I was shown this statement to be a definition. However, I think that the actual statement would be.
A set is open iff its complement is closed.
This way it's a biconditional and it's also true- this is my proof, with C being a continuum that is nonempty, has no first or last point and has an ordering (<): Suppose $U$ is open. Let $x$ be a limit point of $C \setminus U$. It should follow that if $C \setminus U$ is closed, then $x \in (C \setminus U)$. Now we know that $\forall$ regions $R$ containing $x$, there is a nonempty intersection with $C \setminus U$. Also, $x$ cannot be a point of the interior of $U$ because any intersection with a region containing it would be empty, which wouldn't be in accordance to our assumption that $x$ is a limit point of $C \setminus U$. So $x$ is not in the interior of $U$ or $x \notin U$. This would mean that $x \in (C \setminus U)$ and thus $C \setminus U$ is closed.
I was wondering if this proof was reasonable, if there were any gaps in logic, and if this modification of the definition I presented was viable and justified.
Best Answer
This is simply a slightly confusing, but very widespread, convention of mathematical wording.
In statements, “if” is interpreted just as one direction of implication. To specify bidirectional implication, as you say, one needs to write “iff”, or “if and only if”, or “exactly if”, or similar.
In definitions, however, “if” is used for by-definition equivalence, which in particular gives the biconditional. (Quite arguably, it’s exactly the same as a biconditional; logicians can and do split hairs over this issue, but in standard foundations of mathematics, there’s essentially no difference.)