Ultraproduct is Well-Defined

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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:

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

if and only if:

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


Proof

Let:

\(\ds I_k\) \(:=\) \(\ds \set {i \in I: m_{k, i} = m'_{k, i} }\)
\(\ds I^*\) \(:=\) \(\ds \set {i \in I: \tuple {m_{1, i}, \dots, m_{n, i} } = \tuple {m'_{1, i}, \dots, m'_{n, i} } } = \bigcap^n_{k = 1} I_k\)

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:

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

$\blacksquare$


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:

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

$\blacksquare$


Proposition 2

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

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

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

if and only if:

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


Proof

Let:

\(\ds S\) \(:=\) \(\ds \set {i \in I: \tuple {m_{1, i}, \dotsc, m_{n, i} } \in R^{\MM_i} }\)
\(\ds S'\) \(:=\) \(\ds \set {i \in I: \tuple {m'_{1, i}, \dotsc, m'_{n, i} } \in R^{\MM_i} }\)
\(\ds I^*\) \(:=\) \(\ds \set {i \in I: \tuple {m_{1, i}, \dotsc, m_{n, i} } = \tuple {m'_{1, i}, \dots, m'_{n, i} } }\)
\(\ds T\) \(:=\) \(\ds I^* \cap S\)
\(\ds T'\) \(:=\) \(\ds I^* \cap S'\)

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:

$S' \in \UU \implies S \in \UU$

$\blacksquare$