Multiindices under Addition form Commutative Monoid

Theorem
Let $Z$ be the set of multiindices.

Let $+$ denote the addition of multiindices.

Then $\left({Z, +}\right)$ is a commutative monoid.

Proof
We check each of the axioms in turn.

Let $k = \left \langle {k_j}\right \rangle_{j \in J}$, $\ell = \left \langle {\ell_j}\right \rangle_{j \in J}$ and $m = \left \langle {m_j}\right \rangle_{j \in J}$ be multiindices.

Closure
Trivially we have that:
 * $j \mapsto \left({ k_j + \ell_j }\right)$

is a sequence of integers indexed by $J$.

We know that finitely many of the $k_j$ are non-zero, and finitely many of the $\ell_j$ are non-zero.

Therefore finitely many of the $k_j + \ell_j$ are non-zero.

This shows that $k + \ell$ is a multiindex and so $+$ is closed.

Associativity
For all $j \in J$, we have:

Thus $+$ is shown to be associative.

Commutativity
For all $j \in J$, we have:

Thus $+$ is shown to be commutative.

Identity Element
Let $0_Z$ be the multiindex defined by:
 * $\left(0_Z\right)_j = 0$

for all $j \in J$.

Then we have, for all $j \in J$:

Since we have seen that $\left(Z,+\right)$ is commutative, this shows that $0_Z$ is an identity element for $Z$.