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Prove that the set $C:=\{ x : x+S_2 \subset S_1 \}$, with $S_1,S_2\subset \Bbb R^n$ is convex if $S_1$ is convex.

I understant that a vectorial space is a convex set. So $S_1$ and $S_2$ are both convex sets. But I do not understand how to continue with the idea.

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Suppose $x, y \in C$ and $0 \le t \le 1$. You want to show $tx + (1-t)y \in C$, i.e. $tx + (1-t)y + S_2 \subset S_1$. If $s_2 \in S_2$, $tx + (1-t) y + s_2 = t (x + s_2) + (1-t) (y + s_2)$ with $x + s_2 \in S_1$ and $y + s_2 \in S_1$. Since $S_1$ is convex, you are done.

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Here is another way of looking at it:

Note that $C = \{ x | x +y \in S_1 \ \forall y \in S_2 \} = \{ x | x \in S_1-\{y\} \ \forall y \in S_2 \} = \cap_{ y \in S_2} (S_1 - \{y\})$.

Each set $S_1 - \{y\}$ is convex since it is the translate of a convex set and since the intersection of convex sets is convex, we see that $C$ is convex.

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