Show that there is a smallest closed set $\overline A$ which contains ${A}$ I'm a beginner student in real analysis and I don't how to start the question below:
Let $A \subseteq \mathbb{R}^n$ be given. Show that there is a smallest closed set ,$\overline{A}$, which contains $A$, i.e., $\overline{A}$ is closed set containing $A$ and if $C$ is a closed set containing $A$ then $A\subseteq\overline{A}\subseteq C$.
How can I start to show this statement? Any clues , hint...?
Thanks
 A: There's two approaches:


*

*Bottom-up. If I give you $A$, what do you need to add to get a closed set? For instance, if $(n=1$ here) we have $A=(0, 1)$, what should $\overline{A}$ be, and how did we get it?

*Top-down. Let's say I have two closed sets $B, C$ containing $A$. What sort of thing is $B\cap C$? Can you generalize this to answer the question?
Both approaches are things you need to understand. The fact that they give the same result is a good exercise!
A: HINT: Consider the set of all closed sets containing $A$ and then remember that open sets are closed under arbitrary unions.
A: Hint.  What happens if you add all the limit points of A (Let B = A $\cup$ {limit points of A})?  What happens if you don't add all the limit points of A (A $\subset$  B but B doesn't contain all the limit points of A)?
Oh, and of course, if A itself is closed then ....
A: $\mathbb{R}^n$ is clopen (both open and closed), and it contains $A$. Take the intersection of all closed sets that contain $A$ (there is at least one: $\mathbb{R}^n$). This is, by definition, the smallest closed set containing $A$ and we denote it $\bar{A}$.
A: To prove this you can use that the intersection of any number of closed sets is also closed.
Let $F_i, i \in I$ be closed sets such that $A \subset F_i \quad \forall i \in I$. Then $\bar{A}=\cap_{i \in I}$ is a closed set which contains $A$ and it's by the construction of it it's trivial that it is the smallest closed set containing A.
