Let $X$ be a topological space with topology $T$ described by open sets, defining the interior of a subset $A$ of $X$ to be the set whose is the union of all open sets that are subsets of A, then we should have an equivalence between be open and a set be equal to it own interior.
In order to prove this is necessary to show that if the set $A$ is contained in its own interior then it is open. I did a proof using the already know statement that $A$ contains it interior and then since the interior is union of open sets in a topologicla space it should be open and since $A$ is equal to an open set it is also an open set.
My problem with this is because there is no intuition here, and it as a mechanical proof, i am seaking a meaning on this, what does it mean that a set $A$ is subset of its interior? Why should A be open??