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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 Caicedo Jan 22 '11 at 6:52

1 Answer 1

up vote 4 down vote accepted

First, note that if Grandma Helen has only finitely many children, and each of those children have only finitely many children, then Grandma Helen has only finitely many grandchildren.

For the problem, however, there is the subtlety that perhaps some but not all of the children of Helen have only finitely many children. But you can still use the same idea: If Grandma Helen has only finitely many children, and one of those children has at most 10 children (perhaps some others have infinitely many), then the collection of grandchildren of Helen whose parents have at most 10 children is finite.

Those grandchildren (whose parents have at most 10 children) satisfy a first-order expressible property, with Helen as a parameter, that is satisfied by only finitely many other people.

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This is a very nice way to explain what's happening without the formulas getting in the way. Thanks! – Lost In Math Jan 26 '11 at 7:28

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