What is the interior of the set of even numbers in the topological space $(\mathbb{N}, Cofin)$?

Question

What is the interior of the set of even numbers in the topological space $(\mathbb{N}, Cofin)$, where $Cofin$ is the topology on $\mathbb{N}$ consisting of the cofinite subsets of $\mathbb{N}$?

Let $E$ be the set of even numbers and $Int$ the interior operator. Then $Int E = \bigcup \lbrace M \subseteq \mathbb{N} | M \text{ is open } \wedge M \subseteq E\rbrace$.

It seems to me the only open set contained in $E$ is $\emptyset$. $E$ itself is not cofinite, so it won't be in the interior. But I don't know how to show that there are no other cofinite subsets of $\mathbb{N}$ contained in $E$. My guess would be that any cofinite subset of $E$ would be infinite and hence it contains only even numbers and is infinite and so it is isomorphic to $E$. But that is quite informal. I'd appreciate a hint :)

Answer 1

Try using proof by contradiction:

Let $Y$ be in the cofinite topology. This means the either $Y=\emptyset$ or $\Bbb{N}\setminus Y$ is finite.

Assume $Y$ is in $E$. That means $Y$ is some set of even numbers. Since we know that $\Bbb{N}\setminus Y$ is not finite ,therefore it cannot be in the co finite topology. Contradiction. Hence it cannot be in $E$ . Therefore any set of the cofinite topology cannot be in $E$ and hence cannot be in the interior.

Also only the empty set argument works and hence $\operatorname{Int}(E)$ contains only the empty set

Answer 2

You're making this harder than it is.

You want to show that $E$ has no cofinite subsets. So suppose $X\subseteq E$; can you find an infinite set $Y$ which is disjoint from $X$? This would imply $X$ is not cofinite.

Hint: you shouldn't look too far to find $Y$ . . .