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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.

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Does that upvote mean I'm right? –  Casquibaldo Oct 16 '12 at 5:11
    
What I see here is your proposed definition: open iff closed complement, followed by a proof that open only if closed complement. Your proof does establish the latter claim, but you haven't proven that sets with closed complement are always open. I don't see what definition you've modified: your title matches the statement at the beginning of your question. –  Kevin Carlson Oct 16 '12 at 5:18
    
OH my mistake, I meant to write if without the two f's on top. Now it should make sense. Anyway, I just wanted to see if the latter claim was proven, which I am glad it does. Thank you @KevinCarlson –  Casquibaldo Oct 16 '12 at 5:23
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Usually definitions are implicitly taken to be biconditional. So if you have been given the definition of an open set to be a set whose complement is closed, there is no further proof needed. If, however, you have been given a different definition of an open set, proof may be necessary (for the above stated property of open sets). –  Arthur Fischer Oct 16 '12 at 5:27
    
@ArthurFischer That would have been really helpful to know before I took it upon myself to make it that way. Thanks –  Casquibaldo Oct 16 '12 at 5:28
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up vote 5 down vote accepted

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.)

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Thanks, this was exactly my problem. –  Casquibaldo Oct 16 '12 at 5:46
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