If A is a subset of a topological space, then $Bd(A) \subseteq Cl(A)$. Prove.
I know this statement is true. I am now trying to prove it. I am in a basic topology class and to do a lot of set proofs we start by letting an element be included in one side and show it is in the other.
I was given the hint to use this fact $Cl(A)= A \cup Bd(A)$. However I have to prove that statement first. For this statement I started with Let $x \in Cl(A)$ then by defintion of closure $x \in A$ since the closure is the smallest closed set containing $A$ and therefore $x \in A \cup Bd(A)$. Now let $x \in A \cup Bd(A)$. How do I show $x \in$ the $Cl(A)$ and then use that prove my first statement?