Rabbit hole of infinite intersection of open sets.

While reading Real Analysis book concept of closed and open sets of real numbers comes up. Closed real set is defined as set which every accumulation point of the set is in the set. More intuitive every point which you can go as close you like and find member of the set has to be in the set. Open real set is intuitive defined that every point in the set is surrounded by the other points of the set. They are interesting specially when you consider that they aren't each others opposited (for example $\mathbb{R}$ is both open and closed) and how both of them can be generalized to other number sets/spaces (metric space, topology).

First theorems proofed from these defenitions are that intersection and union of open/closed set is also open/closed. $$\begin{array}{c}\mathit{open}\left(A\right)\wedge \mathit{open}\left(B\right)\Rightarrow \mathit{open}\left(A\cup B\right)\wedge \mathit{open}\left(A\cap B\right)\\ \mathit{closed}\left(A\right)\wedge \mathit{closed}\left(B\right)\Rightarrow \mathit{closed}\left(A\cup B\right)\wedge \mathit{closed}\left(A\cap B\right)\end{array}$$

It is easy to see (by induction) that intersections or unions of finete amount of open/closed sets are also open/closed simple because first $N-1$ sets intersection or union is one open/closed set so we just use statement above to conclude that $N$ set intersection or union of is open/closed. So this rules are true for infinite amount of sets? NO!

For infinite sequence of open sets union of these sets is always open but intersection isn't. For infinite sequence of closed sets intersection is always closed but union isn't. These specially open set one dug rabbit hole for me. After couple minutes puzzling over this there is has to be away that infinte intersection results to set that has single number in it. Very simple conclusion was to use sequence of open ranges $\left]-\frac{1}{n}\frac{1}{n}\right[$ certainly any set I point to in this sequence of sets would have zero in it. But then I started to doubt my self what if "at" infinite both sides would be zero which since it is open interval would cause it to be null set. Null set is open so that would cause whole infinite intersection to open since intersection with null set is null set. After while of brain heating I realized I should probably go look how sequence of infinite sets are handled anyway. I never before had to deal with it. So I started to google about $\underset{N\to \mathrm{\infty }}{\mathit{lim}}\bigcap _{i=1}^N\left(U_i\right)$ and disproof of infinite intersection of open sets is open.

From proof wiki for metric spaces proof used $\bigcap _{i\in \mathbb{N}}$ which avoids the infinite sign. But yeah proof is just interval $\left]-\frac{1}{n}\frac{1}{n}\right[$ which means zero always belong to every set and more sets you intersect closer you get to set with zero so no other number can belong to set. However is $\underset{N\to \mathrm{\infty }}{\mathit{lim}}\bigcap _{i=0}^N\left(U_i\right)=\bigcap _{i\in \mathbb{N}}\left(U_i\right)$ true?

By definition given in Billingsley's Probability and Measure anniversary edition (page 55): $$\begin{array}{c}\underset{N\to \mathrm{\infty }}{\mathit{lim\; inf}}{U}_N=\underset{N\to \mathrm{\infty }}{\mathit{lim\; sup}}{U}_N\Rightarrow \underset{N\to \mathrm{\infty }}{\mathit{lim}}{U}_N=\underset{N\to \mathrm{\infty }}{\mathit{lim\; inf}}{U}_N\\ \underset{N\to \mathrm{\infty }}{\mathit{lim\; inf}}{U}_N=\bigcup _{i=0}^{\mathrm{\infty }}\bigcap _{k=N}^{\mathrm{\infty }}{U}_k\\ \underset{N\to \mathrm{\infty }}{\mathit{lim\; sup}}{U}_N=\bigcap _{i=0}^{\mathrm{\infty }}\bigcup _{k=N}^{\mathrm{\infty }}{U}_k\end{array}$$

This definition seem to need us defining $\bigcap _{i=0}^{\infty }\left(U_i\right)≔\bigcap _{i\in \mathbb{N}}\left(U_i\right)$. There is probably other methods to defining for example metamath just defines mass intersection as elements forall sets from set of sets that are in every set. There isn't even index set. We note that monotone set sequences have limit. Also sequences that are constant set have the constant as limit (as proof union and intersection of same set is set it self). If set sequence is monotone then limit is ether $\bigcap _{i=0}^{\mathrm{\infty }}$ or $\bigcup _{i=0}^{\mathrm{\infty }}$ or the sequence depending on is sequence nonincreasing or nondecreasing. So for nonincreasing set we would be just intersecting set that we already intersected ones so there our intersection question just holds. How about when we are taking union then? Well it is just $\mathit{lim\; sup}$ definition again so it should reduce down same way... at least there isn't meaning full difference so if infinite limit of intersection or unions exists it is probably equal to infinite intersection or union.