Hausdorff Space and finite complement topology

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The finite complement topology on the reals R is not Hausdorff because any two points do not have disjoint neighborhoods. In this topology, open sets are defined as those whose complements are finite, meaning neighborhoods cannot be standard open intervals. For example, if two points like {0, 2} are chosen, their neighborhoods must be rays rather than intervals. This leads to the conclusion that it is impossible to separate points with disjoint neighborhoods in this topology. Thus, the finite complement topology fails to meet the Hausdorff condition.
Pippi
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I want to come up with examples that finite complement topology of the reals R is not Hausdorff, because by definition, for each pair x1, x2 in R, x1 and x2 have some disjoint neighborhoods.

My thinking is as follows: finite complement topology of the reals R is a set that contains open sets of the form, (- inf, a1) U (a1, a2) U ... U (an, inf). The complement of an open set is the finite elements {a1, a2, ... an}. However, any two points I pick out of any open set have disjoint neighborhoods. How is it possible?
 
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OK, pick two elements, how do they have disjoint neigborhoods?
 
Ok, I see where my reasoning was wrong. The neighborhoods have to be open sets in this topology. If I pick two points, {0, 2}, in the open set (-inf, 1) U (1 inf), their neighborhoods can't be any open intervals and must be rays.

Cheers!
 

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