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. 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 a finite amount of open/closed sets are also open/closed simple because the first $N-1$ sets intersection or union is one open/closed set so we just use the statement above to conclude that $N$ set intersection or union of is open/closed. So these rules are true for an infinite amount of sets? NO!

For an infinite sequence of open sets union of these sets is always open but the intersection isn't. For an infinite sequence of closed sets intersection is always closed but the union isn't. Especially open set one dug a rabbit hole for me. After a couple of minutes of puzzling over this, there has to be a way that an infinite intersection results in to set that has a single number in it. A very simple conclusion was to use a sequence of open ranges $]-\frac{1}{n},\frac{1}{n}[$. Certainly, any set I point to in this sequence of sets would have zero in it. But then I started to doubt myself what if "at" infinite both sides would be zero which since it is an open interval would cause it to be a null set. The null set is open so that would cause the whole infinite intersection to open since the intersection with the null set is the null set. After a while of brain heating, I realized I should probably go look at how a sequence of infinite sets is 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 the infinite intersection of open sets is open.

From the proof wiki, for metric spaces proof used $\bigcap _{i\in \mathbb{N}}$ which avoids the infinite sign. But yeah, the proof is just interval $]-\frac{1}{n},\frac{1}{n}[$ which means zero always belongs to every set and the more sets you intersect closer you get to a set with only zero. 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 seems to need us to define $\bigcap _{i=0}^{\infty }\left(U_i\right)≔\bigcap _{i\in \mathbb{N}}\left(U_i\right)$. There are probably other methods to define it. For example, Metamath just defines mass intersection as elements for all sets from a set of sets that are in every set. There isn't even an index set. We note that monotone set sequences have a limit Also, sequences that are a constant set have the constant as limit (as proof union and the intersection of the same set is set itself). If the set sequence is monotone then the 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 the nonincreasing set, we would be just intersecting sets that we already intersected so there our intersection question holds. How about when we are taking union then? Well, it is just $\mathit{lim\; sup}$ definition again so it should reduce down the same way... at least there isn't meaning full difference. So, if an infinite limit of intersection or unions exists, it is probably equal to infinite intersection or union.n.