User:DingChao/Sandbox

Definition
Let $\mathcal L$ be a first-order language and let $I$ be an infinite set.

Let $\mathcal U$ be an ultrafilter on $I$.

Let $\mathcal M_i$ be an $\mathcal L$-structure for each $i\in I$

The ultraproduct:
 * $\mathcal M := \displaystyle \left({\prod_{i \mathop \in I} \mathcal M_i }\right) / \mathcal U$

is an $\mathcal L$-structure defined as follows:


 * $(1): \quad$ The universe of $\mathcal M$:

Let $X$ be the cartesian product:
 * $\prod_{i \mathop \in I} \mathcal M_i$

Define an equivalence relation $\sim$ on $X$ by:
 * $\left({a_i}\right)_{i \mathop \in I} \sim \left({b_i}\right)_{i \mathop \in I}$ $\left\{ {i \in I: a_i = b_i}\right \} \in \mathcal U$

The universe of $\mathcal M$ is the set of equivalence classes of $X$ modulo $\sim$.

These are essentially sequences taken modulo the equivalence relation above, and are sometimes denoted $\left({m_i}\right)_\mathcal U$.


 * $(2): \quad$ Interpretation of non-logical symbols of $\mathcal L$ in $\mathcal M$:

For each constant symbol $c$, we define $c^\mathcal M$ to be $\left({c^{\mathcal M_i} }\right)_\mathcal U$.

For each $n$-ary function symbol $f$, we define $f^\mathcal M$ by setting:
 * $f^\mathcal M \left({\left({m_{1, i} }\right)_\mathcal U, \dotsc, \left({m_{n, i} }\right)_\mathcal U}\right)$

to be:
 * $\left({f^{\mathcal M_i} \left({m_{1, i}, \dotsc, m_{n, i} }\right)}\right)_\mathcal U$

For each $n$-ary relation symbol $R$, we define $R^\mathcal M$ to be the set of $n$-tuples:
 * $\left({\left({m_{1, i} }\right)_\mathcal U, \dots, \left({m_{n, i} }\right)_\mathcal U}\right)$ from $\mathcal M$

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

Lemma
Following the definitions above
 * $(m_{k, i})_\mathcal U = (m'_{k, i})_\mathcal U$, $k = 1, \dotsc, n$

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

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
 * $(m_{k, i})_\mathcal U = (m'_{k, i})_\mathcal U$ 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 closed under getting supersets
 * $I_k \in \mathcal U$ for $k = 1, \dotsc, n$

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

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

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


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

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

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

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

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

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

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

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


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

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

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 the lemma imples
 * $I^* \in \mathcal U$

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 $S \in \mathcal U$