# What is the relationship between strict convex combination and convex hull of a set?

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?

$$\sum \alpha_i x_i$$ 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 $$W$$.
• We are talking about finite dimension $(n)$, $n$ is a finite number. How would that possible to let $n$ goes to infinity?
• Replace $n$ above by $k$. Unfortunate collision of variables... Jun 8, 2019 at 22:29
• @copper.hat: Why such a choice for $\alpha_i^k$ is the case?