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I understand the multiple meanings of infinity, per example, the difference between $\aleph_0$ and the $\infty$ in calculus limits as explained here: The Aleph numbers and infinity in calculus.

However, in the analysis book I am studying, in the chapter about the real number field, there is the standard question about the intersection of nested intervals (link1), which (unlike in link1) is presented as the intersection of infinitely many intervals: ${\cap}_1^{\infty}$ with $A_i$ being the closed intervals.

The chapter before this one was about set theory, but nowhere was this use of infinity really defined, nor the symbol itself, in an otherwise very rigorous and meticulous treatment of all other concepts. So my question is: what kind of infinity are we talking about here formally?

Community
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bluemoon
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  • Note: if you are taking the intersection or union of sets $X_\alpha$ with indices ranging over some arbitrary set (say $A$), you will typically just see something like $\alpha \in A$ under the intersection/union symbol... the cardinality of the index set could be anything! – Forever Mozart Dec 20 '12 at 19:00
  • So I suppose it was written like that because it is an elementary text where you are not supposed to know about cardinality on any formal level. – bluemoon Dec 20 '12 at 19:06
  • My main issue with it was, that I first tried to prove it by induction, which is very simple and possible, but it turns out that a proof by induction proves something else, namely that this holds up to any natural $n$, but it does not prove what is requested, the (for me) subtle difference between the two points is what made me wonder about what the symbol here really means, and why a proof by induction does not prove this. Could anyone to comment about the difference between the two and what exactly a proof by induction would miss? – bluemoon Dec 20 '12 at 19:07

4 Answers4

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Take a family of sets $\left\{A_i\right\}_{i\in I}$. We define $$\bigcap_{i\in I}A_i=\left\{x:\forall i\in I\ x\in A_i\right\}$$ If you take the index set $I$ to be $\mathbb{N}$ we have $$\bigcap_{i\in \mathbb{N}}A_i$$ which is commonly denoted by $$\bigcap_{i=0}^{\infty}A_i\text{ or }\bigcap_{i=1}^{\infty}A_i$$ depending on whether your definition of $\mathbb{N}$ contains $0$ or not. Therefore, your intersection is: $$\bigcap_{i=1}^{\infty}A_i=\left\{x\in \mathbb{R}:\forall i\in \mathbb{N}\ x\in A_i\right\}$$

Nameless
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I would try not to think of the $\infty$ in $\cap_1^\infty$ as anything more than a symbol. The notation $\cap_1^\infty A_i$ (usually actually written $\cap_{i=1}^\infty A_i$) means the same thing as $\cap_{i\in \mathbb N} A_i$, that is the intersection of a collection of sets indexed by $\mathbb N$. If you really want to interpret $\infty$ as some infinity, since you are indexing over $\mathbb N$ (in a specific order) it should be interpreted as the ordinaltype of $\mathbb N$, which is $\omega_0$.

Alex Becker
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Usually the notation $\bigcap_{i=1}^\infty A_i$ would suggest that $i$ is an index running over the natural numbers from $1$ and it is unbounded, i.e. all the natural numbers (except $0$).

Asaf Karagila
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The $\infty$ that they're talking about should be: $\displaystyle\lim_{n\in \mathbb{N}} n $ , since the $A_i$'s are indexed by the natural numbers.

Rustyn
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