Let $A$ be a subset of a topological space. Prove that $Cl(A) = Int(A) \cup Bd(A)$ Let $A$ be a subset of a topological space.  Prove that $Cl(A) = Int(A) \cup Bd(A)$
Here are my defintions: 
Closure:    Let $(X,\mathfrak T)$ be a topological space and let $ A \subseteq X$ . The closure of $A$ is $Cl(A) = \bigcap \{U \subseteq X: U$ is a closed set and $A \subseteq U\}$  Based on this I know $A \subseteq Cl(A)$ 
Interior:Let $(X, \mathfrak T)$ be a topological space and let $A \subset X$ is the set of all points $x \in X$ for which there exists an open set $U$ such that $x \in U \subseteq A$.  
My definition of boundary is: Let $(X,\mathfrak T)$ be a topological space and let $A \subseteq X$. A point $x \in X$ is in the boundary of A if every open set containing $x$ intersects both $A$ and $X−A$.
My proofs start by picking an element to be in each side then showing it must be in the other side.  I have tried to start that here.  
Let $x \in Cl(A)$ then 
Let $x \in Int(A) \cup Bd(A)$.  Then $x\in Int(A)$ or $x\in Bd(A)$ If $x \in Int(A)$ then  

The closure of $A$ is the smallest closed subset of $X$ containing $A$. The interior of $A$ is the largest open subset contained in $A$. So we have that $Int(A) \subseteq A \subset Cl(A)$. Let $x \in Bd(A)$. Suppose $ x \notin Cl(A)$. $X \setminus Cl(A)$ is open, and $x \in X \setminus Cl(A)$, so by your definition of boundary, $(X \setminus Cl(A)) \cap A \neq \emptyset$, which contradicts that $ A \subseteq Cl(A)$. So $Cl(A) \supseteq Int(A) \cup Bd(A)$.
For the converse we show that $D=Int(A) \cup Bd(A)$ is closed and contains $A$. Suppose $a \in A$ but $ a \notin D$. In particular $a \notin Bd(A)$ so there is an open set $U_{a} \subseteq A$ which does not intsersect $X \setminus A$. But then $U_{a}$ is an open set contained in $A$, so $U_{a} \subseteq Int(A)$ a contradiction. So $A \subseteq D$. For each $x \in X \setminus D$, there is an open set $U_{x} \subseteq X\setminus D \subseteq X \setminus A$ containing $x$, because otherwise $x \in Bd(A)$. Taking the union of these open subsets for each $x \in X\setminus D$ gives us that $X \setminus D$ is open, so $D$ is closed. $\square$
