Similar to my earlier question I try to understand the characterization of topology by different approaches. There is the "classical" one with open sets, the one with closed sets, the neighborhood characterization and one with closure and interior operator, respectively. (In addition, there is also a possibility to define a topology with respect to filter convergence but I did not dig into that yet. Are there further approaches besides the ones mentioned?). Since $\mathrm{boundary}(A) = \mathrm{closure}(A)\setminus\mathrm{interior}(A)$ it should be possible to characterize a topology by a boundary operator only. My thoughts so far:
Theorem. Suppose that $\mathcal T\subseteq2^X$ is a family of sets and $\partial:2^X\rightarrow 2^X$. Then
- $\mathcal T$ is a topology
- $\partial A := \mathrm{closure}(A)\setminus\mathrm{interior}(A)$
iff
- $\partial$ satisfies
- $\partial\partial A\subseteq\partial A$,
- $\partial(A\cup B)\subseteq\partial A\cup\partial B$,
- $A\subseteq B$ implies $\partial A\subseteq \partial B\cup B$,
- $\partial A = \partial(X\setminus A)$ and
- $\mathcal T = \{U\subseteq\Omega : \partial U\cap U = \emptyset\}$.
I managed to show "$\Rightarrow$". I struggle with the reverse direction however. That $\emptyset,X\in\mathcal T$ is easy. But I could not manage to show the arbitrary unions and finite intersections. My attempt:
Let $A_i\in\mathcal T$ for $i\in I$, $I$ finite. Then $$\partial\left(\bigcap_{i\in I}A_i\right) \cap \left(\bigcap_{i\in I}A_i\right) \subseteq \left(\bigcap_{i\in I}A_i\right)\cap(\partial A_i\cup A_i)$$ for each $i\in I$ by the third property. But Since $A_i\in\mathcal T$ it follows that $\partial A_i\cap A_i =\emptyset$. So we have $\partial\left(\bigcap_{i\in I}A_i\right) \cap \left(\bigcap_{i\in I}A_i\right) =\emptyset$. Is this correct?
Let $A_i\in\mathcal T$ now for all $i\in I$ with arbitrary $I$. Then $\emptyset = A_i\cap\partial A_i$ for all $i$ by assumption. Thus, $$\begin{align*}\emptyset &= \bigcup_{i\in I}(A_i\cap\partial A_i)\\&=\bigcup_{i\in I}(A_i\cap\partial A_i)\cap\partial\left(\bigcup_{i\in I}A_i\right)\\&\supseteq\bigcup_{i\in I}A_i\cap\partial\left(\bigcup_{i\in I}A_i\right)\end{align*}$$ because intersection with emptyset is emptyset. So, $\bigcup_{i\in I}A_i\in\mathcal T$.
Note In the other questions for which my question was flagged as duplicate answer this questions by showing that there exists a one-to-one relation between boundary and closure. I however wanted to prove that boundary characterizes a topology straight.