Let $K$ be a compact non-empty subset of $\mathbb{R}^N$. If $x \in \mathbb{R}^N$ and if $(k_1, k_2, \dots, k_N), k_i \in \mathbb{Z}_+$, consider the function $$x \mapsto x_1^{k_1}x_2^{k_2}...x_N^{k_N}.$$

I want to show that the set $\mathcal{A}$ of all linear combinations of such functions is a $\mathbb{R}$-subalgebra of $C(K)$, but I'm struggling specifically with proving that the sum and the product of elements of $\mathcal{A}$ belongs to $\mathcal{A}$ cause I've very much confused with manipulating the indexes.

$\textbf{My attempt:}$ Consider the set $$\mathcal{A} = \Big\{ \sum_{i=1}^{N}\alpha_i\prod_{i=1}^{N} x_i^{k_i}: x_i, \alpha_i \in \mathbb{R}, k_i \in \mathbb{Z}_+ \Big\}$$ 1. $p, q \in \mathcal{A} \Rightarrow pq \in \mathcal{A}$

Take $p, q \in \mathcal{A}$ such that $p = \sum_{i=1}^{N}\alpha_i\prod_{i=1}^{N} x_i^{k_i}$ and $q = \sum_{j=1}^{N}\tilde{\alpha}_j\prod_{j=1}^{N} x_j^{\tilde{k}_j}$, then it follows:

\begin{align*} p(x)q(x) &= \Big(\sum_{i=1}^{N}\alpha_i\prod_{i=1}^{N} x_i^{k_i} \Big) \Big(\sum_{j=1}^{N}\tilde{\alpha_j}\prod_{j=1}^{N} x_j^{\tilde{k}_j} \Big) \\ &= \sum_{l=1}^{2N}c_l \prod_{i=1}^{l}x_i^{k_i}, \\ &= (pq)(x) \end{align*} with $c_l = \sum_{m=1}^{l}\alpha_m\tilde{\alpha}_{l-m}$.

  1. $p, q \in \mathcal{A} \Rightarrow p+q \in \mathcal{A}$

Take the same $p, q \in \mathcal{A}$ written above, then \begin{align*} p(x) + q(x) &= \Big(\sum_{i=1}^{N}\alpha_i\prod_{i=1}^{N} x_i^{k_i} \Big) + \Big(\sum_{j=1}^{N}\tilde{\alpha_j}\prod_{j=1}^{N} x_j^{\tilde{k}_j} \Big) \\ &= \sum_{l=1}^{N} (\alpha_l + \tilde{\alpha}_l)\prod_{l=1}^{N}x_l^{k_l} \\ &= (p + q)(x) \end{align*}


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