Ultraproduct is Well-Defined

Theorem
Ultraproduct is well-defined.

Proof
Specifically, following the definitions on ultraproduct, it is to be proved that:


 * $(1) \quad f^\MM$ is well-defined
 * $(2) \quad R^\MM$ is well-defined

First of all, we need to prove:

Lemma
Following the definitions on ultraproduct:
 * $\eqclass {m_{k, i} } \UU = \eqclass {m'_{k, i} } \UU$, $k = 1, \dotsc, n$


 * $\set {i \in I: \tuple {m_{1, i}, \dots, m_{n, i} } = \tuple {m'_{1, i}, \dots, m'_{n, i} } } \in \UU$
 * $\set {i \in I: \tuple {m_{1, i}, \dots, m_{n, i} } = \tuple {m'_{1, i}, \dots, m'_{n, i} } } \in \UU$

Proof
Let:

Suppose:
 * $\eqclass {m_{k, i} } \UU = \eqclass {m'_{k, i} } \UU$ for $k = 1, \dotsc, n$

We have:
 * $I_k \in \UU$ for $k = 1, \dotsc, n$

Since $\UU$ is closed under intersection:
 * $I^* \in \UU$

On the other hand, suppose:
 * $I^* \in \UU$

Since $\UU$ is upward-closed:
 * $I_k \in \UU$ for $k = 1, \dotsc, n$

Therefore:
 * $\eqclass {m_{k, i} } \UU = \eqclass {m'_{k, i} } \UU$

Proposition 1
The definition of $f^\MM$ is consistent.

that is, for $\eqclass {m_{k, i} } \UU = \eqclass {m'_{k, i} } \UU$, $k = 1, \dotsc, n$


 * $\eqclass {\map {f^{\MM_i} } {m_{1, i}, \dotsc, m_{n, i} } } \UU = \eqclass {\map {f^{\MM_i} } {m'_{1, i}, \dotsc, m'_{n, i} } } \UU$

Proof
Firstly note that:
 * $\set {i \in I: \map {f^{\MM_i} } {m_{1, i}, \dotsc, m_{n, i} } = \map {f^{\MM_i} } {m'_{1, i}, \dots, m'_{n, i} } } \supseteq \set {i \in I: \tuple {m_{1, i}, \dotsc, m_{n, i} } = \tuple {m'_{1, i}, \dotsc, m'_{n, i} } }$

and by $\UU$ is an ultrafilter on $I$:
 * $\set {i \in I: \tuple {m_{1, i}, \dotsc, m_{n, i} } = \tuple {m'_{1, i}, \dotsc, m'_{n, i} } } \in \UU$

implies:
 * $\set {i \in I: \map {f^{\MM_i} } {m_{1, i}, \dotsc, m_{n, i} } = \map {f^{\MM_i} } {m'_{1, i}, \dotsc, m'_{n, i} } } \in \UU$

Therefore:
 * $\eqclass {m_{k, i} } \UU = \eqclass {m'_{k, i} } \UU$, $k = 1, \dotsc, n$

by the lemma, which is equvalent to:
 * $\set {i \in I: \tuple {m_{1, i}, \dotsc, m_{n, i} } = \tuple {m'_{1, i}, \dotsc, m'_{n, i} } } \in \UU$

implies:
 * $\eqclass {\map {f^{\MM_i} } {m_{1, i}, \dotsc, m_{n, i} } } \UU = \eqclass {\map {f^{\MM_i} } {m'_{1, i}, \dotsc, m'_{n, i} } } \UU$

Proposition 2
The definition of $R^\MM$ is consistent.

that is, for $\eqclass {m_{k, i} } \UU = \eqclass {m'_{k, i} } \UU$, $k = 1, \dotsc, n$:


 * $\set {i \in I: \tuple {m_{1, i}, \dotsc, m_{n, i} } \in R^{\MM_i} } \in \UU$


 * $\set {i \in I: \tuple {m'_{1, i}, \dotsc, m'_{n, i} } \in R^{\MM_i} } \in \UU$
 * $\set {i \in I: \tuple {m'_{1, i}, \dotsc, m'_{n, i} } \in R^{\MM_i} } \in \UU$

Proof
Let:

From the lemma:
 * $I^* \in \UU$

therefore:
 * $S \in \UU \implies T \in \UU$

Note that:
 * $\tuple {m_{1, i}, \dots, m_{n, i} } = \tuple {m'_{1, i}, \dotsc, m'_{n, i} }$ for $i \in I^*$

we have:
 * $T = T'$

Hence:
 * $T' \in \UU$

and:
 * $S' \in \UU$ since $S' \supseteq T'$

So far we have proved:
 * $S \in \UU \implies S' \in \UU$

By symmetry:
 * $S' \in \UU \implies S \in \UU$