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

Question

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.

Answer 1

The answer is negative.

Consider the set $\mathbb{Z}$, which is neither open nor closed as subset of real numbers with cofinite topology

Answer 2

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.