Let $X = \{x^1,\dots,x^m\}$ be a set of linearly independent vectors in $\mathbb{R}^n$. (Superscripts are indices, not powers)
Prove that its conic hull $C=\text{cone}(X)$ is closed.
ATTEMPT
I am attempting to use Caratheodory's theorem for cones in $\mathbb{R}^n$:
Every nonzero vector $x \in \text{cone}(X)$ can be written as a positive combination of at most $n$ elements of $X$.
I interpreted this symbolically as, $$ \forall (x \neq 0)\in \text{cone}(X), \;\; x = \sum_{i\,=\,1}^{m} \alpha_i x^i, \;\;\; (\alpha_i>0). $$
I chose $m$ to be the smallest integer for which $x$ is obtained (making $x^i$ linearly independent).
I am not sure how to prove closedness. I have attempted defining a sequence $x_k \in C$ and tried proving that $x_k \to x \in C$. But I am failing. I am not allowed to use polyhedral cones. So I am really stuck on how to prove closedness.
I am aware that any nonnegative combination of $x^i$ will be contained in $C$, but I am not sure how this would help.
Please, any insight is grealy appreciated.
You define $m$ in two different ways. I will use $m$ as the cardinality of the linearly independent set you are given.
Let $x^k := \sum_{i=1}^m \alpha_i^k x^i \in \text{cone}(X)$ (where $\alpha_i^k \ge 0$) and suppose $x^k \to x$.
Since the $x^k$ lie in the subspace $\text{span}\{x^1, \ldots, x^m\}$, the limit $x$ must also lie in this subspace, since finite-dimensional subspaces are closed.
Since $x^1, \ldots, x^m$ are linearly independent, there exist unique $\alpha_i$ such that $x = \sum_{i=1}^m \alpha_i x^i$.
It suffices to then show $\alpha_i^k \overset{k \to \infty}{\longrightarrow} \alpha_i$ for each $i$, since that would imply $\alpha_i \ge 0$ for all $i$.
Suppose $\alpha_1^k \not\to\alpha_1$. Let $v$ be the unit vector in $\text{span}\{x^1, \ldots, x^m\}$ that is orthogonal to $x^2, \ldots, x^m$ (e.g., Gram-Schmidt). Then $\|x_k - x\|^2 \ge (\alpha^k_1 - \alpha_1)^2 \langle x^1, v \rangle^2$. Since $\alpha^k_1 \not\to \alpha_1$, we cannot have $\|x_k - x\| \to 0$, a contradiction.
Reply to comment: any vector $w$ can be written uniquely as $w = w_\| + w_\perp$ such that $w_\|$ is parallel to $v$ and $w_\perp$ is orthogonal to $v$. Then $\|w\|^2 = \|w_\|\|^2 + \|w_\perp\|^2 \ge \|w_{\|}\|^2$. Note also that $w_\| = \langle w, v \rangle v$ since $v$ is a unit vector. I'm applying this fact to $w := x_k - x$ and noting that $v$ is orthogonal to $x^2, \ldots, x^m$.