# A set is open if its complement is closed.

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.

• 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
• 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). – user642796 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