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Let $(x_n)_n$ be a countable subset of $C$ that is dense in $C$; For every $n$ let $C_n=conv\lbrace x_1, x_2, . . . , x_n \rbrace $

($C\subset E$nonempty convex set, $E$ a finite-dimensional normed space)

How to prove that $C_n$ is compact and that $\displaystyle\bigcup_{n=1}^{\infty}C_n$ is dense in $C$?

please, thank you.

Vrouvrou
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Hints: In finite dimensional spaces "compact" = "closed and bounded" and $x_n \in \bigcup_{k=1}^\infty C_k$, $n \in \mathbb{N}$.

njguliyev
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  • OK, but how to see that C_n is closed and bounded please ? – Vrouvrou Aug 28 '13 at 04:20
  • is $C_n$ is closed because $\lbrace x_1,...,x_n\rbrace$ is finite set ? – Vrouvrou Aug 28 '13 at 04:40
  • Yes. If $y_k = \alpha_k^{(1)} x_1 + \ldots + \alpha_k^{(n)} x_n \to y \in E$ then using the Bolzano–Weierstrass theorem you can choose a convergent subsequence of coefficients. Its limit $(\alpha^{(1)},\ldots, \alpha^{(n)})$ will also satisfy $\alpha^{(1)}+\ldots+\alpha^{(n)}=1$, $\alpha^{(i)} \ge 0$. Bu uniqueness of the limit $y=\alpha^{(1)} x_1 + \ldots + \alpha^{(n)} x_n \in C_n$. – njguliyev Aug 28 '13 at 07:57
  • and the boundedness? please – Vrouvrou Aug 28 '13 at 08:26
  • $|\alpha_1 x_1 + \ldots + \alpha_n x_n| \le |x_1| + \ldots + |x_n|$. – njguliyev Aug 28 '13 at 08:33