Difference between $\forall n\in\mathbb N$ and $\bigcap_{i = 1}^{\infty}$ Really confused about the difference between $\forall n\in\mathbb N$ and $\bigcap_{i=1}^\infty$.
In Understanding Analysis, I quote from Exercise 1.2.13. that
It is tempting to appeal to induction to conclude $(\bigcup_{i = 1}^\infty A_i)^c = \bigcap_{i=1}^\infty A_i^c$.
but induction does not apply here. Induction is used to prove that a particular statement holds for every value of $n\in\mathbb N$, but this does not imply the validity of the infinite case.
Have done some research on that for a while and understood that eventually the fact that I can point out a $n\in\mathbb N$ means that $n$ is finite. Hence, it cannot apply on the infinite case.
Yes, I understand the rationale. But if $\forall n \in\mathbb N$ doesn't work, then what works on proving infinite case?
Just as I feel comfy about the difference. The confusion is again brought up by the book and I quote in the following, in hopes of making it as short as possible:
The nested interval property assumes that each $I_n$ contains $I_{n+1}$. They are a nested sequence of closed intervals defined as such. $I_n = [a_n, b_n] = \{x\in\mathbb R : a_n\leq x \leq b_n\}$.
The proof focuses on finding a single real number x that belongs to all $I_n$ and it argues it is supA.
In the proof, it said $x\in I_n$, for every choice of $n\in\mathbb N$. Hence, $x\in \bigcap_{n=1}^\infty I_n$ and the intersection is not empty.
Let me know if the missed out details are needed. However, my point is just that:

*

*Why in the infinite de morgan's rule $\forall n\in\mathbb N$ doesn't apply to $\infty$

*Why in the nested interval property $\forall n\in\mathbb N$ applies to $\infty$
 A: De Morgan's Rule does happen to work for infinite sets. But this cannot be proven by inducting on the finite version of De Morgan's Rule, since induction is a tool for proving that a statement is true for an arbitrarily large value of $n$ (but $n$ is still finite).
As for the intersection of a countably infinite number of sets, this follows from the definition. We say that $x \in \bigcap_{n=1}^\infty I_n$ iff $x \in I_n$ for all $n \in \mathbb N$.
A: $\forall n\in\Bbb N$ never applies to $\infty$, because $\infty$ is not an element of $\Bbb N$. In the nested interval theorem there is no $I_\infty$. What we know is that $x\in I_n$ for each $n\in\Bbb N$, and therefore by definition $n$ is in the intersection of the sets $I_n$. You could call this intersection $I_\infty$ if you wanted to do so, but that would be an arbitrary choice altogether independent of the induction argument involving the sets $I_n$; you could just as well call it George. (Many years ago a friend of mine did in fact publish a paper about a mathematical object that he named George.)
As for De Morgan’s law, one proves it for arbitrary families of sets simply by showing that each side of the proposed identity is a subset of the other. This is done for arbitrary indexed families of sets here and in this answer (and probably other places at MSE as well). The proof does not depend on the theorem for finite families of sets and does not involve any kind of induction.
