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I'm currently reading Griffith's Introduction to Algebraic Curves. In chapter 2, he proves the following:

Given $C$ an irreducible algebraic curve, and $S$ the set of singularities in the curve, $C^{*}=C\backslash S$ is connected.

I understand that $S$ is a finite set, but I don't see how intuitively how $C$ stays connected when you remove singularities. For example, when taking the curve defined by $x^3-x^2+y^2=0$. This function clearly has a singularity at the origin. When the origin is removed, how is the resulting curve still connected- there is a section to the left of the y axis and a section to the right. Taking the closure of either could not intersect the other. How is the set $C\backslash S$ still connected, intuitively? What part of what I'm saying doesn't make sense?

Alvin Chen
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    You're only looking at the real picture - you should be thinking about the complex picture. Your curve is complex one-dimensional = real two-dimensional, so removing one point can't disconnect it. – KReiser Jan 16 '20 at 20:45
  • the book talk about curve over complex field these are surfaces when you look at them over real numbers when you remove finite point from a surface it usually remains connected.in your example you are taking about solution of that equation over real numbers you must think about complex solution – ali Jan 16 '20 at 20:46
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    Here are two questions to think about: (1) connectedness depends on the topology; so which topology are you using? (Your intuitive notion of connectedness depends on the usual topology.) (2) if you are working in $ \mathbb{C}^2 $, a picture in $ \mathbb{R}^2 $ does not show you all the points of the curve around the singularity. – Alex Elzenaar Jan 16 '20 at 20:47
  • If this is the case, how are we sure there aren't two parts of the curve that meet at only one point which removing would disconnect it? – Alvin Chen Jan 16 '20 at 20:51
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    If that were the case, the closures of each of those connected components inside the original curve would be distinct irreducible components meeting only at that point, contradicting the assumption that your curve is irreducible. – KReiser Jan 16 '20 at 21:17

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