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  
 A: You can do it directly, after showing that $x\in\operatorname{Cl}(A)$ if and only if, for every neighborhood $U$ of $x$, $U\cap A\ne\emptyset$.
Suppose $x\in\operatorname{Cl}(A)$. If there exists a neighborhood $U$ of $x$ such that $U\subseteq A$, then $x\in\operatorname{Int}(A)$. Otherwise, no neighborhood of $x$ is contained in $A$, so every neighborhood of $x$ interesects $X\setminus A$, which means $x\in\operatorname{Bd}(A)$.
Thus we have proved that $\operatorname{Cl}(A)\subseteq\operatorname{Int}(A)\cup\operatorname{Bd}(A)$.
Conversely, it is clear that $\operatorname{Int}(A)\subseteq A\subseteq\operatorname{Cl}(A)$. Also $\operatorname{Bd}(A)\subseteq\operatorname{Cl}(A)$ is clear, because every neighborhood of a point $x\in\operatorname{Bd}(A)$ intersects $A$ (and $X\setminus A$).
A: 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. $\square$
A: Here's how I would go about it. 
Let $x \in \text{Bd}(A) \cup \text{Int}(A)$. If $x \in \text{Bd}(A)$ then $x$ is a limit point because all boundary points are limit points. Then $x \in \overline{A}$. Else if $x \in \text{Int}(A)$ it is also clear that $\text{Int}(A) \subseteq \overline{A}$ so $x \in \overline{A}$. Hence $\text{Bd}(A) \cup \text{Int}(A) \subseteq \overline{A}$.
Now let $x \in \overline{A}$. We know that either $x \in \text{Int}(A)$ or  $x \notin \text{Int}(A)$. We have interest only in the latter case. Suppose toward contradiction that $x$ is not a boundary point of $A$. Then there exists an open set $U$ containing $x$ such that either $U\cap A = \emptyset$ or $U\cap(X\setminus A) = \emptyset$. We can dismiss the first case, because that would imply $x$ is not a limit point of $A$ and hence $x\notin A$, so it must be that $U\cap(X\setminus A) = \emptyset$. Clearly this means $x \notin \emptyset =U\cap(X\setminus A)$ so $x$ is in the complement, $U^c \cup A$ (you can prove this is the complement with Demorgan's law and a little set algebra). We know $x \notin U^c$ since $x \in U$. Then $x \in U^c \cup A$ means $x \in A$. Further, the complement of the emptyset is $X$, so we know $X = U^c \cup A$. Since $X$ is open and $U^c$ is closed, $X\setminus U^c = A$ is open, meaning $A = \text{Int}(A)$. Hence, $x \in \text{Int}(A)$, a contradiction. We conclude that $\overline{A} \subseteq \text{Bd}(A) \cup \text{Int}(A)$.
