# Let $(X, \mathfrak T)$ be topological space and suppose that A and B are subsets of X such that $A \subsetneq B$. Then $Int(A) \subsetneq Int(B)$.

Let $(X, \mathfrak T)$ be topological space and suppose that A and B are subsets of X such that $A \subsetneq B$. Then $Int(A) \subsetneq Int(B)$. ( $\subsetneq$ means "is a proper subset")

My definition of proper subset is "If $A$ and $B$ are subsets with $A \subseteq B$ and $A \neq B$ then $A$ is called a proper subset of $B$.

I know that it this is a true statement if you just have a subset and not a proper subset. I think I have a counterexample to show that this is false.

Let $X$ be $\mathbb R$ in the usual topology. Let $A = [0,1) \cup (1,2)$ and $B = (0, 1) \cup (1,2)$ Then the interior of $A =(0,1) \cup (1,2)$ and the interior of $B = (0,1) \cup (1,2)$

Am I correct?

You are correct, and generic couterexamples can be constructed as follows. Let $B$ be any set not equal to its interior $A$. Then $A \subsetneq B$, but $Int(A) = Int(Int(B)) = Int(B)$.