Marker exercise 1.4.10c

$dcl(A)=\{x \in M : x$ definable from $A \}$, Show $dcl(dcl(A)=A$

I assume "$x$ definable from $A$" means "$\{x\}$ is $A$-definable."

Definition 1.3.1: Let $\mathcal{M} = (M, . . .)$ be an $\mathcal{L}$-structure. We say that $X \subseteq M^n$ is definable if and only if there is an $\mathcal{L}$-formula $\phi(v_1 , . . . , v_n , w_1 , . . . , w_m )$ and $b \in M^m$ such that $X = \{a \in M^n : M \models \phi(a, b)\}$. We say that $\phi(v, b)$ defines $X$. We say that $X$ is $A$-definable or definable over $A$ if there is a formula $\phi(v, w_1 , . . . , w_l )$ and $b \in A^l$ such that $\phi(v, b)$ defines $X$.

We can define every $\{a\} \subseteq A$ by a formula $\phi(x,a) \mathrel{\mathop:}=x=a$ since $a \in A$. Plus there are possibly other $\emptyset$-definable singletons depending on our language which are $A$-definable. If we have a language of abelian groups $\mathcal{L}=(0,+)$ and although $0 \notin A$, we can define $0$ by the following formula: $\phi(x,a) \mathrel{\mathop:}=x=0$.

So for the example I have given $dcl(A)$ would be a superset of $A$, $0 \in dcl(A)$, consequently $0 \in dcl(dcl(A))$.

What is wrong with my reasoning?

  • 3
    $\begingroup$ There is a typo in the exercise. It should read $\mathrm{dcl}(\mathrm{dcl}(A))=\mathrm{dcl}(A)$. $\endgroup$ – Andrés E. Caicedo Aug 4 '13 at 23:34

As Andres Caicedo noted, there is a typo in the exercise. It should read $dcl(dcl(A))=dcl(A)$.

  • $\begingroup$ This does not provide an answer to the question. To critique or request clarification from an author, leave a comment below their post. $\endgroup$ – Dan Rust Aug 5 '13 at 1:52
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    $\begingroup$ @DanielRust I disagree entirely. This is the answer to the question. The question is "how to prove blah?" The OP noticed themselves that blah is false, and they are confused. The reason it is being asked in the book is because it is a typo. If the OP or someone else is further interested in why the correct identity holds, this should be asked as a separate question. $\endgroup$ – Andrés E. Caicedo Aug 5 '13 at 1:55
  • $\begingroup$ Wow you're right. This was my fault for reviewing too quickly. At a glance, such a sentence seems like a comment. I'll try and be more careful reviewing in future. $\endgroup$ – Dan Rust Aug 5 '13 at 1:57

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