Proof the conic hull of a finite set of vectors is closed

Question

I want to prove that given a family of vectors $(u_1,...,u_m)$ in $\mathbb{R}^n$ : $$C = \Big\{ \sum_{i=1}^{m} \alpha_i u_i,\; \alpha_i\geq 0 \; \: \forall i \in\{1,..,m\} \Big\} $$ is closed. It is easy to prove when $(u_1,...,u_m)$ are linearly independent. Next, if we take $I \subset \{1,...m\} $ s.t $(u_i)_{i\in I}$ are a linearly independent subfamily, I'd want to show that: $$\bigcup_{I} C_I = C $$ where $C_I := \Big\{ \sum_{i \in I} \alpha_i u_i,\; \alpha_i\geq 0 \; \: \forall i \in I \Big\} $. The inclusion $\bigcup_{I} C_I \subset C $ is clear but it is the reciprocal inclusion that I am having trouble with.

Answer 1

Suppose $x$ is in the convex conical hull, then $x = \sum_k \alpha_k u_k$ for some $\alpha_k \ge 0$. We can presume that none of the $u_k$ are zero.

If $\sum_k \alpha_k = 0$ then we can write $x = 0 = 0 u_1$, so we can suppose that $\alpha_k > 0$ for all $k$.

If the $u_k$ are linearly dependent, there are $\beta_k$ not all zero such that $\sum_k \beta_k u_k = 0$ and so $\sum_k \alpha_ku_k = \sum_k (\alpha_k + s \beta_k) u_k$ for any $s$.

Let $k_* \in \operatorname{argmin}_k \{ | { \alpha_k \over \beta_k} | | \beta_k \neq 0 \}$ and $s_* = -{\alpha_{k_*} \over \beta_{k_*}}$. Then $\alpha_{k_*} + s \beta_{k_*} = 0$, $\alpha_k + s \beta_k \ge 0$ for all $k$ and so $x$ lies in the convex conic hull of $u_k$, $k \neq k_*$. This can be repeated as long as the remaining $u_k$ are linearly dependent.

In particular, if $x \in \operatorname{cone} \{ u_\alpha \}$, it lies in the convex conical hull of a linearly independent subset of $ \{ u_\alpha \}$.