Final answer:
To show that the set of interior points of any set E, int(E), is open, we use the definition of interior points and neighborhoods. Any point x in int(E) has a neighborhood entirely within E. Since this argument holds for every point in int(E), it follows that int(E) itself is open.
Step-by-step explanation:
The question asks to show that the set of interior points of any set E, denoted as int(E), is open. By definition, an interior point of a set E is a point that has a neighborhood completely contained within E. To show that int(E) is open, we need to take any point x in int(E) and prove that there exists a neighborhood around x that is entirely contained within int(E).
Consider a point x in int(E). By the definition of an interior point, there exists an epsilon neighborhood Vϵ(x) such that Vϵ(x) is entirely contained in E. Now, take any point y in Vϵ(x). Because y is within this epsilon neighborhood, it, too, will have a neighborhood completely within E, thus making y an interior point of E. Therefore, the original neighborhood Vϵ(x) is a subset of int(E). Since this criteria holds for any point x within int(E), we can conclude that int(E) is open because every point in int(E) has a neighborhood that lies within int(E).