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Let $X=\mathbb{N}$, and $\tau=\tau_{cof}$, the co-finite topology. Let $E_1=${1,2,3,4,5,6} and $E_2=$ {odd numbers}. Find Closure and Interior of $E_1$ and $E_2$.

$E_1$ is not open because $C_X(E_1)$ is an infinite set. But $E_1$ is closed because $C_X(X\setminus E_1)=E_1$, so $Cl(E_1)=E_1$. And the interior of $E_1 $ is $Int(E_1)=\emptyset$, because if there is an open set in $E_1$, its complementar set wouldn't be finite.

$E_2$ is not open and not closed. It's interior is $Int(E_2)=\emptyset$ and it's closure is $Cl(E_2)=X$.

Is it ok?

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  • $\begingroup$ What does $C_X$ mean? $\endgroup$ – M. Winter Aug 17 '17 at 12:16
  • $\begingroup$ With $C_X(A)$ I mean $X \setminus A$ $\endgroup$ – VoB Aug 17 '17 at 13:15
  • $\begingroup$ Please always use standard notation and if you are not sure if it is standard, then define it. $\endgroup$ – M. Winter Aug 17 '17 at 13:17
  • $\begingroup$ Sorry, I thought it was a standard notation. $\endgroup$ – VoB Aug 17 '17 at 13:18
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Yes, but perhaps that you could explain why is it that you say that ${E_2}^\circ=\emptyset$ and that $\overline{E_2}=\mathbb N$.

By the way, since your space is $\mathbb N$, I don't understand why you keep calling it $X$.

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  • $\begingroup$ ${E_2}^\circ=\emptyset$ because if I take an open set, say $A$, made by odd numbers, this is automatically not open because $X \setminus A$ is not finite. So the only open set which is contained in $E_2$ is $\emptyset$. $\overline{E_2}=\mathbb N$ because in the co-finite topology, every closed set must be formed from a finite number of elements, since $A$ is closed $<=>$ $C_X(A)$ is open $<=>$ $C_X(X \setminus A)$ is finite $<=>$ $A$ is finite. But there are infinite odd numbers, so the only closed set that contain all of them is $\mathbb{N}$ $\endgroup$ – VoB Aug 17 '17 at 13:30

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