Contrapositive of a Definition I have a problem (in real analysis class) that states "What is the contrapositive of the definition of "closed"?"
The definition in our class of closed is: "a set E is closed iff the set contains all of its accumulation points". Another way of stating this would be "a set E is closed iff when p is an accumulation point of E then p is in E".
So I basically have: A iff B$\implies$ C
So my question is whether the contrapositive is one of 2 things:


*

*Not(B$\implies$ C) iff not(A)

*A iff Not(C)$\implies$Not(B)
EDIT: As a note for potential readers looking for the same info I was looking for: The professor in our class said that he meant it as 2 not 1 because the "E closed iff" is not the definition, but part of the definition in sentence form. 
So the answer was : "E is closed iff if $p\not\in E$ then p not an acc. pt."
 A: I think that it makes little sense to "contrapose" a definition ...
But formally, we have here a bi-implication :

$Closed (E) \leftrightarrow \forall p (p \in Accumul(E) \rightarrow p \in E)$.

We have that : $(A \leftrightarrow B) \leftrightarrow ((A \rightarrow B) \land (B \rightarrow A))$; thus, applying contraposition, we have : $(A \leftrightarrow B) \leftrightarrow ((\lnot B \rightarrow \lnot A) \land (\lnot A \rightarrow \lnot B))$.
I.e. : $(A \leftrightarrow B) \leftrightarrow (\lnot A \leftrightarrow \lnot B)$, as expected. 
Thus we can re-write the above definition as follows :

not-$Closed (E) \leftrightarrow \lnot \forall p (p \in Accumul(E) \rightarrow p \in E)$

i.e.


not-$Closed (E) \leftrightarrow \exists p (p \in Accumul(E) \land p \notin E)$.


A: You are correct that the statement is $A$ iff $B\implies C$ where A is the statament (The set E is closed), B is the statement (p is an accumulation point of E) and C is the statement (p is in E). It will be clearer if you consider it with one way implication first. $A \implies (B \implies C)$ and $(B\implies C)\implies A$. The contrapositive of the first is $\neg(B \implies C) \implies \neg(A)$ and the contrapositive of the second is $\neg(A) \implies \neg(B \implies C)$ so combining we get $\neg(A)$ iff $\neg(B \implies C)$
A: In usual terminology, we talk of a contrapositive of conditional statement.
The statement you have has the structure of a bicionditional
$$A \Leftrightarrow (B \to C)$$
Now since both LHS and RHS hold, we have

$\neg A \Rightarrow \neg (B \to C)$ and $\neg (B \to C) \Rightarrow \neg A$

Naturally, we have the following too
$$ \neg A \Leftrightarrow \neg (B \to C)$$
