0
$\begingroup$

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, as not sure of the proof validity.

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?

Not clear about the practical significance. Just repeated the conclusion of both parts below.

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

$\endgroup$
  • $\begingroup$ I'm having a hard time understanding your proof of Def. 1 $\implies$ Def. 2. You want to show that $s\leq s^\prime \leq s$, so the addition of the variable $k$ is not necessary. Furthermore, your explanation is not very clear. $\endgroup$ – Hossmeister May 9 at 2:20
  • $\begingroup$ @Hossmeister If I do not add variable $k$, then it would be difficult to show formally how the lower bound of $(s-\epsilon,s]$ need be equal to $s'-\epsilon$. Please detail your method that formally achieves the same without introducing variable $k$. $\endgroup$ – jiten May 9 at 2:26
1
$\begingroup$

I noticed that in your statement of Def 2 you neglected to mention that $s$ had to be an upper bound of $X$. I have written the proof with this assumption in mind.

To show that Def. 1 $\implies$ Def. 2, let $s$ be the supremum of $X$ defined as per Def 1. Let $s^\prime$ be an upper bound of $X$ such that $X\cap(s^\prime - \epsilon, s^\prime]\ne \emptyset, \,\, \forall \epsilon\gt 0$. Since $s^\prime$ is an upper bound of $X$, $s\leq s^\prime$. Now suppose for contradiction that $s<s^\prime$. Let $\epsilon:=s^\prime-s$. Then $X\cap(s,s^\prime]\neq \emptyset$, which contradicts the assumption that $s$ is an upper bound for $X$. So $s\geq s^\prime$, so it must be true that $s=s^\prime$.

To show that Def. 2 $\implies$ Def. 1, let $s$ be an upper bound of $X$ such that $X\cap(s - \epsilon, s]\ne \emptyset, \,\, \forall \epsilon\gt 0$. Suppose there is another upper bound, say $t$, such that $s>t$. Let $\epsilon:=s-t$. Then $X\cap (t,s] \neq \emptyset$, which contradicts the assumption that $t$ is an upper bound for $X$. So $s$ is both an upper bound for $X$, and its least upper bound.

Showing that these definitions are logically equivalent makes it easier to prove the Least Upper Bound Property of the real numbers (and equivalently proving the Greatest Lower Bound Property), as well as demonstrating other related proofs.

$\endgroup$
  • $\begingroup$ Please see my related post at : math.stackexchange.com/q/3219295/424260 $\endgroup$ – jiten May 9 at 5:04
  • $\begingroup$ I am sorry, but taking $s\gt t$ is a very easy assumption to use, in part (ii). A better proof would be one that considers $t$ to be a still upper bound than $s$. But then $\epsilon:= t - s$. So, $X \cap (s -\epsilon, s] \ne \emptyset \implies (2s-t, s]\ne \emptyset$. This implies $s \gt 2s-t \implies s\lt t$, but it is just a restatement of the assumption. It leads to no proof. $\endgroup$ – jiten May 10 at 2:08
  • $\begingroup$ I request help on my last comment, as want to solve similar question for infimum, but want in part (ii) the different approach, & it leads nowhere like above & gives only a restatement. There took $t\lt i$, But then $\epsilon:= i-t.$ So, $X\cap [i, i +\epsilon) \ne \emptyset \implies [i, 2i-t)\ne \emptyset$. Hence, need $2i -t \gt i$. This implies $i \gt t $, but there is no proof by contradiction here, just a restatement of the assumption made. $\endgroup$ – jiten May 10 at 2:21
  • $\begingroup$ I have a feeling that the failure is due to attempt to prove the axioms: i.e. the supremum is the $lub$ among the bounds for a non-empty set. $\endgroup$ – jiten May 10 at 15:00
  • 1
    $\begingroup$ @jiten I don't understand what you're trying to do here. What do you mean by $s>t$ is a very "easy" assumption to use? In part (ii), you know that $s$ is an upper bound of $X$, so you just need to prove that it is the least upper bound. That is why I take $t<s$. For the related problem regarding the infimum, you know that $i$ is a lower bound of $X$; you want to show it is the greatest lower bound. So take $t>i$. $\endgroup$ – Hossmeister May 11 at 2:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.