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Hausdorff Space and finite complement topology |
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| Apr28-12, 07:48 PM | #1 |
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Hausdorff Space and finite complement topology
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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| Apr28-12, 07:52 PM | #2 |
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OK, pick two elements, how do they have disjoint neigborhoods???
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| Apr28-12, 08:09 PM | #3 |
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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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