TITLE: Why Zariski topology is not Hausdorff
QUESTION [8 upvotes]: I am reading the book about Algebraic geometry. I am confused about the following 
two things the book mentioned:
Zariski topology is
1. different from the topology studied in real and complex analysis.
2. not Hausdorff.
Well, I roughly now about Hausdorff and some def. in real and complex analysis. However,
I cannot know why still.    
A hint or direction for thinking is ok. 
Thanks,

REPLY [15 votes]: The Zariski topology on $K$, where $K$ is a finite field, is indeed Hausdorff. However, as soon as $K$ is infinite, it is not Hausdorff. This might be interesting, if you are reading a book on algebraic geometry. 
The Zariski topology is the coarsest topology which satisfies the $T_1$ separation axiom, which means that singleton sets are closed. 
So let $K$ be an infinite field. Then The Zariski topology over $K$ is not Hausdorff. Indeed, if we have any two nonempty open sets
$U$ and $V$ , then $U^c$ and $V^c$ are finite, so $(U \cap V )^c = U^c \cup V^c$ is finite as well. In particular, $U \cap V$ is nonempty. Thus, open neighborhoods of any two points $x$ and $y$ will intersect.