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 $\le$ denote the ordering on integers.

Let $\preceq$ denote the ordeing on multiindices.

Recall that if $k = \family {k_j}_{j \mathop \in J}$ and $\ell = \family {\ell_j}_{j \mathop \in J}$ 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 $m = \family {m_j}_{j \mathop \in J} \in Z$.

From Integers under Addition form Totally Ordered Group, the integers are totally ordered by $\le$.

Hence the ordering on the integers is reflexive, antisymmetric and transitive.

Reflexivity

By reflexivity of $\le$, we have for all $j \in J$:

$k_j \le k_j$

This shows that $\preceq$ is reflexive.

$\Box$

Antisymmetry

Suppose that $k \preceq \ell$ and $\ell \preceq k$.

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

$k_j \le \ell_j \le k_j$

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

This shows $\preceq$ to be antisymmetric.

$\Box$

Transitivity

Suppose that $k \preceq \ell$ and $ell \preceq m$.

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

$k_j \le \ell_j$ and $\ell_j \le m_j$

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

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

$\blacksquare$