I am having a problem with one aspect of the following proof I came across in "An Easy Path to Convex Analysis and Applications" by Mordukhovich and Nam.
It is Proposition 3.9 and it is the line marked by * that is giving me troubles
Proposition 3.9 $\;\;\;\;\;\;\;$Let $a_1,\ldots,a_m$ be elements of $\mathbb{R}^n.$ Then the convex cone $K_{\Omega}$ generated by $\Omega:=\{a_1,\ldots,a_m\}$ is closed in $\mathbb{R}^n$.
Proof $\;\;\;\;\;$ Assume first that the elements $a_1,\ldots,a_m$ are linearly independent and take any sequence $\{x_k\}\subset K_{\Omega}$ converging to $x.$ By construction of $K_{\Omega}$ find numbers $\alpha_{ki}\geq0$ for each $i=1\ldots m$ and $k\in\mathbb{N}$ such that $$x_k=\sum_{i=1}^{m}\alpha_{ki}a_i$$ *Letting $\alpha_{k}:=(\alpha_{k1},\ldots,\alpha_{km})\in\mathbb{R}^n$ and arguing by contradiction, it is easy to check that the sequence $\{\alpha_{k}\}$ is bounded$\ldots$
I am not convinced to is easy to check that the sequence is bounded. My initial thoughts are as follows:
Assume the sequence $\{\alpha_k\}$ is not bounded. It then follows that $\{\alpha_k\}$ divergess to infinity. i.e. as $k\rightarrow \infty, \{\alpha_k\}\rightarrow\infty$. But then $\sum_{i=1}^{m}\alpha_{ki}a_i\rightarrow\infty$ which means $\{x_k\}\rightarrow\infty$. However, this contradicts our choice of $\{x_k\}$ which converges to $x$. Thus, $\{\alpha_k\}$ is bounded.
Does this seem like I am on the correct path?