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Let $x_i \in \mathbb{R}^n $ where $i=1,...,l$ and $1\leq l \leq n$. Also, let $$ W = \{y = \sum_1^l \alpha_i x_i \mid \alpha_i > 0 \sum_1^l \alpha_i =1 \} $$ be the set of all strict convex combination of $x_i$'s.

We know that the following $$ U = \{y = \sum_1^l \alpha_i x_i \mid \alpha_i \geq 0 \sum_1^l \alpha_i =1 \} $$ is the convex hull of the set $\{x_i\}_{i=1}^{l}$.

What is the closure of $W$, i.e. $cl(W)$? Can we just say if we relax strictly greater than zero of $\alpha_i$ in $W$ we can get its closure? Please give a counter example if it is not true or prove it if it is true?

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$\sum \alpha_i x_i$ is the limit of $\sum \alpha_i^{k} x_i$ as $k \to \infty$ where $\alpha_i^{k}= (\alpha_i+\frac 1 k)/( \sum_j( \alpha_j+\frac 1 k))$ so $U$ is the closure of $W$.

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  • $\begingroup$ We are talking about finite dimension $(n)$, $n$ is a finite number. How would that possible to let $n$ goes to infinity? $\endgroup$
    – user494522
    Jun 8, 2019 at 18:35
  • $\begingroup$ Replace $n$ above by $k$. Unfortunate collision of variables... $\endgroup$
    – copper.hat
    Jun 8, 2019 at 22:29
  • $\begingroup$ @Saeed Sorry about the typo. Answer edited. $\endgroup$
    – Kavi Rama Murthy
    Jun 8, 2019 at 23:10
  • $\begingroup$ @copper.hat: Why such a choice for $\alpha_i^k$ is the case? $\endgroup$
    – user494522
    Jun 9, 2019 at 3:52
  • $\begingroup$ The multipliers are strict. $\endgroup$
    – copper.hat
    Jun 9, 2019 at 3:54

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