$\cap_{n=1}^{\infty}A_n$ and infinity

Question

One question:

1. If the definition of $\cap_{n=1}^{\infty}A_n=\{x\in A_i\forall n\in N\}$ and it is nonempty, then does it mean that its elements belong to the infinite intersection of $A_n$ or any finite intersection of $A_n$ for all natural numbers?

To further elaborate, I would like to show how I feel towards this confusing notation $\cap_{n=1}^{\infty}A_n$.

Understanding Analysis Steven Abbott

Example 1.2.2 in which it defines $A_i = \{x\in N: x\geq i\}$. By induction, it is nonempty for each finite intersections. But a proof by contradiction can show that when it goes to infinite case, which uses the notation $\cap_{n=1}^{\infty}A_i$, it is a null set. In other words, in this example, this notation is used for infinite intersection.

Theorem 1.4.1 in which it proves the nested interval property. $I_n = \{x\in R: a_n\leq x\leq b_n\}$. Here, it doesn't specify whether this is infinite intersection or not. Instead, it said, $\exists x\forall n\in N x\in I_n$. Hence, that $x\in\cap_{n=1}^{\infty}A_n$. In other words, in this example, this notation is used for every finite natural number

Theorem 1.5.8 says If $A_n$ is a countable set for each $n\in N$, then $\cup_{n=1}^{\infty}A_n$ is countable. In other words, in this example, this notation is used for infinite intersection.

I am confused by this notation in a sense that the notation includes the infinity sign but its definition means every natural number. Hence, whenever I see it, I just don't know which one to apply.

Say if I go for the direction in which it is applicable $\forall n\in N$, then induction should work because induction is doing exactly the same thing! Though, this post suggests otherwise by saying the notation is about infinity.

Fine, I switch direction in which it is about infinite intersection. But then in some cases, for example, the one I listed above, somehow if something is applicable for all natural numbers, it is fine to be part of this notation.

So in short, I feel that this notation has 2 conflicting meanings

1. $\forall n\in N$
2. Infinity

I have done researches and asked questions before but I still don't understand. So I guess I got something every wrong and confused in some definitions.

Answer 1

$\bigcap_{n=1}^\infty A_n$ is a set. What set? The set of all things that belong to every one of the sets $A_n$ for $n\in\Bbb Z^+$. Let $\mathscr{A}=\{A_n:n\in\Bbb Z^+\}$; then $\bigcap\mathscr{A}$ means exactly the same thing. $\bigcap_{n=1}^\infty A_n$ is simply a customary notation that means neither more nor less than $\bigcap_{n\ge 1}A_n$, $\bigcap\mathscr{A}$, and $\bigcap\{A_n:n\in\Bbb Z^+\}$. There is no $A_\infty$: the $\infty$ is just a signal that the index $n$ is to assume all positive integer values.

Suppose that for each positive real number $x$ I let $I_x$ be the open interval $(-x,x)$. Then $\bigcap_{x\in\Bbb R^+}I_x$ is the set of all real numbers that belong to every one of these open intervals. If $\mathscr{I}=\{I_x:x\in\Bbb R^+\}$, then

$$\bigcap\mathscr{I}=\bigcap_{x\in\Bbb R^+}I_x=\bigcap_{x\in\Bbb R^+}(-x,x)=\{0\}\,.$$

How do I know? If $y\in\Bbb R\setminus\{0\}$, then $y\notin(-|y|,|y|)=I_{|y|}$, so there is at least one member of $\mathscr{I}$ that does not contain $y$, and therefore by definition $y$ is not in the intersection of the sets in the family $\mathscr{I}$. On the other hand, $0\in(-x,x)=I_x$ for every $x\in\Bbb R^+$, so $0$ is in the intersection $\bigcap\mathscr{I}$.

In neither case have we used induction anywhere. In the case of the sets $A_n$ we might be able to use induction on $n$ to show that each of the sets $A_n$ has some property $P$, but we could not extend that induction to show that $\bigcap\mathscr{A}$ has $P$. We might somehow be able to use the fact that each $A_n$ has property $P$ to show that $\bigcap\mathscr{A}$ also has $P$, but that would require a separate argument; it would not be part of the induction. The induction argument in that case would prove that

$$\forall n\in\Bbb Z^+(A_n\text{ has property }P)\,;$$

the separate argument would then show, using that result and other facts, that the single set $\bigcap\mathscr{A}$ has property $P$. You could call this set $A_\infty$ if you wished to do so, but that would just be a label; you could equally well call it $A$, or $X$, or even $A_{-1}$, though offhand I can’t imagine why you’d want to use that last label.

In the case of the sets $I_x$ there is no possibility of using induction to show that each $I_x$ has some property: these sets cannot be listed as $I_1,I_2,I_3$, and so on, because there are uncountably many of them. We can still prove things about the set $\bigcap\mathscr{I}$, however. And we could give it any convenient label. $\bigcap\mathscr{I}$ is informative but perhaps a little inconvenient; I might choose to give it the handier label $I$.

In the case of $\mathscr{A}$ there happens to be a customary notation that uses the symbol $\infty$, but that is simply a consequence of the fact that the sets $A_n$ are indexed by integers. We’re doing exactly the same sort of thing in the example with $\mathscr{I}$, but in that case there is no possibility of using a limit of $\infty$ on the intersection, because there is no way to index the uncountably many sets $I_x$ by integers.