Questions based on $\epsilon$ based definition of Supremum.

Need help in vetting my answers for questions in sec 2.5 in chap. 2 (page 8) in CRM series book by MAA: Exploratory Examples for Real Analysis, By Joanne E. Snow, Kirk E. Weller. The question refers also to "new definition" introduced in Q. 2.3 (in answer for part $$4$$) in my last post, that am repeating below for ease of reference:

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

Q. 2.5:

1. If $$X$$ is a nonempty set for which the supremum exists, that is, $$sup(X)=s$$, what is the minimal number of elements of $$X$$, if any, that must lie in $$(s-\epsilon, s]$$ for every choice of $$\epsilon\gt 0$$. Carefully, explain your response. Is your response consistent with the definition you wrote in Section 2.3?

Not sure what the question is asking as for reals cannot count elements in any interval. However, my answer is : Minimal number of elements of $$X$$ in the interval $$(s-\epsilon, s]$$ is $$= (s-\epsilon, s]$$, if $$s\in X$$, else $$(s-\epsilon, s)$$ if $$s\notin X$$.

1. Use the "new" definition to prove that
$$\sup(\{(1- \frac1{3^n})\,\,: n \in N \})= 1.$$

$$s=1$$, for $$n\rightarrow \infty$$. Below is table for set members for few values of $$n$$: $$\begin{array}{c|c|} & \text{n\in \mathbb{N}}& \text{(1- \frac1{3^n})}\\ \hline a & 1& \frac23=0.\overline{6}\\ \hline b & 2& \frac89=0.\overline{8}\\ \hline c & 3& \frac{26}{27}=0.\overline{962}\\ \hline d & 4& \frac{80}{81}=0.\overline{987654320}\\ \hline \vdots\\ \hline \infty & n\rightarrow \infty& \approx 1\\ \hline \end{array}$$ $$s\not\in \{(1- \frac1{3^n}\,\,: n \in N \})$$, as $$s$$ does not lie in the set.

$$(1- \frac1{3^n}) \in [\frac23, 1)$$.

$$\epsilon \in\mathbb{R}, \epsilon\gt 0\implies \forall \epsilon \gt 0: n\lt \infty$$, as $$s - \epsilon \lt 1$$.

For interval $$(s-\epsilon,s]$$, the lower bound implies for $$\epsilon\gt 0$$ that $$s-\epsilon \lt 1\implies n \lt \infty$$, & the upper 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.

1. Identify the supremum of the set given below, & use the "new" definition of supremum to prove your claim: $$\{(-\frac12)^n\,\,: n \in N \}$$

$$\sup(\{(-\frac12)^n\,\,: n \in N \})= 1.$$

$$s=\frac14$$, for $$n=2$$. Below is table for set members for few values of $$n$$: $$\begin{array}{c|c|} & \text{n\in \mathbb{N}}& \text{(-\frac12)^n}\\ \hline a & 1& -\frac12=-0.5\\ \hline b & 2& \frac14=0.25\\ \hline c & 3& -\frac 18=-0.125\\ \hline d & 4& \frac{1}{16}=0.0625\\ \hline \vdots\\ \hline \infty & n\rightarrow \infty& \approx 0\\ \hline \end{array}$$

As $$-(\frac12)^n\in \{-\frac12, \cdots, \frac14\}$$, & as $$n \rightarrow \infty, -(\frac12)^n \approx 0$$.
For interval $$(s-\epsilon,s]$$, the lower bound implies for $$\epsilon\gt 0$$ that $$s-\epsilon \lt \frac14$$, & the upper bound implies the max. value is $$s=\frac14$$. All possible values of $$n$$ are covered in the interval $$(s-\epsilon,s]$$ with $$n=2$$ at the upper bound.

1. In considering the "new" definition of supremum, state a condition, or set of conditions, by which a set would fail to have a supremum? In other words, what is the negation of the definition that you came up with in Section 2.3?

The "new" definition is based on premises: a nonempty set $$X$$, a positive real number $$\epsilon$$, & the half-open interval $$(s−\epsilon,s]$$, where $$s$$ denotes the supremum.

Out of this only the first premise can be negated, as then the two further assumptions of : (i) positive real non-negative $$\epsilon$$ does not hold, (ii) similarly, the assumption of having interval of values: $$(s-\epsilon, s]$$ cannot hold.

Hence, the negation of "new" definition is:
"new definition": Let there be a nonempty set $$X$$ with supremum $$s$$, then $$X\cap(s - \epsilon, s]\ne \emptyset, \,\, \forall \epsilon\gt 0$$.

"negation": $$\forall X =\emptyset$$, supremum $$=-\infty$$, & there is no possible $$\epsilon\gt 0$$.

1. If $$s$$ is the supremum of a nonempty set $$X$$ of real numbers, under what condition(s) do we find an infinite number of elements of $$X$$ in the interval $$(s-\epsilon, s]$$ for every choice of $$\epsilon\gt 0$$. Justify your reasoning.

