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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.

Syd Amerikaner
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  • How did you get $$\partial\left(\bigcap_{i\in I}A_i\right) \cap \left(\bigcap_{i\in I}A_i\right) \subseteq \left(\bigcap_{i\in I}A_i\right)\cap(\partial A_i\cap A_i)$$when $A_i\in \mathcal{T}$ and $I$ is finite? Can you explain in detail? –  Jun 28 '18 at 05:39
  • There was a typo, thanks for pointing it out. Since $\bigcap A_i\subseteq A_i$ for each $i$ it follows from the third property that $\partial\bigcap A_i \cap \bigcap A_i \subseteq \bigcap A_i \cap \partial A_i\cup A_i = \bigcap A_i \cap \partial A_i$ (since $A_i\cup A_i = A_i$) and by assumption $A_i\cap \partial A_i = \emptyset$, so the intersection includes the emptyset which means that the intersection yields the emptyset. – Syd Amerikaner Jun 28 '18 at 15:12
  • @Paul Frost This question was my starting point. I tried to prove it as an exercise but I got stuck at the arbitrary union step. The linked question is not helping in that regard – Syd Amerikaner Jun 28 '18 at 22:02
  • In a comment to this question Tony Piccolo gives a link to https://arxiv.org/pdf/math/0605259v1.pdf which contains the complete story. It shows that boundary operators and closure operators are equivalent. If you want to prove directly that arbitrary unions and finite intersections of open sets are open, you have to incorporate the proof that closure operators are equivalent to topologies. – Paul Frost Jun 28 '18 at 22:20

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