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Sorry, but I have to ask a dumb question:

Algebraically, a hyperbola has only one irreducible component (given by an irreducible polynomial).

Why, then, does the real image of a hyperbola show two components?

Better yet: How should I interpret, visually, the components of an algebraic variety?

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    Short and rather dumb answer: because the Zariski topology is much coarser than the Hausdorff topology. (every polynomial vanishing on one Hausdorff component must already vanish entirely on both). However, each of the Hausdorff components is what is sometimes called [semi-algebraic](http://en.wikipedia.org/wiki/Semialgebraic_set) (that is, if you allow polynomial *inequalities*, then you can $x + y \geq 0$.2011-05-27
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    The real points are the intersection of the complex points in $\mathbb{C}^2$ with the real plane, and there's no reason for an intersection to have the same number of components as the original thing.2011-05-27

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