Prove that: If $X$ is a topological space and $A$ and $B$ are two subsets of $X$ then,$Cl(A) \cup Cl(B) = Cl(A \cup B) $ Prove that: If $X$ is a topological space and $A$ and $B$ are two subsets of $X$ then,$Cl(A) \cup Cl(B) = Cl(A \cup B) $ where $Cl(H)$ means the closure f the subsets $H $of $X$.
I was able to prove that $Cl(A) \cup Cl(B) \subset Cl(A \cup B) $. I don't know how to prove the other way. The problem is:
If $x\in Cl(A \cup B) $ then every neighborhood $U$ of $x$ intersect $A \cup B$ so every neighborhood $U$ of $x$ intersects either $A$ or $B$ or both. But it may happen that some neighborhoods intersect $A$ but not $B$ and the rest of them intersect $B$ but not $A$, in which case we get $Cl(A \cup B) \not\subset Cl(A) \cup Cl(B)$. How to proceed?
This problem is from Munkres topology. 

A very useful lemma here is the fact that $Cl(A)$ is the minimal closed set containing $A$ (Verify for yourself!).
Assume for the sake of contradiction that the two sets are not equal. We find that, just as Michael said, $Cl(A) \cup Cl(B)$ is closed. In addition, we must have $A \cup B \subset Cl(A) \cup Cl(B)$. Combining this with $Cl(A) \cup Cl(B) \subset Cl(A \cup B)$ gives us the desired result since otherwise we arrive at a contradiction due to the lemma. 
