I want to prove that $cl(cl(A)) = cl(A)$ for a subset $A$ of topological space $X$. Given a subset $A$ of a topological space $X$, I have at hand the following two equivalent definitions about the closure of $A$ (denoted $cl(A)$):
Definition 1: The closure of $A$ is the intersection of all closed sets containing $A$.
Definition 2: $x \in cl(A)$ if and only if every open set $U$ containing $x$ intersects $A$.
Based on the first definition, I can prove the statement by first showing that $A = cl(A) \Leftrightarrow A \textrm{ is closed}$. The statement $cl(cl(A)) = cl(A)$ follows from the fact that $cl(A)$ is closed.
However, I failed to prove the statement using the second definition directly, especially the $cl(cl(A)) \subset cl(A)$ part.
My trial of $cl(cl(A)) \subset cl(A)$: Take any $x \in cl(cl(A))$, we have (by definition 2) every open set $U$ containing $x$ intersects $cl(A)$. Suppose that $x \notin cl(A)$ by contrapositive, then there exists an open set $V$ containing $x$ that does not intersect $A$. Therefore, we get an open set $S$ containing $x$ such that (1) $S \cap cl(A) \neq \emptyset$ and (2) $S \cap A = \emptyset$.
So, how should I obtain the contradiction?
Best Answer
The basic fact that you might be missing is the following: If $\newcommand{\cl}{\mathrm{cl}}U \subseteq X$ is open and $A \subseteq X$ is arbitrary, then $U \cap A = \varnothing$ implies that $U \cap \cl ( A ) = \varnothing$. (This is due to the fact that $U$ is an open neighbourhood of each of its elements which is disjoint from $A$, and so no element of $U$ can be an element of $\cl ( A )$. Also worth noting is that $X \setminus U$ is a closed set and $A \subseteq X \setminus U$, and so $\cl ( A ) \subseteq X \setminus U$ by definition of $\cl ( A )$.)
You can also do this pretty directly. Suppose that $x \in \cl ( \cl ( A ) )$, and let $U$ be any open neighbourhood of $x$. Then $U \cap \cl ( A ) \neq \varnothing$, so pick some $y \in U \cap \cl ( A )$. Note that $U$ is an open neighbourhood of $y$ and $y \in \cl ( A )$, and so $U \cap A \neq \varnothing$.