# Prove that acl(acl(A))=acl(A), in model theory

Can someone give me an elementary proof of this fact?

Edit: This is an exercise in Marker's text, right after he defines

$$\text{acl}(A)=\{x:x \text{ is algebraic over }A\}.$$

The question and the full definition is here.

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What properties of $\operatorname{acl}$ do you have at your disposal? E.g., if you are working through a book could you please provide that as a reference? – Jonas Meyer Jan 22 '11 at 6:06
Thanks for the comment. I've added to the question, so that hopefully it's more accessible. – Lost In Math Jan 22 '11 at 6:31
Can you do this in particular cases? Say, do you see how to prove that the span of the span of a set of vectors is just the span of the set of vectors, or that the result holds if the algebraic closure is in the usual field theoretic sense? – Andrés E. Caicedo Jan 22 '11 at 6:52