# Closed set contains the set and its closure proof check

The problem is as following: $A\subset X$. Show that IF $C$ is closed set of $X$ and $C$ contains $A$, then $C$ contains the closure of $A$

Here is my proof, but I dont know whether I have the right idea or not. A comment about its truth would be appreciated. It is not a homework problem, I am doing them for the hell of it.

Since $C$ is closed $\implies C$ contains its limit points. Since $A\subset C$, then $C$ contains the limit points of $A$ too. The definition of $\bar A$ is that $\bar A$ is the set $A$ and its limit points. Then, $C$ contains $\bar A$ too.

Is this proof true?

I don't think the second sentence of the line "since $C$ is closed $\implies C$ contains its limit points. Since $A\subset C$, then $C$ contains the limit points of $A$" is inherently obvious. While it is true, I think this should be proven. Towards contradiction, suppose there is a limit point $p$ of $A$ such that $p \notin C$. Since $C$ contains all its limit points, $p$ cannot be a limit point of $C$, so there is an open set $U \ni p$ such that $U \cap [C - \{ p\}] = \emptyset$ Since $A \subset C$ it follows that $U \cap [A - \{ p\}] \subset U \cap [C - \{ p\}]$ which means $U \cap [A - \{ p\}] = \emptyset$, a contradiction. This is simply a preference of mine; you are definitely on the right track either way.
• Alternatively, if $p$ is the limit point of $A$ such that $p \not\in C$, then any deleted neighbourhood of $p$ will contain a point of $A$ which is, in turn, a point of $C$ and hence a limit point of $C$. Since $C$ contains all its limit points (by def. of a closed set), we arrive at a contradiction. – charlesh Oct 10 at 22:40
The closure of $A$ in $X$ is usually defined as the smallest closed subset containing $A$. This set exists because it is the intersection of all closed sets containing $A$ and your statement is trivial.