genericness and the Zariski topology

What does it mean (in a mathematically rigorous way) to claim something is "generic?" How does this coincide with the Zariski topology?

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have you checked out these wikipedia pages? en.wikipedia.org/wiki/Generic_property, en.wikipedia.org/wiki/Generic_point – Eric O. Korman Jul 23 '10 at 14:56
If you give an example or context in which you have seen it, or possibly elaborate on your question some more, you might have better luck. Other than that, Eric's links seem useful... – BBischof Jul 24 '10 at 19:54

The term generic usually applies to something which happens in an open dense set of some space. The idea is that open dense sets are large subsets of the space. Indeed, they are closed under finite intersections and thus form a base for a filter of subsets of the space.

Sometimes, the term is applied more generally to something which happens in a countable intersection of open dense sets (a dense Gδ set) of some space. Usually, this is in the context of a complete metric space or a locally compact Hausdorff space for which the Baire Category Theorem applies.

The generic point is a (sometimes fictitious) point which lies in every open dense set of the space. Irreducible sober spaces always have a generic point, it is the unique point whose closure is the whole space. The only Hausdorff space with a generic point is the one-point space.

Fictitious generic points have a variety of uses. Usually one means a point which lies in all open dense sets which are considered in the current argument, without specifying the open dense sets in question. This is fine because the intersection of finitely many (or even countably many in the case of Baire spaces) open dense sets is guaranteed to be nonempty.

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In general*, if something is "generic", it means it happens or is true "almost all of the time" or "almost everywhere".

In measure theory, for example, when you say "$P(x)$ is true for almost all $x$", this has the precise meaning that the set of $x$'s for which $P(x)$ does not hold has measure zero.

One can relate this to the Zariski topology via the fact that Zariski closed subsets of $\mathbb{C}^n$ have Lebesgue measure zero: http://mathoverflow.net/questions/25513/zariski-closed-sets-in-cn-are-of-measure-0