A set $A$ is said to be an affine set if for any two distinct points, the line passing through these points lie in the set $A$.
Note −
$S$ is an affine set if and only if it contains every affine combination of its points.
Empty and singleton sets are both affine and convex set.
For example, solution of a linear equation is an affine set.
Let S be the solution of a linear equation.
By definition, $S=\left \{ x \in \mathbb{R}^n:Ax=b \right \}$
Let $x_1,x_2 \in S\Rightarrow Ax_1=b$ and $Ax_2=b$
To prove : $A\left [ \theta x_1+\left ( 1-\theta \right )x_2 \right ]=b, \forall \theta \in\left ( 0,1 \right )$
$A\left [ \theta x_1+\left ( 1-\theta \right )x_2 \right ]=\theta Ax_1+\left ( 1-\theta \right )Ax_2=\theta b+\left ( 1-\theta \right )b=b$
Thus S is an affine set.
If $C$ is an affine set and $x_0 \in C$, then the set $V= C-x_0=\left \{ x-x_0:x \in C \right \}$ is a subspace of C.
Let $x_1,x_2 \in V$
To show: $\alpha x_1+\beta x_2 \in V$ for some $\alpha,\beta$
Now, $x_1+x_0 \in C$ and $x_2+x_0 \in C$ by definition of V
Now, $\alpha x_1+\beta x_2+x_0=\alpha \left ( x_1+x_0 \right )+\beta \left ( x_2+x_0 \right )+\left ( 1-\alpha -\beta \right )x_0$
But $\alpha \left ( x_1+x_0 \right )+\beta \left ( x_2+x_0 \right )+\left ( 1-\alpha -\beta \right )x_0 \in C$ because C is an affine set.
Therefore, $\alpha x_1+\beta x_2 \in V$
Hence proved.