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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

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