# Let $(X, \mathfrak T)$ be a topological space and suppose that $A$ is a subset of $X$. Then 1. $Cl(A) = Cl(Int(A))$ 2. $Int(A) = Int(Cl(A))$

Let $(X, \mathfrak T)$ be a topological space and suppose that $A$ is a subset of $X$. Then

1. $Cl(A) = Cl(Int(A))$
2. $Int(A) = Int(Cl(A))$

I believe both of these statements are false and I think I have two counterexamples and I just want to double check that I am correct. Both these counterexamples are in the usual topology.

1. Let $A= [0,1] \cup \{2\}$ then $Cl(A) = [0,1] \cup \{2\}$ and the $Int(A) = (0,1)$ therefore the $Cl(Int(A))= [0,1]$ which does not equal the Cl(A).

2. Let $A= (0,1) \cup (1,2)$ then the $Cl(A) = [0,2]$ and the $Int(A)= (0,1) \cup (1,2)$ therefore the $Int(Cl(A))= (0,2$) which does not equal the interior of A.

My definition of closure is:
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 $A \subseteq Cl(A)$

Am I correct? After typing this I am starting to doubt my calculations for the closure of both sets.

Your work looks correct to me! You could have gone with an even easier example for $(1)$ like letting $A = \{0\}$, but what you did is certainly fine. For your second problem I think you picked the perfect set to disprove the claim.
Your examples are correct. Something that I personally like to use for questions like these are the rational numbers $\mathbb{Q}$. Note that it has empty interior and its closure equals the whole real line. So it serves as a counter example for both $1.$ and $2.$