April 5th, 2015, 04:51 AM  #1 
Member Joined: Mar 2015 From: Brasil Posts: 90 Thanks: 4  Finite and Infinite Sets
Note \[A, B\subset \mathbb{R}\] nonempty sets limited, consider the set \[A+B=\left \{ x+y\setminus x\in A, y\in B \right \}.\] Show that: a) \[A+B\] is a limited set; b)\[\sup(A+B)=\sup A+\sup B\] c) \[\inf(A+B)=\inf A+\inf B\] Last edited by skipjack; April 5th, 2015 at 10:15 PM. 
April 5th, 2015, 12:48 PM  #2 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,617 Thanks: 2608 Math Focus: Mainly analysis and algebra 
1) Consider $a$, an upper bound of $A$, and $b$, an upper bound of $B$ (these exist because $A$ and $B$ are both bounded). Then $x \lt a$ for all $x \in A$ and $y \lt b$ for all $y \in B$. So $x + y \lt a + b$ for all $x \in A$ and $y \in B$ and thus for all $(x +y) \in A+B$. Thus $a + b$ is an upper bound for $A+B$. A similar argument can be constructed for lower bounds. For 2) assume that $\sup{(A+B)} \ne \sup A + \sup B$ and come to a contradiction. For example: suppose that $\sup{(A+B)} \lt \sup A + \sup B$ then $\sup{(A+B)} = \sup (A) + \sup (B)  4c$ for some $c \gt 0$. Thus we may take $a = \sup (A)  c \in A$ and $b = \sup (B)  c \in A$. Thus $a + b \in A+B$ by the definition of $A+B$, but $a+b \gt \sup{(A+B)}$. This contradiction means that $\sup{(A+B)} \not \lt \sup A + \sup B$. A similar argument shows that $\sup{(A+B)} \not \gt \sup A + \sup B$. Further similar arguments can be employed for part 3). 
April 5th, 2015, 04:54 PM  #3 
Member Joined: Mar 2015 From: Brasil Posts: 90 Thanks: 4 
Can you show the rest ?????????

April 5th, 2015, 05:29 PM  #4 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,617 Thanks: 2608 Math Focus: Mainly analysis and algebra 
Why don't you have a go? And we'll correct your work if there are errors.

April 8th, 2015, 05:56 AM  #5 
Member Joined: May 2013 Posts: 31 Thanks: 3 
a) $\displaystyle A, B \subset \mathbb{R}$ limited $\displaystyle \implies$ $\displaystyle \exists \ a := sup(A), \ \exists \ b:=sup(B), \ a,b \in \mathbb{R} \ \ \implies \ \ x \le a, \ y\le b \ \ \ \forall x\in A, \ \forall y \in B \ \ \implies \ \ x + y \le a + b, \ \ \ \forall x+y \in A+B$ $\displaystyle \therefore z:=x+y, \ \ z \le a + b$, $\displaystyle \forall z \in A+B$ b) i) upper bound: item a) $\displaystyle \implies z \le a + b, \ \ \ \forall z \in A+B$ ii) least upper bound: $\displaystyle \forall \epsilon >0:\ \ \ \exists x_0\in A, \ \exists y_0 \in B : a  \frac{\epsilon}{2}<x_0, \ b  \frac{\epsilon}{2}<y_0 \ \ \implies \ \ a + b  \epsilon < x_0+y_0\in A+B $ $\displaystyle \therefore$ i, ii $\displaystyle \implies \ \sup(A+B) = a + b, \ \ a=\sup A, \ b= \sup B$ c) Proposition: $\displaystyle \sup(A) =  \inf (A)$ Proposition $\displaystyle \implies \inf(A+B) =  \sup((A+B)) =  \sup((A) + (B)) = (\ \sup(A) + \sup(B) \ ) =  \sup(A)  \sup(B)$ $\displaystyle \therefore \inf(A+B) =  \sup(A)  \sup(B) = \inf(A) + \inf(B)$ Last edited by Prokhartchin; April 8th, 2015 at 06:00 AM. 

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