Is if a set is not open and its complement not closed, is the set closed and its complement open? I'm working in baby Rudin. I think understand the ideas of interior and limit points, open and closed sets. 
We have the theorem that $E$ is closed iff $E^c$ is open. 
This gives me some idea about the relationship between a set and its complement. However, I find myself stuck when it comes to determining whether the following statements are true, and, if so, how to prove them:
1. If $E$ is not open then it is closed, and vice-versa.


*If $E$ is not open and $E^c$ is not closed then $E$ is closed and $E^c$ is open. 


Is this statement equivalent to saying that a set is closed if its complement is open? 


*Proving the following (This one is a question from Rudin):



Let $E^o$ be the set of all interior points of a set $E$. Prove $E^o$ is always open.

(This is my first time using this markup language so if anyone can suggest improvements or good practices I'd be happy to adopt them).
Thanks
 A: "Opposite" as a term will only mislead you. The entire space $\mathbb{R}$ is both open and closed, as is the empty set $\emptyset$; neither is the "opposite" of itself. The complement of an open set is closed, and vice versa.

*

*If $E$ is not open then it is closed, and vice-versa.
This is false. Counterexample: $[0,1)$.


*if $E$ is not open and $E^c$ is not closed then $E$ is closed and $E^c$ is open.
False. Saying that $E^c$ is not closed is equivalent to saying that $E$ is not open, so the hypothesis is just "$E$ is not open". From this, as noted in 1., it doesn't follow that $E$ is closed – as mentioned, they are not "opposites".


*Let $E^o$ be the set of all interior points of a set $E$. Prove $E^o$ is always open.
$x$ is an interior point of $E$ iff some open neighborhood $U$ of $x$ is contained in $E$. So if $y \in E^o$, then $y \in U \subseteq E$ for some open neighborhood $U$ of a point $x \in E$. But then $y$ is by definition an interior point of $E$, as this same $U$ is an open neighborhood of $y$ contained in $E$.
A: In general a set may be neither open nor closed,and if this the case,then its complement is also neither closed nor open. The old joke is that "If X is not closed then X is open" is true if X is a door. As for Q.3. I don't  know how Rudin defines $E^o$. Topologists define $E^o$ as the union of all open subsets of $E$  so of course it's open because the union of any family  of open sets is an open set....... We may define $p\in E^o \iff \exists \text { open } V  (p\in V\subset E). $ Now suppose  $p\in E^o$.  Choose an open set $V(p)$ where $p\in V(p)\subset E$. Observe that  $\forall q\in V(p)$ we have [ $q\in V(p)\subset E \wedge V(p)$ is open.] So by the def'n of $E^o$,we have $\forall q\in V(p) (q\in  E^o)$. That is, $V(p)\subset E^o$......... So if we let $F=\cup \{V(p) : p\in E^o\}$, we have $F\subset E^o$ . But $E^o=\{p :p\in E^o\}\subset \cup \{V(p) :p\in E^o\}=F.$ Therefore $F=E^o$. 
