# Consider cofinite topology on an infinite set X. Is every subset of X either open, or closed, or clopen?

First, the only clopen sets are $X$ and $\phi$, because if a proper subset $A$ is clopen, then $A$ and $X-A$ are open, so X-A and A are closed and hence finite, giving as $X=A\cup (X-A)$ is finite, which is a contradiction.

Now, as X is infinite, there must be an infinite set $A$ such that its complement $X-A$ is also infinite. So $A$ is neither open nor closed. [A subset is open if its complement is finite, and closed if it is finite. $A$ is neither.]

Is this proof correct? I have intuitively guess the existence of such an $A$ and have not been able to actually prove it.

• You can just give an example. Let $X=\Bbb Z$ and $A=\Bbb Z^+$. Then $A$ is infinite and $A^c$ is infinite. Hence $A$ is neither open nor closed. Proof done. – ThePortakal Mar 20 '18 at 12:55
• Why do you doubt your proof? – Jochen Mar 20 '18 at 12:55
• Do you have a definition of "infinite set"? – aschepler Mar 20 '18 at 12:57
• @Jochen because I could not given a concrete proof of the existence of such an A – Diya Mar 20 '18 at 12:58
• @ThePortakal That proves the suggested theorem is not true, but it would be a stronger result to say any infinite set has a subset that is neither open nor closed in the cofinite topology. – aschepler Mar 20 '18 at 12:58

Consider the set $\mathbb{Z}$, which is neither open nor closed as subset of real numbers with cofinite topology
You are indeed correct: if $A \subseteq X$ exists with $A$ infinite and $X\setminus A$ infinite, then $A$ is not open (as it's not empty and its complement is not finite) and same holds for its complement, so $A$ is not closed either.
That such a set exists is clear from set theory: e.g. $X$ infinite means that there is a bijection $f$ between $X$ and $X \times \{0,1\}$ and then $A = f^{-1}[X \times \{0\}]$ is such a set.
• Well, perhaps a bit simpler: If $X$ is infinite it contains distinct $x_n$ for $n\in\mathbb N$. Then $A=\{x_{2n}:n\in\mathbb N\}$ is infinite as well as $X\setminus A$. – Jochen Mar 20 '18 at 14:56