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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 to 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 indirectly 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 $\sup A$.

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:

  1. Why in the infinite de morgan's rule $\forall n\in\mathbb N$ doesn't apply to $\infty$
  2. Why in the nested interval property $\forall n\in\mathbb N$ applies to $\infty$
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  • $\begingroup$ Maybe a counterexample could help to understand: the intervals $U_n = (-\frac1n,\frac1n) \subset \mathbb R$ are all open. Moreover, $U_n \cap U_m = U_{\min(n,m)}$ is again open. So every intersection of a finite number of the $U_n$ is again open. But if you take the intersection of all the $U_n$, you end up with $\{0\}$, which is not open. $\endgroup$ Commented Dec 27, 2020 at 0:14
  • $\begingroup$ The nested interval property does not use induction. $\endgroup$
    – Mike
    Commented Dec 27, 2020 at 0:15
  • $\begingroup$ You say you're quoting, but there is no indication of which parts of what you write are quoted. $\endgroup$ Commented Dec 27, 2020 at 3:39
  • $\begingroup$ I wouldn't use induction in the finite case; rather I would prove it the same as as in the infinite case. What I would do would not depend on whether the index set is finite or infinite. $\endgroup$ Commented Dec 27, 2020 at 3:41

2 Answers 2

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$\forall n\in\Bbb N$ 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.

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  • $\begingroup$ LOL about George $\endgroup$ Commented Dec 27, 2020 at 1:45
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    $\begingroup$ @J.W.Tanner: We all got a kick out of that one. It’s a topological space inspired by R.H. Bing’s well-known (in some circles) Example G, and George also happens to be my friend’s middle name. $\endgroup$ Commented Dec 27, 2020 at 1:47
  • $\begingroup$ In particular, when I see pishy things like $\cap_{i=1}^{\infty}$, should I first apply induction then check whether infinite case holds? $\endgroup$
    – Andes Lam
    Commented Dec 27, 2020 at 8:23
  • $\begingroup$ Or should I put it in this way, if $\infty$ is not necessarily referring to $P(\infty)$ and may refer to $\forall n\in N$, why in first place do people not differentiate the 2 clearly? Though I know natural number set is infinite either. But it is confusing, isn't it? $\endgroup$
    – Andes Lam
    Commented Dec 27, 2020 at 8:49
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    $\begingroup$ @AndesLam: I’m afraid that I honestly don’t understand what it is that you find confusing. The $\infty$ in $\bigcap_{i=1}^\infty A_i$ does not mean that there is a set $A_\infty$. You could just as well write $\bigcap_{i\ge 1}A_i$ or $\bigcap\{A_i:i\in\Bbb Z^+\}$. Similarly, writing $\sum_{n=1}^\infty a_n$ instead of $\sum_{n\ge 1}a_n$ for an infinite series does not mean that there is an $a_\infty$. $\endgroup$ Commented Dec 27, 2020 at 19:32
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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 $n \in \mathbb N$.

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