Let $I_A: (X, \mathcal{T}) \to \mathbb{R}$ be the indicatorfunction of $A$. I.e., it takes the value 1 in $A \subseteq X$ and 0 elsewhere. Show that $x \notin \delta A\implies I_A$ is continuous at $x$.
Attempt:
Wlog, $x \notin cl(A)$. Then $I_A(x) = 0$ and we can pick a nbh $V$ of x s.t. $A \cap V$ is empty.
Let $W$ be a neighborhood of 0. It suffices to show that $I_A^{-1}(W)$ is a neighborhood of x. But this is clear, since $V \subseteq I_A^{-1}(W)$.
Indeed, if $v \in V$, then $v \notin A$ and hence $I_A(v) = 0 \in W$.
Is this correct?
You have to consider 2 regions.
Consider first the open region $B=X\backslash(A\cup∂A)$. There you have $I_A|_B=0$ everywhere. OTOH consider the open region $C=A\backslash∂A$. There you have $I_A|_C=1$ everywhere. That is, in both cases you just have to show that the constant function (with respective value) with an open supporting region will be continuous, which is immediate.
--- rk