Ordering on Multiindices is Partial Order

Theorem
Let $Z$ be the set of multiindices indexed by a set $J$.

The ordering on $Z$ is a partial ordering.

Proof
Let $\preceq$ denote the ordering on integers.

Let $\le$ denote the ordeing on multiindices.

Recall that if $k = \left \langle {k_j}\right \rangle_{j \in J}$ and $\ell = \left \langle {\ell_j}\right \rangle_{j \in J}$ are multiindices, then $k \le \ell$ if $k_j \preceq \ell_j$ for all $j \in J$.

Let $k$ and $\ell$ be as above, and let $m = \left \langle {m_j}\right \rangle_{j \in J} \in Z$.

Since the integers are totally ordered by $\preceq$, we know that the ordering on the integers is reflexive, antisymmetric and transitive.

Reflexivity
By reflexivity of $\preceq$, we have for all $j \in J$:
 * $k_j \preceq k_j$

This shows that $\le$ is reflexive.

Antisymmetry
Suppose that $k \le \ell$ and $\ell \le k$.

Then for all $j \in J$, we have:
 * $k_j \preceq \ell_j \preceq k_j$

By antisymmetry of $\preceq$, we have $k_j = \ell_j$ for all $j$, that is, $k = \ell$.

This shows $\le$ to be antisymmetric.

Transitivity
Suppose that $k \le \ell \le m$.

Then for all $j \in J$, we have:
 * $k_j \preceq \ell_j \preceq m_j$

By transitivity of $\preceq$, this shows that for all $j \in J$, $k_j \preceq m_j$.

Therefore $k \le m$, and $\le$ is shown to be transitive.