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[parent] countable algebraic sets (Theorem)

An algebraic set over an uncountably infinite base field $\mathbb{F}$ (like the real or complex numbers) cannot be countably infinite.

Proof: Let $S$ be a countably infinite subset of $\mathbb{F}^n$ By a cardinality argument (see the attachment), there must exist a line such that the projection of this set to the line is infinite. Since the projection of an algebraic set to a linear subspace is an algebraic set, the projection of $S$ to this line would be an algebraic subset of the line. However, an algebraic subset of a line is the locus of zeros of some polynomial, hence must be finite. Therefore, $S$ could not be algebraic since that would lead to a contradiction.




"countable algebraic sets" is owned by rspuzio. [ full author list (2) ]
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lemma on projection of countable sets (Theorem) by rspuzio
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Cross-references: contradiction, finite, polynomial, locus, algebraic, linear subspace, projection, line, argument, cardinality, subset, proof, countably infinite, complex numbers, real, base field, infinite, algebraic set

This is version 5 of countable algebraic sets, born on 2006-03-08, modified 2006-10-10.
Object id is 7696, canonical name is CountableAlgebraicSets.
Accessed 968 times total.

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AMS MSC14A10 (Algebraic geometry :: Foundations :: Varieties and morphisms)

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