How can $S+a$ convex, if $S \in \mathbb{R}^n$ and $a \in \mathbb{R}^n$

Question

I'm reading a book on convex optimisation and understood the definition of convexity. However, I'm not able to understand the following statement.

If $S \subseteq \mathbb{R}^n$ is convex, and $a \in \mathbb{R}^n$, then the set $S+a = \{x+a \mid x \in S \}$ is convex.

How can we guarantee that every point on the line segment between $a$ and an element of $S$ will be in $S+a$?

Here, $S$ is a convex set, and $S+a$ is a translate of the set $S$. For example, $S$ is the unit ball centered at $0$, and $a$ is a vector, then $S+a$ is the unit ball centered at $a$. — GEdgar, Jul 20 '19 at 11:13
Have you tried taking $x+a$ and $y+a$ for $x,y\in S$ and considering $t(x+a)+(1-t)(y+a)$? — WoolierThanThou, Jul 20 '19 at 11:14

score 1 · Accepted Answer · answered Jul 20 '19 at 11:16

$S + a $ is convex because if you take any two points in $S + a$, they can be written as $x + a$ and $y + a$ where $x$ and $y$ are in $S$ and then, because the line segment $[x, y]$ is contained in $S$ by the convexity of $S$, the line segment $[x + a, y + a]$ will be contained in $S + a$. $a$ will not be a member of $S + a$ in general.

How can $S+a$ convex, if $S \in \mathbb{R}^n$ and $a \in \mathbb{R}^n$

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