3
$\begingroup$

Show that $a \notin acl_{\mathbb{M}}(A) \Longleftrightarrow$ exists an infinite indiscernible sequence over $A$ containing $a$.

$ \Rightarrow) a \notin acl_{\mathbb{M}}(A)$ then $\forall \phi(x,b) \in L(A)$ such that $\mathbb{M} \models\phi(a,b)$ we have that $\mid \phi(\mathbb{M},b) \mid $ is infinite, so let $(I, <)$ a set linearly ordered with $\mid I \mid=\mid\phi(\mathbb{M}) \mid$ where $\{ a_{i} \mid i \in I\}=\phi(\mathbb{M},b)$ ( every set can be well-ordered). Then $\{ a_{i} \mid i \in I\}$ is sequence indiscernible over $A$ since $\mathbb{M} \models \phi(a,b) \Leftrightarrow \phi(c,b), \forall c \in \{ a_{i} \mid i \in I\} $ (I think if this can be done for each element you can be done for each finite set in $\{ a_{i} \mid i \in I\}$) so $ \{ a_{i} \mid i \in I\}$ is sequence indiscernible over $A$ and $a \in A$. I'm not sure if my reasoning is valid. In the other direction I have no idea how it could be. I appreciate any hint.

$\endgroup$
3
$\begingroup$

Most importantly, note that your proof of the $(\Rightarrow)$ direction is incorrect, as it seems to use a weaker (and incorrect) definition for indiscernible sequence. It is not enough to show that every element in the sequence has the same type over $A$. I'll recall the definition of indiscernible sequence, then give a rough sketch of the proof of each direction.

Definition. Let $M$ be a model, $A \subseteq M$, $I$ linearly ordered, $\{b_i\}_{i\in I}$ distinct tuples from $M$. Say $\{b_i\}_{i\in I}$ is indiscernible over $A$ if, for all $n$, for all formulas $\phi(x_1,\ldots,x_n)$ with parameters from $A$, whenever $i_1<\cdots<i_n$ and $j_1<\cdots<j_n$ are from $I$, $$M\models \phi(b_{i_1},\ldots,b_{i_n}) \leftrightarrow \phi(b_{j_1},\ldots,b_{j_n})\text{.}$$

That is, whether a formula possibly involving more than one element of the sequence holds doesn't change if we pick different elements from the sequence, as long as we pick them in the same order. (Note that sometimes we may want to consider the definition without the word "distinct", but the theorem you're trying to prove requires it.)

Of the two directions, $(\Leftarrow)$ is actually the easy one. If we know we have an indiscernible sequence, even only looking at $n = 1$, that means we have infinitely many distinct tuples, all with the same type as $a$ (over $A$). That type is therefore not algebraic.

The proof of $(\Rightarrow)$ is a little tricky if you haven't seen something similar before. We'll be making a sequence indexed by natural numbers. Also, note that we may have to pass to a sufficiently saturated, sufficiently homogeneous elementary extension to find the sequence. The key steps are as follows:

  1. Note that we can characterize being an indiscernible sequence (indexed by natural numbers) as an infinite conjunction of formulas, even if we insist that every member of the sequence has the same type as $a$ over $A$. By compactness, we can find such a sequence in an elementary extension just if that conjunction is finitely consistent. In our context, that means that it is sufficient to find a sequence that satisfies the definition for some fixed, arbitrary $n$ and $\phi(x_1,\ldots,x_n)$.
  2. We know that $\operatorname{tp}(a/A)$ is not algebraic, so find infinitely many realizations of this type; index them by the naturals in whatever way you want. Use the Infinite Ramsey Theorem to find an infinite subset that meets the definition with respect to $\phi$. (I'm omitting some of the details of how exactly we apply the theorem, but it's nothing too hard.)
  3. Apply compactness as described in step 1 to find an indiscernible sequence in our larger model, all of whose members have the same type as $a$ over $A$. By homogeneity, find an automorphism $f$ of the big model, fixing $A$ pointwise, sending the first element of the sequence to $a$. The image of the sequence under $f$ is as desired.

Last word: you should pay attention to the process of using Ramsey's theorem to build an indiscernible sequence, and work out the details for yourself. That technique forms an important tool for working with indiscernibles.

$\endgroup$
0
$\begingroup$

As stated above, the direction $(\Longleftarrow$) is quite simple. An alternative way to prove $(\Longrightarrow)$ can be the following:

Work in some monster model $\mathfrak{C}$ and note that $\mathrm{acl}_\mathbb{M}(A)=\mathrm{acl}_\mathfrak{C}(A)$. Let $M$ be such that $A\subset M\prec\mathfrak{C}$ and $a \notin M$ (as shown here, such $M$ exists) and consider $\mathrm{tp}(a/M)$. As a type over $M\prec\mathfrak{C}$ it is finitely satisfiable over $M$ and can therefore be extended to a complete (and $M$-invariant) type $p\in S_x(\mathfrak{C})$.

Consider the Morley sequence generated by $p$ over $M$, that is a sequence such that $a_0 \models p|_M$, $a_1 \models p|_{Ma_0}$, $a_2 \models p_{Ma_0a_1}$ and so on. Since $a$ realizes $p|_M$ one can set $a_0=a$. This is a sequence of indiscernibles over $M$ (and therefore over $A$). Furthermore, this sequence is non-constant: if it were to be constant then the formula $x=a$ would have been a part of $p$ and since $p$ is $M$-invariant every automorphism of $\mathfrak{C}$ would fix $a$. This implies that $\mathrm{tp}(a/M) \models x=a$ and by compactness a finite set of formulas $\phi_1(x),...,\phi_n(x)$ (over $M$) from $\mathrm{tp}(a/M)$ proves that $x=a$. Since $M\prec \mathfrak{C}$ and $$\mathfrak{C} \models \exists x \bigwedge _{i=1} ^n \phi_i(x)$$ such $x$ must also exist in $M$, but this implies that $a=x \in M$, a contradiction.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.