# Thread: Real Analysis - Open/Closed Sets

1. ## Real Analysis - Open/Closed Sets

Given A is a subset of R, let L be the set of all limit points of A.

a. Show that the set L is closed

b. Argue that if x is a limit point of A U L, then x is a limit point of A.
Use this observation to furnish a proof of the following theorem:

For every A is a subset of R, the closure of A bar is a closed set and is the smallest closed set containing A.

2. If I could figure out how to do notation on here it would be easier, but basically if you assume an x not in the set of all limit points then there is an epsilon neighborhood around x such that the complement of L is open. Thus if L complement is open then L is closed. Where L is the set of all limit points.

3. For (a) you can use the fact that a set is closed if and only if it contains all its boundary points. The complement of a boundary is open since it does not contains its boundary. Since the complement of the boundary is open then the boundary is a closed set.
For (b) you need to show that the boundary of a boundary is the boundary itself.