I am trying to solve part d of problem 9 in chapter 2 of Rudin's Principles of Mathematical Analysis. The problem is: Let E* denote the set of all interior points of a set E (in a metric space X). Prove the complement of E* is the closure of the complement of E.
I will use E^c to denote the complement of E. It might be helpful to pull out Rudin and follow along; my notation is poor and the problem will make more sense.
2. The attempt at a solution
First, we show that the complement of the interior points of E is a subset of the closure of the complement of E. Because the interior of E lies in E, the complement of E lies in the complement of the interior of E. Then because the complement of the interior of E is closed, the closure of the complement of E lies in the complement of the interior of E.
Now we need to show (E*)^c is in the closure of E^c. Pick an arbitrary point p in (E*)^c, Then it is not an interior point of E, and all neighborhoods around it contain a point of E^c. Then p is a limit point of E^c, and lies in the closure of E^c.
3. The error!
I have just shown that (E*)^c= closure(E^c). Now E*^c is obviously closed, because the interior of E is always open (this is part a of the problem). The second paragraph shows that any point in E*^c is a limit point of E^c. Now if E is open, E* and E are the same, so this would show that any point in E^c is a limit point of E^c. This would mean E^c is always a perfect set (closed and every point is a limit point). Surely this is much too strong: the complement of an arbitrary open set is not always perfect!?