Boundary and intersection identity $(\bigcap S_k)\cup(\bigcap \partial S_k)=(\bigcap S_k)\cup \partial(\bigcap S_k)$

Question

I would like to prove the following identity $$\tag{*} \left(\bigcap S_k\right)\cup\left(\bigcap \partial S_k\right)=\left(\bigcap S_k\right)\cup \partial\left(\bigcap S_k\right) $$ using only set theory and the following properties of the map $\partial:\mathcal{P}(X)\to \mathcal{P}(X)$, and $S,T\in\mathcal{P}(X)$:

1. $\partial \varnothing=\varnothing$
2. $\partial S=\partial(X-S)$,
3. $\partial\partial S\subseteq \partial S$,
4. $S\cap T\cap \partial(S\cap T)=S\cap T\cap (\partial S\cup \partial T)$.

Is (*) correct? Some hint?

I have spent some hour on it, and I'm geometrically (intuition) convinced that it is correct. But although I've solved more similar problems in the last days, this one seems harder to me.

Answer 1

It's not true.

With the usual topology on $\Bbb R$ let $ S_1=(0,2)$ and $S_2=(1,3).$ Then $$(S_1\cap S_2)\cup ((\partial S_1)\cap (\partial S_2))=$$ $$=(1,2)\cup (\{0,2\}\cap \{1,3\})=$$ $$=(1,2)\cup \emptyset=(1,2).$$ But $$(S_1\cap S_2)\cup \partial (S_1\cap S_2)=$$ $$=(1,2)\cup \partial (1,2)=(1,2)\cup \{1,2\}=[1,2].$$

Another counter-example is $S_1=\Bbb Q$ and $S_2=\Bbb R\setminus \Bbb Q.$ Then $S_1\cap S_2=\partial (S_1\cap S_2)=\emptyset,$ but $\partial S_1=\partial S_2=\Bbb R.$