A convex set plus a constant is still convex

Question

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?

Answer 1

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.

Answer 2

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?