Let $a,b,c,d\in \mathbb{R}$ with $a\leq b $ and $c\leq d$. If we consider the Minkowski sum $A+ B = \{x+y : x\in A,y\in B \},$ is true in general that $[a,b] + [c,d] = [a+c, b+d] $?
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1um yea, why not? – mathworker21 Jun 18 '19 at 23:50
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Yes it is always true. – GReyes Jun 18 '19 at 23:50
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2$\max (A+B) = \max A + \max B$, and similarly for $\min$. Both $A,B$ are convex & compact, hence so is $A+B$. Convex subsets of the real line are intervals. – copper.hat Jun 18 '19 at 23:52
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1If $a\le x\le b$ and $c\le y\le d$ then $a+c\le x+c\le x+y \le b+y\le b+d$ – J. W. Tanner Jun 18 '19 at 23:54
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Yes. If $a\le x\le b$ and $c\le y\le d,$ then $a+c\le x+c\le x+y \le b+y\le b+d,$
using the property of the ordered field of real numbers that
if $r,s,t\in\Bbb R$ and $s\le t$ then $r+s\le r+t$.
J. W. Tanner
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The set $X = [a, b] + [c, d]$ clearly lies within the closed interval $[a + c, b + d]$ and contains its endpoints. Since $X$ is the image of the connected set $[a, b]\times [c, d]$ under the continuous map $(x, y) \to x + y$, it is itself connected, and thus must be exactly $[a+c, b + d]$.
anomaly
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Let $x \in [a,b] + [c,d],$ then there exists $x_1 \in [a,b]$ and $x_2 \in [c,d]$ s.t. $x = x_1 + x_2.$ Clearly $a + c \leq x_1 + x_2 \leq b + d$ so $x \in [a+c,b+d].$ To complete this simply prove the reverse containment, $x \in [a+c, b+d]$ implies $x \in [a,b] + [c,d].$
Fady Nakhla
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