Boundary of set equal the same set Let $(X, \tau)$ be a topological space and $A$ is a subset of $X$.
When we have $\delta(A)=A$, where $\delta(A)$ is the boundary of $A$.
I suspect that $A$ is the empty set or a meagre set.
Any help is welcome
 A: The boundary $\delta(A)$ is defined to be $\overline{A}\setminus A^\circ$, the closure of $A$ minus the interior of $A$. In particular, $\delta(A)$ is closed, so if $\delta(A) = A$, then $A$ must be closed. But if $A$ is closed, then $\overline{A} = A$, so $\delta(A) = A\setminus A^\circ$. If $\delta(A) = A$, this implies that $A^\circ = \emptyset$. 
So we can conclude that $\delta(A) = A$ if and only if $A$ is closed with empty interior, i.e. $A$ is a closed nowhere dense set. 
A meager set is by definition a countable union of nowhere dense sets. So if $\delta(A) = A$, we've shown that $A$ is meager, as you suspected, simply because it is nowhere dense (explicitly listing the possibility $A = \emptyset$ is redundant, since the empty set is meager). 
But the converse is not true: Not every meager set is equal to its boundary. For example, $\mathbb{Q}$ is a meager subset of $\mathbb{R}$, but $\delta(\mathbb{Q}) = \mathbb{R}$. Of course, this is just because $\mathbb{Q}$ is not closed in $\mathbb{R}$. 
In a space $(X,\tau)$ where the Baire category theorem holds, every closed meager set is equal to its boundary. (Proof: Suppose $C$ is closed and meager. By the Baire category theorem, no nonempty open set is meager, so $C$ does not contain any nonempty open set. In particular $C^\circ = \emptyset$, so $\delta(C) = C$.) But if the Baire category theorem fails, it's possible that $X$ itself is meager (e.g. $\mathbb{Q}$ with its usual topology is meager). Then $X$ is a closed meager set with empty boundary. 
