# 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:

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

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

### 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: {{explain|Check whether the above link should be Upper Section or Upper Closure}

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

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

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

### 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:

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

$\blacksquare$