Translation of Convex Set in Vector Space is Convex

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Theorem

Let $\Bbb F \in \set {\R, \C}$.

Let $X$ be a vector space over $\Bbb F$.

Let $C \subseteq X$ be convex.

Let $x \in X$.


Then $C + x$, the translation of $C$ by $x$, is convex.


Proof

Let $t \in \closedint 0 1$ and $u, v \in C + x$.

Then there exists $u', v' \in C$ such that:

$u = u' + x$

and:

$v = v' + x$

Then, we have:

\(\ds t u + \paren {1 - t} v\) \(=\) \(\ds t u' + t x + \paren {1 - t} v' + \paren {1 - t} x\)
\(\ds \) \(=\) \(\ds \paren {t u' + \paren {1 - t} v'} + \paren {t + \paren {1 - t} } x\)
\(\ds \) \(=\) \(\ds \paren {t u' + \paren {1 - t} v'} + x\)

Since $C$ is convex, we have:

$t u' + \paren {1 - t} v' \in C$

so:

$t u + \paren {1 - t} v \in C + x$

So we have that $C + x$ is convex.

$\blacksquare$