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. $s'$.

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 $(s - \epsilon, s]=s'$.

Answer 1

1. Whether $s$ is positive, zero, or negative are irrelevant.

The question is asking whether $X \cap (s-\epsilon, s]$ is always non-empty.

What you did is describe that $(s-\epsilon, s]$ is an interval and then restate the question. Hence you are not answering the question at all.

Also, I dont' think the term continuous set is well defined.

2. The question is asking whether you can always find element of $X$ in $(u-\epsilon, u]$ for any positive $\epsilon$. You are describing a particular choice of $\epsilon$. Again, you are not answering the question.

You should consider $\epsilon$ in general. If it is false, just show an example.

3. supremum is the least upper bound (that may lie in the set). Remove the bracket content, those are redundant information.

4. What is supremum? Does every element that satisfy the condition that you stated can be classified as supremum. You have to state that the supremum is an upper bound that satisfy that property.

I think the intention of the question is to assume that suppose a point $s$ satify the first definition, verify that it has the second definition and the opposite holds too. But let me ignore the instruction and verify your proof.

$$s-s'=k\epsilon, k<1$$

what do you mean? if $s$ is the least upper bound $s-s'$ is non-positive, $k\le0$.

I don't understand what do you mean by "this keep lower bound $<s'$."

Perhaps you might want to say that $\exists x \in X \cap \text{Something}$.

What is the rationale to conclude that $X \cap (s-\epsilon, s]\neq \emptyset$? Perhaps you are trying to say $(s-\epsilon, s'] \neq \emptyset$?

If you have two equivalent definition, you can use whichever that is easier to show that a particular value is a supremum.