The members of set $$X$$ lie in reals, so for infinite number of elements $$\forall \epsilon \gt 0$$, need $$s = \infty$$. Then the interval $$(s - \epsilon, s]$$ for any positive value of $$\epsilon$$ is having infinite number of values.

1. If $$X$$ is the empty set, does the supremum of $$X$$ exist? Explain your answer.

supremum $$\ge$$ larger than any value in the set, & is the $$lub$$. For an empty set, any value can act as upper bound, & supremum is the least of any possible value. so $$s=-\infty$$.

• Concentrate on the set $X$, not on the intervals $(s-\epsilon,s]$. Those intervals always have infinitely many elements when $\epsilon>0$. The set $X$ on the other hand... – Jyrki Lahtonen May 9 at 5:21
• Hmm. Consider the set $X=\{0,1\}.$ It has a supremum $s$, right? What is it? What can you say about the number of elements of $X$ in the interval $(s-\epsilon,s]$ for various choices of $\epsilon>0$? – Jyrki Lahtonen May 9 at 5:26
• For the purposes of Q2.5. you are free to select the set $X$ in any way that exhibits the possible behaviors. I simply picked $X=\{0,1\}$. As it has a supremum, it can be used as an example. There is nothing to "derive" here. – Jyrki Lahtonen May 9 at 7:52
• What I was getting at is that my answer to Q.5 would read: For every $\epsilon>0$ there are infinitely many elements of $X$ in the interval $(s-\epsilon,s]$ if and only if $\sup X=s=\sup(X\setminus\{s\}$. For example, when $X=\{0,1\}$ there are only finitely many elements of $X$ in the said interval for any $\epsilon$. On the other hand, for $X=(0,1)$ there will be infinitely many. Note: the example is only about the statement of my answer to Q.5. Justification is longer (and might be worth a separate question). – Jyrki Lahtonen May 9 at 9:09
• Anyway, you are right. When $s\notin X$, we have $X=X\setminus\{s\}$. Implying that $\sup X=\sup (X\setminus\{s\})$. So one direction is to show that when $s=\sup X$ and $s\notin X$, then there will always be infinitely many elements in $X\cap (s-\epsilon,s]$. – Jyrki Lahtonen May 9 at 9:12

1. Answer is $$1$$. Suppose for some $$\epsilon>0$$, $$X \cap (s-\epsilon, s]=\emptyset$$, then this will contradict that $$s$$ is a supremum. Let $$X=\{0\}$$, and $$s=0$$, then we can easily see that the answer is $$1$$.

2. You want to show two thing, $$1$$ is an upper bound and also $$\forall \epsilon > 0, (1-\epsilon, 1] \neq \emptyset$$. To show the first point, since $$\frac1{3^n}>0$$, $$1-\frac1{3^n} < 1$$, hence $$1$$ is an upper bound. Also, clearly, we have $$1 \in (1-\epsilon, 1], \forall \epsilon >0$$.

3. The odd terms are negative, the even terms are positive. Hence we can focus on the even terms. Also the even subsequence forms a decreasing subsequence. Hence $$\frac1{2^2}=\frac14$$ is an upperbound. Since $$\frac14$$ is in the set. It is the supremum.

4. The new definition of supremum is: Given a non-empty set $$X$$, $$s$$ is a supremum of $$X$$ if $$s$$ is an upperbound of $$X$$ and $$\forall \epsilon >0, (s-\epsilon, s] \cap X \neq \emptyset$$. A non-empty set $$X$$ doesn't have a supremum if the set has no upper bound or $$\exists \epsilon >0, (s-\epsilon, s] \cap X = \emptyset$$.

5. A sufficient condition to ensure that there are infinitely elements of $$X$$ in $$(s-\epsilon, s] \cap X$$ is when $$s \notin X$$. For any $$\epsilon >0$$, We can construct a sequence of distinct elements by first drawing the first element, $$x_1$$ from $$(s-\epsilon, s]$$, since $$s \notin X$$, $$x_1 < s$$, afterwhich, we pick $$x_2 \in (x_1, s]$$, and so on.

6. Your book did not define supremum for emtpy set.

• I disagree with answer for 4, as do not know of a set where there is no upper bound; as for me $\infty$ is a possible upper bound & hence can be supremum. I saw at :math.stackexchange.com/a/1425792/424260, that : If you consider the real numbers as a subset of itself, there is no supremum. However, do not know its meaning at all, or how to derive this result. – jiten May 9 at 16:08
• when we disagree with each other, we check the definition. Check in your book, does supremum have to be a real number? is $\infty$ a real number? – Siong Thye Goh May 9 at 16:10
• Please tell me of cases where supremum is not in a continuous range of values (like reals), & does not exist too. – jiten May 9 at 16:15
• $\mathbb{N}, \mathbb{Z}, \mathbb{Q}$. – Siong Thye Goh May 9 at 16:16
• My request that the values are not in a continuous range, like reals, was wrongly worded. Should have stated: values are not unbounded. But, then the issue is solved. But, then my earlier questions (in earlier posts, for questions from this book) stated $s=\infty$ for unbounded sets. So, are those all answers wrong? – jiten May 9 at 16:22