# countable algebraic sets

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^{}.

Title | countable algebraic sets |
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Canonical name | CountableAlgebraicSets |

Date of creation | 2013-03-22 15:44:41 |

Last modified on | 2013-03-22 15:44:41 |

Owner | rspuzio (6075) |

Last modified by | rspuzio (6075) |

Numerical id | 8 |

Author | rspuzio (6075) |

Entry type | Theorem |

Classification | msc 14A10 |