Interior of a set

Q9. Basic Topology (Rudin)

Let $E^0$ denote the set of all interior points of a set E. 

(a) Prove that  $E^0$ is always open.

Solution: Suppose x is an arbitrary point in $E^0$. We will show that x is an interior point of $E^0$.

By definition of $E^0$,  x is an interior point of E. Hence there exists a neighborhood N of radius r of x, such that $N \subset E$.

Since N is a neighborhood therefore N is open. Hence all points of N are interior points of N i.e. they have neighborhoods that are subsets of N. Since N is a subset of E, hence those neighborhoods (of the points of N which are subsets of N) are also subsets of E. Hence all the points in N are interior points of E. Thus all the points of N are member of $E^0$ by definition (since $E^0$ contains all the interior point of E). Thus N is a subset of $E^0$. Since x is an arbitrary element of $E^0$, and it has a neighborhood which is a subset of $E^0$, hence x is interior point of $E^0$ and $E^0$ is open.

(b) Prove that E is open if and only if $E^0 = E$

Solution: If E is open, all point of E are interior points of E (by definition). Hence all points of E are in $E^0$ and no other (as $E^0$ contains only the interior points of E and all points of E are interior points). Hence $E^0 = E$

Again if $E^0 = E$ , E contains only the points present in $E^0$ and $E^0$ contains only the interior points of E. Thus all points in E are interior points. Hence E is open.

(c) If $G \subset E$ and G is open, prove that $G \subset E^0$

Solution: Since G is open, all points of G are interior points of G. Hence each point of G has neighborhood N which is a subset of G. As G is a subset of E, hence these N's are also subsets of E. Therefore for each point in G, there exist neighborhood N which is a subset of E. Hence all points in G are interior points of E. Since $E^0$ contains all interior points of E, $E^0$ contains all points of G. Hence $G \subset E^0$.

(d) Prove that the complement of $E^0$ is the closure of the complement of E.

Solution: Complement of $E^0$ has the points which are not interior points of E. Either they are not members of E which makes them member of $E^c$ or even if they are members of E, they are points whose   every neighborhood contains some elements which are not member of E. In other words complement of \whose every neighborhood contains some elements from $E^c$. Hence they are limit points of $E^c$.
Thus the complement of $E^0$ contains elements of $E^c$ and limit points of $E^c$ which makes it by definition closure of the complement of E.

(e) Do E and $\bar{E}$ always have same interiors?

Solution: No. Counterexample: E= Set of Rational Numbers. Interior of this set is Null Set for if q be any rational number, any neighborhood of q will contain irrationals hence it cannot be totally inside E. Limit Points of this set E consists of rationals as well as irrationals. Hence closure of the set of rational numbers is $\bar{E}$ (set of real numbers). $\bar{E}$ has plenty of interior points (in fact all points are interior points). Hence E and $\bar{E}$ does not have same interior.

(f) Do E and $E^0$ always have the same closures?

Solution: