1

Taken from sec. 1.4.1 of the book by Mary Hart, titled: Guide to Analysis.

Let $A, B$ be two non-empty sets of real numbers with supremums $\alpha, \beta$ respectively, and let the sets $A + B$ and $AB$ be defined by :
$A + B = {a + b: a\in A, b\in B}$,
$AB= {ab:a\in A, b\in B}$.

Show that $\alpha+ \beta$ is the supremum of A + B.

My attempt :

If a non-empty set of reals $A$ is having supermum $\alpha$, & other set $B$ is the supremum $\beta$; then two cases are possible for each supermum:
(i) supremum is in set, and then it equals the maximum; as say $(0.1,0.2]$,
(ii) supremum is not in set, & maximum does not exist; as say $(0.1, 0.2)$.

In case of combination of such two sets possibilities arise based on larger value. However, its inclusion, or exclusion (i.e. $],)$) does not affect supermum, as it always exists.
Say, set $A$ is $(a,b)$ & set $B$ is $(c,d)$, then the sum has upper limit decided by the relative magnitudes of $a,b,c,d$. So, if $a\gt b \gt c \gt d$; then supermum is $b$, & no maximum.

But, it is not clear how to prove the question.

P.S. There are few other questions from the same book, & request link to their answers; as could not find (even, could not find answer to this!).
Till now, I resisted stating these 'other' questions, as wanted to post seperate questions (to prevent clutter), but if links were provided then no need.
These questions are stated below:

  1. Give an example to show that AB need not have a supremum.
  2. Prove also that even if AB has a supremum, this supremum need not be equal to $\alpha \beta$.

  3. Show that if $A$ be set of positive reals with supremum $\alpha$, & let $Y = {x^2 : x\in X}$; then $\alpha^2$ is supremum of Y.

jiten
  • 4,524
  • 1
    https://math.stackexchange.com/questions/2605515/supremum-of-sumset-proof-writing/2605614#2605614 – Peter Szilas Apr 21 '19 at 07:19
  • @PeterSzilas Thanks a lot. But, there are other questions as stated in edited OP. Please help for those too. – jiten Apr 21 '19 at 07:58
  • I'd guess your question will be closed, so maybe you should ask a new question anyway, without the clutter of the duplicate. You'll also have space to show your work on questions 2,3,4 – Calvin Khor Apr 21 '19 at 08:14
  • jiten.Welcome. As suggested perhaps ask another question, also try a bit, yourself, and you will get more answers, and I shall try, too,:) – Peter Szilas Apr 21 '19 at 09:27
  • @PeterSzilas Being new to the topic, it seems will take lot of time before placing new post on that. Other hurdles include 1. variations in defn. : as Kaczor states in answer to 1.1.11 (a) infinity as supremum, while usually for bounded intervals (in range) the supremum is defined; 2. Difficulty in understanding theoretical answers on mse, but the referred book (by me, here) giving terse hints using quad. eqn. for part 3. Better I be in chat as then flow is better. Also, it is possible then to evaluate different theoretical answers on mse, & how to derive examples from them as in the book. – jiten Apr 22 '19 at 00:11
  • @CalvinKhor Kindly see my last comment. – jiten Apr 22 '19 at 00:18
  • 1
    (i) Who is Kaczor? What is 1.1.11(a)? (ii) I think "need not have a supremum" should be understood as "need not have a finite supremum, even if $\alpha,\beta<\infty$. (iii) For 3, try to violate question 4's assumptions. Above all, try some concrete examples. If this is not enough, please ask a new question. – Calvin Khor Apr 22 '19 at 05:39
  • @CalvinKhor Meant book by authors: 'Kacjor, Nowak', titled : 'Problems in mathl. analysis I: Real numbers, sequences & series'. Book gives solutions after all problems. For exercise 1.1.1 (a) Find $ \sup { x \in \mathbb{R} : x^2 +x+1 \gt 0 } $, it states in soln. : $+\infty $, which is in direct contrast to what is stated elsewhere, i.e. only for bounded intervals do supremum exists. Anyway, nowhere would infinity be stated as supremum. Sorry, tried to understand your comment under (ii), but couldn't grasp. You stated in (iii) to violate Q. 4 assumptions, but which ones? Please elaborate. – jiten Apr 22 '19 at 06:41
  • @CalvinKhor I think of asking new question but for Q.3 there is given hint in book (here) as: quadratic $3x^2-10x+3$ with roots $x=\frac13, 3$. Am only able to elaborate upon it, with no further theoretical gain preventing me to post : Form a bounded set taken by $x \in \mathbb{R}$, in which range $\lt 0, \frac13<x<3$. So, no max., but supermum exists $= 0$. Coming to linear factors: $(3x−1)(x−3)$ in bounded domain, with supremum $8,0$ resp. at $x=3$. While at $x=\frac 13$, have supremum values in range as $0, \frac{−2}3$. The product of supermum at each end of domain $= 0$ too as needed. – jiten Apr 22 '19 at 06:51
  • @CalvinKhor Unable to develop any / theoretical knowledge for Q. 2,4. This is preventing me to ask new question on it. Only on hint (using a quad. eqn.) for Q. 3 am able to elaborate somewhat. This does not help me theoretically enough to post. If could help by elaborating how the Q. 4 assumptions can be violated for Q. 3, then might be it helps. – jiten Apr 22 '19 at 06:56
  • @CalvinKhor Regarding my 3rd last comment, I want to elaborate as to why could not grasp your comments under (ii). The reason is how it is possible to have no finite supremum for product of sets $AB$, if their respective both supremum $\alpha, \beta$ are finite (i.e. $\lt \infty$). – jiten Apr 22 '19 at 08:07
  • 1
    For Q4, violate the positivity. You're looking at the wrong region for Q3. For Q2, I am telling you to take "$\sup A = \infty$" as equivalent to "$\sup A$ does not exist". Sorry, but I'm not going to reply further on this post because its too messy. This is enough to ask a new question, please do so if you need more help. – Calvin Khor Apr 22 '19 at 08:09
  • @CalvinKhor Please find the post at : https://math.stackexchange.com/q/3196871/424260 – jiten Apr 22 '19 at 09:14
  • @PeterSzilas Please find the post for rest questions at : https://math.stackexchange.com/q/3196871/424260 – jiten Apr 22 '19 at 09:18
  • @CalvinKhor Kindly elaborate as comment or answer in my new post, your statement of violation of condition of Q.4 for Q. 3. – jiten Apr 23 '19 at 01:32

0 Answers0