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I was reading David Marker’s Model Theory and came upon the following problem in Chapter 3.

Setting

Let $ \mathcal{M} $ be a saturated $ \mathcal{L} $-structure.

  • A definable subset $ X \subseteq M $ is said to be strongly minimal if and only if $ X $ is defined by an $ \mathcal{L}_M $-formula $ \phi(v) $ and every subset of $ X^{\mathcal{M}} $ definable in $ \mathcal{L}_{\mathcal{M}} $ is either finite or co-finite.
  • $ {\text{acl}^{X}}(A) \stackrel{\text{df}}{=} X \cap \text{acl}(A) $ is the set of all $ b \in X $ such that $ b $ is algebraic over $ A $, i.e., there is an $ \mathcal{L}_{A} $-formula $ \psi $ such that $ \mathcal{M} \models \psi(b) $ and $ |\{ c \mid \mathcal{M} \models \psi(b) \}| < \infty $.
  • $ A \subseteq X $ is called independent if and only if $ a \notin \text{acl}^{X}(A \setminus \{ a \}) $ for all $ a \in X $.

Problem

I would like to show that if $ b \in {\text{acl}^{X}}(A \cup \{ c \}) $ and $ b \notin {\text{acl}^{X}}(A) $, then $ c \in {\text{acl}^{X}}(A \cup \{ b \}) $, and that for any $ A \subseteq X $, there is an independent $ B \subseteq A $ so that $ A = {\text{acl}^{X}}(B) $.

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  • $\begingroup$ Hmm... Maybe the Exchange Lemma helps? My model theory is a little rusty. $\endgroup$ – Berrick Caleb Fillmore May 1 '15 at 16:28
  • $\begingroup$ The first statement is the exchange lemma. Once you have the exchange lemma, the second statement follows. (But the proof is not short.) $\endgroup$ – Primo Petri May 1 '15 at 17:12
  • $\begingroup$ @Primo: Hi Primo. LOL... I just realized that it is the Exchange Lemma. $\endgroup$ – Berrick Caleb Fillmore May 1 '15 at 17:15
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Strongly minimal sets are taken up in detail later in Marker (in Section 6.1). Part (a) of this exercise (exchange) is Lemma 6.1.4, and part (c) (any two bases have the same cardinality) is Lemma 6.1.9. Part (b), the existence of a basis (an independent set $B$ such that $X \subseteq \text{acl}(B)$) is mentioned in passing just after Definition 6.1.8.

If you want to solve the exercise yourself (and you should!) instead of looking up the proofs in Chapter 6, here are some hints.

(a) This is the trickiest one. To get you started: If $b\in \text{acl}(A\cup\{c\})\setminus \text{acl}(A)$, look at the formula $\phi(y,c)$ which algebraizes $b$ (so $M\models\phi(b,c)$ and there are exactly $n$ realizations of $\phi(y,c)$ in $X$). Now turn things around, by looking at $\exists^{!n}y\,\phi(y,z)$ as a formula that's true of $c$. Think about what strong minimality means about this formula...

(b) Try Zorn's lemma. (Contrary to Primo Petri's comment, the proof is very short, using Zorn. I would say it's part (c) that isn't short.)

(c) Do the finite and infinite cases separately. In the finite case, given two bases $B$ and $B'$, use exchange to replace elements of $B$ by elements of $B'$ one by one.

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    $\begingroup$ @chibro2 We want an independent set $B$ which is large enough so that $X\subseteq \text{acl}(B)$. OK, let's use Zorn's Lemma to get a maximal one. Check that the poset of independent subsets of $X$, ordered by inclusion, satisfies the hypotheses of Zorn's Lemma (i.e. it's nonempty and the union of a chain of independent sets is independent). Ta-dah, Zorn gives us a maximal $B$. Now if $X\not\subseteq \text{acl}(B)$, show that we could make $B$ larger (add some $x\notin \text{acl}(B)$ to $B$), contradicting maximality. $\endgroup$ – Alex Kruckman May 5 '15 at 16:02

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