Let $A\subset$ $\mathbb{R}$$^{n}$ be given. Show that there is a smallest closet set $\bar{A}$, which contains A: I.e., $\bar{A}$ is a closed set such that $A\subset$$\bar{A}$ and if C is a closed se such that $A\subset C$, then $A\subset \bar{A}\subset C$. The set $\overline{A}$ is called the closure of A.
I have some exercise that has been left to the reader in my math class. We will not be tested in this, but I would like to know how to solve this example problem from the text. Can anyone give a step by step breakdown of the problem with the resultant solution? I think it may be interesting to analyze and see how the answer is derived.
You can take the collection of all closed subsets of $\mathbb{R}^n$ which contain $A$. This collection is not empty, since $\mathbb{R}^n$ belongs to it. Any intersection of closed subsets is closed, and by definition of intersection it is the smallest closed subset of $\mathbb{R}^n$ containing $A$.
Equivalently, you can consider the set $A$ together with all the limit points of $A$, that is the union $A\cup A'$, where $A'$ is called sometimes the derived set of $A$. Since closed subsets contain their limit points, you can see that any closed subset $F$ of $\mathbb{R}^n$ containing $A$ must contain also the closure of $A$. Indeed, let $F$ be a subset of $\mathbb{R}^n$ which is closed and which contains $A$. Since it is closed, it contains all its limits points, and since it contains $A$, it also contains all limits point of $A$ (because they are limit point of $F$), hence it contains the union $A\cup A'$.