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April 5th, 2015, 04:51 AM   #1
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Finite and Infinite Sets

Note \[A, B\subset \mathbb{R}\] non-empty 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.
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April 5th, 2015, 12:48 PM   #2
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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).
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April 5th, 2015, 04:54 PM   #3
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Can you show the rest ?????????
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April 5th, 2015, 05:29 PM   #4
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Why don't you have a go? And we'll correct your work if there are errors.
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April 8th, 2015, 05:56 AM   #5
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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 $\displaystyle \implies \inf(A+B) = - \sup(-(A+B)) = - \sup((-A) + (-B)) = -(\ \sup(-A) + \sup(-B) \ ) = - \sup(-A) - \sup(-B)$

$\displaystyle \therefore \inf(A+B) = - \sup(-A) - \sup(-B) = \inf(A) + \inf(B)$
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Last edited by Prokhartchin; April 8th, 2015 at 06:00 AM.
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