Ultraproduct is Well-Defined

Theorem
Definition:Ultraproduct is well-defined.

More specificly, following the definitions on Definition:Ultraproduct,

we are going to prove that:


 * (1) $f^\mathcal M$ is well-defined
 * (2) $R^\mathcal M$ is well-defined

Proof
First of all, we need to prove

Lemma
Following the definitions on Definition:Ultraproduct
 * $\left(m_{k, i}\right)_\mathcal U = \left(m'_{k, i}\right)_\mathcal U$, $k = 1, \dotsc, n$


 * $\left\{ i : \left({m_{1, i}, \dots, m_{n, i} }\right) = \left({m'_{1, i}, \dots, m'_{n, i} }\right) \right \} \in \mathcal U$

Proof
Let
 * $I_k := \left\{ i \in I : m_{k, i} = m'_{k, i} \right\}$
 * $I^* := \left\{ i : \left({m_{1, i}, \dots, m_{n, i} }\right) = \left({m'_{1, i}, \dots, m'_{n, i} }\right) \right \}= \displaystyle \bigcap^n_{k = 1} I_k $

Suppose
 * $\left(m_{k, i}\right)_\mathcal U = \left(m'_{k, i}\right)_\mathcal U$ for $k = 1, \dotsc, n$

we have
 * $I_k \in \mathcal U$ for $k = 1, \dotsc, n$

Since $\mathcal U$ is closed under intersection
 * $I^* \in \mathcal U$

On the other hand, suppose
 * $I^* \in \mathcal U$

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

Therefore
 * $\left(m_{k, i}\right)_\mathcal U = \left(m'_{k, i}\right)_\mathcal U$

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

i.e. for $\left(m_{k, i}\right)_\mathcal U = \left(m'_{k, i}\right)_\mathcal U$, $k = 1, \dotsc, n$


 * $\left(f^{\mathcal M_i}\left(m_{1, i}, \dots, m_{n, i}\right)\right)_\mathcal U = \left(f^{\mathcal M_i}\left(m'_{1, i}, \dots, m'_{n, i}\right)\right)_\mathcal U$

Proof
Firstly note that:
 * $\{i : f^{\mathcal M_i}\left(m_{1, i}, \dots, m_{n, i}\right) = f^{\mathcal M_i}\left(m'_{1, i}, \dots, m'_{n, i}\right) \} \supseteq \{i : \left(m_{1, i}, \dots, m_{n, i}\right) = \left(m'_{1, i}, \dots, m'_{n, i}\right) \} $

and by $\mathcal U$ is an ultrafilter on $I$, we have
 * $\{i : \left(m_{1, i}, \dots, m_{n, i}\right) = \left(m'_{1, i}, \dots, m'_{n, i}\right) \} \in \mathcal U$

implies
 * $\{i : f^{\mathcal M_i}\left(m_{1, i}, \dots, m_{n, i}\right) = f^{\mathcal M_i}\left(m'_{1, i}, \dots, m'_{n, i}\right) \} \in \mathcal U$

Therefore,
 * $\left(m_{k, i}\right)_\mathcal U = \left(m'_{k, i}\right)_\mathcal U$, $k = 1, \dotsc, n$, by, which is equvalent to $\{i : \left(m_{1, i}, \dots, m_{n, i}\right) = \left(m'_{1, i}, \dots, m'_{n, i}\right) \} \in \mathcal U$

implies
 * $\left(f^{\mathcal M_i}\left(m_{1, i}, \dots, m_{n, i}\right)\right)_\mathcal U = \left(f^{\mathcal M_i}\left(m'_{1, i}, \dots, m'_{n, i}\right)\right)_\mathcal U$

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

i.e. for $\left(m_{k, i}\right)_\mathcal U = \left(m'_{k, i}\right)_\mathcal U$, $k = 1, \dotsc, n$


 * $\left\{i \in I: \left({m_{1, i}, \dots, m_{n, i} }\right) \in R^\mathcal M_i\right\} \in \mathcal U$


 * $\left\{i \in I: \left({m'_{1, i}, \dots, m'_{n, i} }\right) \in R^\mathcal M_i\right\} \in \mathcal U$

Proof
Let
 * $S := \left\{i \in I: \left({m_{1, i}, \dots, m_{n, i} }\right) \in R^\mathcal M_i\right\}$
 * $S' := \left\{i \in I: \left({m'_{1, i}, \dots, m'_{n, i} }\right) \in R^\mathcal M_i\right\}$
 * $I^* := \left\{ i : \left({m_{1, i}, \dots, m_{n, i} }\right) = \left({m'_{1, i}, \dots, m'_{n, i} }\right) \right \}$
 * $T := I^* \cap S$
 * $T' := I^* \cap S'$

As implies
 * $I^* \in \mathcal U$

therefore
 * $S \in \mathcal U$ implies $T \in \mathcal U$

Note that
 * $\left({m_{1, i}, \dots, m_{n, i} }\right) = \left({m'_{1, i}, \dots, m'_{n, i} }\right)$ for $i \in I^*$

we have
 * $T = T'$

Hence
 * $T' \in \mathcal U$

and
 * $S' \in \mathcal U$ since $S' \supseteq T'$

So far we have proved
 * $S \in \mathcal U$ implies $S' \in \mathcal U$

By symmetry,
 * $S' \in \mathcal U$ implies $S \in \mathcal U$