Say we have a collection $\Omega$ of sets, and a statement $P(s)$ which can be either true or false depending on the input set $s$. Say we take two arbitrary sets $a$ and $b$ in $\Omega$ and that for all $a,b\in\Omega$, $P(a) \land P(b) \implies P(a \cap b)$.
Does this then imply that $\displaystyle \forall a\in\Omega\; P(a) \implies P\left(\bigcap_{s\in\Omega}s\right)$? Intuitively I can see how it would be true for collections with a finite or even countably infinite sizes, as you could just keep appending intersections with another element until you "got them all", like say if $\Omega = \{s_0,s_1,s_2,\cdots\}$ then you can do $$\displaystyle P(s_0) \Rightarrow P(s_0 \cap s_1) \Rightarrow P(s_0 \cap s_1 \cap s_2) \Rightarrow \cdots \Rightarrow P\left(\bigcap_{i\in\mathbb{N}}s_i\right)$$ but I don't necessarily know if this generalises to collections with uncountably infinite sizes since there really isn't that sort of inductive reasoning you can do to recursively append intersections when you're in the uncountably infinite. Does this intuitive implication hold true for collections of any size?
Best Answer
This is not true, even for countable intersections. For instance, let $$\Omega=\left\{\left(-\frac{1}{n},\frac{1}{n}\right)\mid n\in\mathbb{N}\right\}$$ be a collection of open intervals in $\mathbb{R}$ which contain $0$. For $s\subset\mathbb{R}$ let $P(s)$ be true if $s$ is open and false otherwise. Hence $\left\{0\right\}$ is the intersection of all elements of $\Omega$ and $P(\left\{0\right\})$ is false, but $P(s)$ is true for all $s\in\Omega$.