Linked Questions

3
votes
1answer
6k views

Proving $\sup(A+B)=\sup(A)+\sup(B)$ [duplicate]

Possible Duplicate: How can I prove $\sup(A+B)=\sup A+\sup B$ if $A+B= \lbrace a+b\mid a\in A, b\in B\rbrace $ here's a homework question I'm currently working on: Let $A,B \subset \mathbb{R}$ ...
2
votes
2answers
398 views

Prove that $\sup{(A+B)} = \sup{A}+\sup{B}$ [duplicate]

Let $A,B$ be subsets of $\mathbb{R}$. Prove that $\sup{(A+B)} = \sup{A}+\sup{B}$. I think in order to solve this we are going to have to use the mathematical definition of supremum. Maybe we can ...
1
vote
3answers
438 views

proving a inequality about sup [duplicate]

Possible Duplicate: How can I prove $\sup(A+B)=\sup A+\sup B$ if $A+B=\{a+b\mid a\in A, b\in B\}$ I want to prove that $\sup\{a+b\}\le\sup{a}+\sup{b}$ and my approach is that I claim $\sup a+ \...
2
votes
1answer
476 views

Prove $\inf (A + B) = \inf A + \inf B$ [duplicate]

Possible Duplicate: How can I prove $\sup(A+B)=\sup A+\sup B$ if $A+B=\{a+b\mid a\in A, b\in B\}$ Prove: If $A$ and $B$ are nonempty subsets of $\mathbb{R}$ and are both bounded above, ...
4
votes
0answers
612 views

Proving that $\sup(A+B)=\sup(A)+\sup(B)$, $A,B\subseteq\mathbb{R}$. [duplicate]

Possible Duplicate: How can I prove sup(A+B)=supA+supB if A+B={a+b} How would I go about proving that the supremum of A + B (where A and B are each subsets if $\mathbb{R}$) is equal to the ...
0
votes
0answers
95 views

Proving $\sup(A) + \sup(B) = \sup(A + B)$ without using $\epsilon$ [duplicate]

Suppose sets $A$ and $B$ are not empty and are bounded from above. Prove that $\sup(A) + \sup(B) = \sup(A + B)$ without using $\epsilon$. This is exercise 1.3.6 from Understanding Analysis by Abbot, ...
0
votes
1answer
48 views

Taking suprema on a set-equality [duplicate]

Suppose we have the following: $X = A + B $ where $X, A,$ and $B$ are any sets. $A + B = \{ a + b : a \in A , \; \; \; b \in B \} $ Can we conclude that $\sup X = \sup A + \sup B $ ?
1
vote
0answers
44 views

Proving $x+y =\sup\left\{ r+s: r,s \in \mathbb{Q}, r\leq x, s\leq y\right\}$ [duplicate]

I am trying to show that the following holds: $$x+y =\sup\left\{ r+s: r,s \in \mathbb{Q}, r\leq x, s\leq y\right\}.$$ I have been able to show that $$x=\sup\{q\in \mathbb{Q}: r\leq x\}$$ and ...
6
votes
2answers
18k views

Prove that $\sup(A+B) = \sup(A) + \sup(B)$ and why does $\sup(A+B)$ exist?

We want to show that $\sup(A)+\sup(B)$ is the least upper bound of the set $A + B$. First, we need to show that $\sup(A) + \sup(B)$ is an upper bound for the set $A + B$. Indeed, if $z\in A + B$, then ...
2
votes
1answer
5k views

Supremum of set of sum of elements

Given the sets $A, B \subseteq \mathbb R$ and $A, B \ne \varnothing,$ $A,B$ upper bound, with $$C = \{a + b | a \in A, b \in B\},$$ I am asked to show that $\sup C = \sup A + \sup B$. I have a few ...
7
votes
2answers
373 views

1-Variable Real Analysis True/False questions on Supremum, Infimum, and Inequalities with them

(S. Abbott. Understanding Analysis 1 ed. pp 18 question 1.3.9) is asking me to answer the following questions without any formal proofs. I have some intuition for them, but I was hoping to get some ...
2
votes
2answers
1k views

Proving sup(A + B) = sup A + sup B

There are two sets $A$ and $B$ which are bound and are not empty. Now we'll define another set as: $A + B = \{a + b | a \in A, b \in B\}.$ I need to prove that $\sup(A + B) = \sup A + \sup B.$ I ...
3
votes
3answers
46 views

What is $sup{ (-1)^n + 1/n} $?

My doubt is that $\sup (A+B) = \sup A + \sup B$. So, if we consider $A =(-1)^n$ and $B=\frac{1}{n},$ then $\sup A = 1 $ and $\sup B = 1$. So will $\sup (A+B) = \sup{ (-1)^n + 1/n} = \sup (-1)^n ...
0
votes
1answer
264 views

Why is $L(P,f)+L(P,g) \le L(P,f+g) \le U(P,f+g) \le U(P,f)+U(P,g)$

Let $R \subset \Bbb{R}^n$ and $f,g$ be two integrable functions from $R\to \Bbb{R}$ , in a proof of a property of integrals Rudin writes the following: $$L(P,f)+L(P,g) \le L(P,f+g) \le U(P,f+g) \le U(...
2
votes
1answer
110 views

The norm of operators in Hilbert space

In this question $\mathcal{H}$ stands for a Hilbert space over $\mathbb{K}=\mathbb{R}$ or $\mathbb{C}$, with inner product $\langle\cdot\;| \;\cdot\rangle$ and the norm $\|\cdot\|$ and let $\mathcal{B}...

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