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I have read from my topology text book that there are few topological space that does not have Hausdorff property...and they may have some exceptional things that the singleton sets are not closed,has not unique limit point of a convergent sequence....The author describes an example by considering the finite set {a,b,c}.Here the singleton sets are not closed....Which is very unexpected to me as we know that every finite sets form discreet topology.So every singleton sets are closed.How do we find other topology on that set to avoid Hausdorff principal to find that crazy result !

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  • $\begingroup$ Finite sets can also be equipped with the indiscrete topology, in which there are only two open sets. It's quite clear this isn't Hausdorff in a non-singleton set. $\endgroup$ – астон вілла олоф мэллбэрг Apr 9 '17 at 10:33
  • $\begingroup$ On any set $X$ you can consider the topology in which the open sets are only the empty set $\emptyset$ and all the set $X$. If the cardinality of $X$ is greater than 1, then it is not an Haussdorf $\endgroup$ – Exodd Apr 9 '17 at 10:33
  • $\begingroup$ A (literally) complex but important example: Let $X=\Bbb C^2$ and say that $U$ is open if ther is a polynomial $f\in\Bbb C[X,Y]$ such that $U=\{\,(x,y)\in\Bbb C^2\mid f(x,y)\ne 0\,\}$. - Simpler: For any infintie set $X$ consider the co-finite topology, i.e., $U$ is open if $U=\emptyset$ or $X\setminus U$ is finite. $\endgroup$ – Hagen von Eitzen Apr 9 '17 at 10:39
  • $\begingroup$ I apologise,I just missed the indiscret case....But does the topology necessarily be indiscret not to satisfy Hausdorff property $\endgroup$ – Subhajit Saha Apr 9 '17 at 10:42
  • $\begingroup$ Another example is the Zariski topology on $X=\{(a_1,\ldots,a_n):a_i\in\mathbb{F}\}$, where open sets are the set of points that take a system of polynomial equations to zero. $\endgroup$ – Santana Afton Apr 9 '17 at 10:42

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