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Define a topology on $X=\mathbb N$:

$A\subset X$ (proper subset) is closed if and only if:

  1. $\:A$ is finite;
  2. $\:n\in A\wedge p\in\mathbb P\wedge p|n \implies p\in A$.

A nonempty open subset is by definition infinite and therefore no finite set includes a nonempty open set. This topology is not Hausdorff since $B=\{6\}\implies\bar{B}=\{2,3,6\}$, and I guess it isn't even $T_0$.

How to prove whether it is $T_0$ or not?

I would also like a list of explicit non trivial criterion for what an open set is.

I might award several answers.

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Let $x \neq y$ be two nonzero natural numbers. WLOG, we can assume that $y$ is not a prime divisor of $x$ (otherwise just swap them). Let $A$ be the set of prime divisors of $x$. Then $A' = A \cup \{x\}$ is closed in your topology: it is finite (because $x$ is nonzero), and if a prime $p$ divides an element of $A'$, it is in $A'$ too. Thus, $X \setminus A'$ is open, and it contains $y$ because we assumed $y$ wasn't a prime divisor of $x$. But of course it doesn't contain $x$, so $x$ and $y$ are topologically distinguishable.

I don't know if your $\mathbb{N}$ contains zero or not, but if it does: let $x \neq 0$ be a natural number, and as before let $A$ be the set of prime divisors of $x$, $A' = A \cup \{x\}$. Then $A'$ is closed and $0 \in X \setminus A'$, so $x$ and $0$ are topologically distinguishable.

Conclusion: your $X$ is $T_0$.

As for characterizing open sets, I'm not sure what you want, but a set is open iff it's cofinite and whenever a number is not in it, then none of its prime divisor are either...

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