Let $C ⊂ R^n$ be a nonempty convex set and let $b ∈ R^n.$ Then the set given by $b + C = {x + b : x ∈ C}$is a convex set. How do we prove this statement?
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Take a convex combination of $x+b$ and $y+b$, where $x,y\in C$, and show that it has the form $z+b$, where $z$ is a convex combination of $x$ and $y$. – Brian M. Scott Oct 08 '20 at 19:44
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Welcome to stackexchange. You are more likely to get help rather than downvotes or votes to close if you [edit] the question to show us what you tried and where you are stuck. Here I suggest you start with the definition of convex applied ot the translated set. – Ethan Bolker Oct 08 '20 at 19:45
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So we'll need to use that the convex combination of $x,y \in C$ then $(1-t)x + ty \in C$ with $t \in [0,1]$ to show that the convex combination of two points in $C+b$ are also in $C+b$. To see this we have that $$(1-t)(x+b)+t(y+b)=(x+b)-t(x+b)+t(y+b)= (x+b) + t((y+b)-(x+b)) = (x+b) + t(y-x) = b + (x + ty-tx) = (1-t)x + ty + b$$ and so the convex combination of $x+b$ and $y+b$ is in $C+b$ therefore it is convex.
CyclotomicField
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Show that, if for any $x,y\in C$, the segment $[x,y]$ is contained in $C$, the segment $x+b,y+b$ is contained in $b+C$.
Do you know how to parameterise a segment?
Bernard
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This is the right idea, but proving it would not actually do what the OP wants. A correct argument should start with two points in the translated set. – Ethan Bolker Oct 08 '20 at 19:57
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@EthanBolker: Having two points in the original set or in the translated set is equivalent. Anyway, I posted only a starting point. – Bernard Oct 08 '20 at 20:01
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Yes, equivalent because the translation is a bijection. But a beginner writing a proof will end up with a formal implication in the wrong direction. I think @BrianMScott 's hint in a comment is better. – Ethan Bolker Oct 08 '20 at 20:08