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Let $T_1=\{U:X-U \text{ is finite for all of } X\}$. Then $T_1$ is the cofinite topolgy on $X$,where $X$ is an arbitrary infinite set. Then $T_1$ is not a Hausdorff space.Is it a regular space or a normal space?

Let $T_2=\{U:X-U \text{ is countable or all of }X\}$,then $T_2$ is the co-countable topology.This also not a Hausdorff space.But the convergent sequences have unique limits.Is it true? Is $T_2$ a regular or normal space?

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What's the difficulty? Do you not know the definitions of regular and normal? do you know the definitions, but there is some specific place where you get stuck trying to solve the problem? Help us help you. –  Gerry Myerson Dec 17 '12 at 5:04
You might want to require that $X$ is uncountable for the second, otherwise you get the discrete topology which is Hausdorff. I would also avoid using $T_1$ and $T_2$ as the topologies as these often denote separation axioms -- which is incidentally the topic of the question! –  Asaf Karagila Dec 17 '12 at 6:56

1 Answer 1

Note that if $X$ is finite, then $T_1$ is just the discrete topology on $X$. Similarly, if $X$ is countable, then $T_2$ is the discrete topology on $X$. I'll assume that $X$ is infinite in the first case, and uncountable in the second.


  • To show that $( X , T_1 )$ is neither regular nor normal for essentially the same reason it is not Hausdorff: any two nonempty open sets must have nonempty intersection. You only need to exhibit (for regularity) a point $x \in X$ and a closed subset $F \subseteq X$ not containing $x$, and (for normality) two disjoint nonempty closed sets.

  • To show that convergent sequences in $( X , T_2 )$ have unique limits, one must first note that the only convergent sequences in $( X , T_2 )$ are the eventually constant sequences. Regularity and normality fail for the same reason as above.

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what is the Pseudocharacter of $(X,T_1)$ at point $x$? the value of $ψ(x,X)$ is undefined? –  TXC Apr 22 '13 at 5:47

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