Is Every Convex Polytope Both Convex and Closed?

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SUMMARY

Every convex polytope is both convex and closed, as established in the discussion. The convexity is proven using the definition of a convex polytope, represented as C=\{ \sum_{j=1}^n x_j a^j | x_j \geq 0, \sum_{j=1}^n x_j = 1\}. To demonstrate that C is closed, the Bolzano-Weierstrass theorem is applied to a bounded sequence \{ x^N \}_{N=1}^\infty, ensuring that every subsequence converges to the same limit b. This confirms that the limit of the sequence lies within the polytope, thus proving its closed nature.

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Homework Statement


Prove that every convex polytope is convex and closed.

Homework Equations


C=\{ \sum_{j=1}^n x_j a^j | x_j \geq 0, \sum_{j=1}^n x_j = 1\} is a convex polytope

The Attempt at a Solution


I've already proven the convexity portion. To prove C is closed, I let \{ b^N \}_{N=1}^\infty \subseteq C and assumed \lim_{N\to\infty} b^N = b.
b=\sum_{j=1}^n x_j a^j, so I have to show \lim_{N\to\infty} x^N = x.

I started with x_j \geq 0, \sum_{j=1}^n x_j = 1\ means |x^N| \leq 1 and the sequence \{ x^N \}_{N=1}^\infty is a bounded sequence. From here, I can use the Bolzano-Weierstrass theorem to show that there exists a subsequence that converges. From here, I'm unsure of what to do because the subsequence converges to some value which may or may not be the right value
 
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zcd said:
the subsequence converges to some value which may or may not be the right value

You know that the whole sequence converges to b, so every subsequence converges to b as well.
 

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