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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.

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    $\begingroup$ 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. $\endgroup$ – ThePortakal Mar 20 '18 at 12:55
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    $\begingroup$ Why do you doubt your proof? $\endgroup$ – Jochen Mar 20 '18 at 12:55
  • $\begingroup$ Do you have a definition of "infinite set"? $\endgroup$ – aschepler Mar 20 '18 at 12:57
  • $\begingroup$ @Jochen because I could not given a concrete proof of the existence of such an A $\endgroup$ – Diya Mar 20 '18 at 12:58
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    $\begingroup$ @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. $\endgroup$ – aschepler Mar 20 '18 at 12:58
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The answer is negative.

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

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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.

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  • $\begingroup$ 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 $X\setminus A$. $\endgroup$ – Jochen Mar 20 '18 at 14:56
  • $\begingroup$ @Jochen yes, that’s another way to see it. $\endgroup$ – Henno Brandsma Mar 20 '18 at 14:57

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