The short answer is Yes.

Sorry for the delay in giving a complete answer to this question.

First, the interior of a set and closure of a set are defined in a metric topological space.  Let X be a metric topological space and let A be a subset of X.   In addition, let d(x,y) be the standard distance between any x,y in X.   Then define

\begin{align*}

D(x,A) &= inf\{ d(x,a): a\in A \} \mbox{(Hausdorff distance)},\\

cl A &= \{x\in X: D(x,A) = 0 \} \mbox{(closure of $A$ )},\\

bdy A &= cl A \setminus A,\\

int A &= cl A \setminus bdy A \mbox{(interior of A)}.

\end{align*}

Let $A^c$ denote the complement of $A$.   Then, we can also view the boundary points of $A$ is the set of points in $X$ that near $A$ and also near the complement of $A$ i.e., near $A^c$.     A subset A in X is an open set if and only if, for each point x in A, there is an open ball $B(x,\varepsilon)$ that is a proper subset of $A$.   The open ball $B(x,\varepsilon)$ is defined by

\[

B(x,\varepsilon) = \{ y\in A: d(x,y) < \varepsilon \} \mbox{(open ball)}.

\]

This leads to a second definition of the interior of a set A in X.    The interior of a subset A in X (int A) is the union of all open sets contained in A. 

Now contrast the structures cl A, int A, and bdy A the structures $A_* $(lower approximation of A) and $A^*$ (upper approximation of A$ in rough set theory introduced by Zdzislaw Pawlak during the early 1980s.   In rough set  theory, it not necessary to define a rough set in a metric topological space.    Instead, let X be some space, A a subset in X, x in X, B a set of attributes of each x in X, a  in B, a(x) an attribute value, and define the indiscernibility relation $\sim_B$  is defined by

\begin{align*}

\sim_B &= \{(x,y) \in X\times X: x \sim_B y \Leftrightarrow a(x) = a(y)\forall a\in B\}.\\

[x]_B &= \[ y\in X: x \sim_B y\} \mbox{(equivalence class)}.\\

X_{/_{\sim_B}} &= \{{[x]_B: x\in X\} \mbox{(partition of X)}.

\end{align*}

Let $A$ be a subset in the partition of $X$ induced by the indiscernibility relation.  Then define the lower approximation of $A$:

A_* = \mathop{\bigcup}\limits_{[x]_B \in X_{/_{\sim_B}} [x]_B \mbox{(lower approximation of $A$)}.

Notice that we do not need to assume that X is a metric topological space to define the lower approximation of a subset A in $$.  If we assume X is a metric topological space.   However, if we let $X$ be a metric topological space, then clearly

\[

A_* \subseteq int A,

\]

which is not the same thing as asserting that $A_* = int A$.   In fact, it is often the case that $A_* \subset int A$.    Let $A^*$ denote the upper approximation of $A$ (see my definition of the upper approximation of $A$ given recently in another RG thread.   Observe that

\[

cl A \subseteq A^*  \mbox{cl $A$ is a subset of the $A^*$}.

\]

It is often the case that $cl A \subset A^*$.  Hence, in general, the lower approximation of a subset A is not the same as the int A and the upper approximation is not the same thing as cl A.

If P is the partition of the universe U, then lower approximation operation is equivalent to interior operation  in the topology generated by the partition, the upper approximation is equivalent to closure operation.

If E is a covering of the universe U, then in general lower and upper operations are not topological.