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
Best Answer
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 $X\setminus A$).