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Define the square (in $\mathbf{R}^2$) $S = \{ x \in \mathbf{R}^2 \mid 0 \leq x_i \leq 1,~i=1,2 \}$, and the disk $D = \{ x \in \mathbf{R}^2 \mid \|x\|_2 \leq 1 \}$.

$S \cap D$ is convex.

• Correct! Intersection of convex sets is convex.
• Incorrect.

$S \cup D$ is convex.

• Correct! In general, the union of convex sets is not convex, but here, it is.
• Incorrect.

$S \setminus D$ is convex.

$C = \{ (1,0), (1,1), (-1,-1), (0,0) \}$.

$(0,-1/3) \in \mathbb{conv}\; C$.

$(0,1/3) \in \mathbb{conv}\; C$.

$(0,1/3)$ is in the conic hull of $C$.

Consider the set $S = \{ (0,2),~ (1,1),~ (2,3),~ (1,2),~ (4,0) \}$.

$(0,2)$ is the minimum element of $S$.

$(0,2)$ is a minimal element of $S$.

$(2,3)$ is a minimal element of $S$.

$(1,1)$ is a minimal element of $S$.

$\{Ax + b \mid Fx = g \}$ is affine.

$S = \{ \alpha \in \mathbf{R}^3 \mid \alpha_1 + \alpha_2e^{-t} + \alpha_3 e^{-2t} \leq 1.1 \mbox{ for } t\geq 1\}$.

$S$ is affine.

$S$ is a polyhedron.

$S$ is convex.

Generalized inequality.

$K = \{(x_1,x_2) \mid 0 \leq x_1 \leq x_2 \}$.

$(1,3) \preceq_K (3,4)$.

$(-1,2) \in K^*$.