- #1
squenshl
- 479
- 4
Homework Statement
Prove that the set $$\Delta = \left\{x \in \mathbb{R}^{n+1}: \sum_{i=1}^{n+1} x_i = 1 \quad \text{and} \quad x_i \geq 0 \; \text{for any} \; i\right\}$$ is a polytope. This polytope is called an ##n##-dimensional simplex.
Prove that the set $$C = \left\{x \in \mathbb{R}^{n+1}: 0 \leq x_1 \leq 1 \; \text{for any} \; i\right\}$$ is a polytope. This polytope is called an ##n##-dimensional cube.
Prove that the set $$O = \left\{x \in \mathbb{R}^{n+1}: \sum_{i=1}^{n+1} |x_i| \leq 1\right\}$$ is a polytope. This polytope is called a octahedron.
Homework Equations
The Attempt at a Solution
For the first one we claim that ##\Delta = D = \text{conv}\left\{e_1, e_2, \ldots, e_{n+1}\right\}##. If ##x = (x_1, x_2, \ldots, x_{n+1}) \in \Delta## then ##\sum_{i=1}^{n+1} x_i = 1## and ##x_i \geq 0## for all ##i## and ##x = \sum_{i=1}^{n+1} x_ie_i## so ##x \in D##. If ##x \in D## then ##\sum_{i=1}^{n+1} \alpha_i e_i = (\alpha_1, \alpha_2, \ldots, \alpha_{n+1})## with ##\sum_{i=1}^{n+1} \alpha_i = 1## and ##\alpha_i \geq 0## for all ##i##. But then ##x \in \Delta##. Thus, ##\Delta## and ##D## are the same set. Hence, ##\Delta## is a polytope.
For the second one we claim that ##C = \text{conv}(S)## where ##S## consists of all vectors ##(x_1, x_2, \ldots, x_n)## where ##x_i \in \left\{-1,1\right\}## for each ##i##. Not sure what to do after that.
For the third one we claim that ##O = P = \text{conv}(\pm e_1, \pm e_2, \ldots, \pm e_{n+1})##. If ##x = (x_1, x_2, \ldots, x_{x+1}) \in O## then ##\sum_{i=1}^{n+1} \left|x_i\right| \leq 1## and ##x_i \geq 0## for all ##i## and ##x = \sum_{i=1}^{n+1} x_ie_i## so ##x \in O##. If ##x \in P## then ##\sum_{i=1}^{n+1} \left|\alpha_i e_i\right| = (\alpha_1, \alpha_2, \ldots, \alpha_{n+1})## with ##\sum_{i=1}^{n+1} \alpha_i = 1## and ##\alpha_i \geq 0## for all ##i##. But then ##x \in O##. Thus, ##O## and ##P## are the same set. Hence, ##O## is a polytope.
For the second and third cases, is it almost identical to the first one. Some help would be awesome!