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0 votes
2 answers
61 views

Whether $\sup(\sum\limits_{i=1}^{\infty}X_i)$ is equal to $\sum\limits_{i=1}^{\infty}(\sup X_i)$

We have $\sup(A+B)=\sup(A)+\sup(B) $.Thus, we have $\sup(\sum\limits_{i=1}^{n}X_i)=\sum\limits_{i=1}^{n}(\sup X_{i})$ for every finite integer $n\in\mathbb{N}$. However, what about the set sequence? ...
MGIO's user avatar
  • 117
0 votes
2 answers
198 views

Following up with a previous question on $\sup(A)+\sup(B) = \sup(A + B)$

The question link is here: Prove that $Sup(A + B) = Sup(A) + Sup(B)$ Can someone look at the answer given and explain why epsilon is introduced and how that whole second part works?
K. Gibson's user avatar
  • 2,391
1 vote
1 answer
4k views

Prove that $Sup(A + B) = Sup(A) + Sup(B)$

Earlier on in the book it showed that to prove $a = b$ it is often best to show that $a \leq b$ and that $b \leq a$. This is the way I want to go about the proof. I am sure there is an easier way but ...
K. Gibson's user avatar
  • 2,391
0 votes
3 answers
137 views

Supremum of Sumset (Proof Writing)

Given $A,b\subseteq\mathbb{R}$, define the set $A+B=\lbrace a + b | a\in A, b\in B\rbrace$. I would like to prove that $\sup(A+B)=\sup(A)+\sup(B)$, but in a specific way. Here is what I have done so ...
高田航's user avatar
  • 2,125
0 votes
2 answers
1k views

Proving $\sup \left( {A + B} \right) = \sup A + \sup B$ using the usual definition.

Prove $\sup \left( {A + B} \right) = \sup A + \sup B$ using the definition given in this problem (link). I was able to prove using lemma of the given definition. My attempt is here, let $s = \sup A$,...
user avatar
1 vote
1 answer
101 views

Question about Schnirelmann Density and Sumset: if $d(A) \ge \frac{1}{2}$ and $d(B) > 0$, wouldn't $d(A+B)=1$

I've been thinking about the Schnirelmann Density and I think that I may still be confused about SumSet and Density. It seems to me that if $d(A) \ge \frac{1}{2}$ and $d(B) > 0$, then $d(A+{B}) = ...
Larry Freeman's user avatar
49 votes
5 answers
90k views

How can I prove $\sup(A+B)=\sup A+\sup B$ if $A+B=\{a+b\mid a\in A, b\in B\}$

If $A,B$ non empty, upper bounded sets and $A+B=\{a+b\mid a\in A, b\in B\}$, how can I prove that $\sup(A+B)=\sup A+\sup B$?
Lona Payne's user avatar