What is the set $\{M \subset \mathbb{R}^n| M^{af} \subsetneq M^f\}$?

Question

I am reading an elementary general topology book now.

I found this formula in the book:

$M^{af} \subset M^f$ for any $M \subset \mathbb{R}^n$

$M^f$ is the set of boundary points of $M$.

$M^a$ is the closure of $M$.

I wonder $M^{af} = M^f$ if $M$ is an "ordinary" set.(I don't know what set is an "ordinary" set.)

What is the set $\{M \subset \mathbb{R}^n| M^{af} \subsetneq M^f\}$ ?

$\mathbb{Q}^n \in \{M \subset \mathbb{R}^n| M^{af} \subsetneq M^f\}$

Example($S = \{1,2,3,4\}$ instead of $S = \mathbb{R}^n$):

$S = \{1,2,3,4\}$ $\mathfrak{O} = \{\{1\}, \{2\}, \{3\}, \{4\}, \{1, 2\}, \{1, 3\}, \{1, 4\}, \{2, 3\}, \{2, 4\}, \{3, 4\}, \{1, 2, 3\}, \{1, 2, 4\}, \{1, 3, 4\}, \{2, 3, 4\}, \{\}, \{1, 2, 3, 4\}\}$

$M = \{2\}, M^a = \{2, 3, 4\}, M^{af} = \{4\}, M^f = \{2, 3, 4\}$

$M = \{3\}, M^a = \{2, 3, 4\}, M^{af} = \{4\}, M^f = \{2, 3, 4\}$

$M = \{2, 4\}, M^a = \{2, 3, 4\}, M^{af} = \{4\}, M^f = \{2, 3, 4\}$

$M = \{3, 4\}, M^a = \{2, 3, 4\}, M^{af} = \{4\}, M^f = \{2, 3, 4\}$

Answer 1

bd for boundary, cl for closure, int for interior. bd M = cl M - int M
bd cl M = cl cl M - int cl M = cl M - int cl M
Since int M subset int cl M, it follows that
bd cl M subset bd M.

$\partial M = \overline M - M^o$
$\partial \overline M = \overline M - \overline M^o$
Since $M^o \subseteq \overline M^o,$
the desired conclusion follows.