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I'm just starting my first course in real analysis next week this semester. I've began reading the first chapter on real numbers and came across this issue I can't understand. I've searched around here and came across Cantor's intersection theorem and its counter-examples. I read here a little about compactness and am still bamboozled from this page in my textbook (the final paragraph):

textbook page

The problem I have digesting this paragraph is that: The argument presented in the text seems to have given 'infinity' a double meaning. First, it allows it to be a 'solid state number' where m could be substituted to be it. Then it allows m+1 to let the argument run through, this would have to assume that 'infinity' is mutable and not a fixed number. Hence creating two different definitions for it. For me, it seems like if we stick to just one single definition of infinity:

If infinity is a solid number (infinitely large, but cannot be changed to be even larger), then we cannot have m+1 into the argument.

If infinity is not a solid number, then we cannot pick a natural number m for it to begin with.

So I'm thinking the infinite intersection of decreasing sequence of closed sets, should still contain a non empty set, and that set should be named $A_{\infty}$

I know I can't be right but why?

Eric Wofsey
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Kenshin
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  • My answer to that question may help you here: https://math.stackexchange.com/questions/2639000/a-question-about-infinity – Arnaud Mortier Feb 20 '18 at 16:30
  • What does "solid" mean? – 5xum Feb 20 '18 at 16:30
  • Recall that $x \in \bigcap A_i$ iff $x \in A_i$ for evry $i$. Assume that the intersection is not empty: this means that there is a number $m$ such that $m \in A_i$, for very $i \in \mathbb N$. But what about $A_{m+1} = { m+1, m+2, \ldots }$ ? – Mauro ALLEGRANZA Feb 20 '18 at 16:32
  • We have no $A_{\infty}$ because $\infty \notin \mathbb N$ (it is not a natural number). Part of the perplexity can be avoided writing: $\bigcap_{n \in \mathbb N} A_n$. – Mauro ALLEGRANZA Feb 20 '18 at 16:35
  • @5xum: I thought the book said: "Suppose we had some natural number m that we thought might actually satisfy m belong to the infinite intersection...." that it's making the assumption that the set of numbers in $A_{m}$ is a finite, solid number of numbers. – Kenshin Feb 20 '18 at 16:39
  • @Kenshin Why would you think that? I'm honestly asking. What in the sentence you wrote makes you think the book is assuming that $A_m$ is finite? – 5xum Feb 20 '18 at 16:39

2 Answers2

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First, it allows it to be a 'solid state number' where m could be substituted to be it.

No it doesn't. Nowhere in the text does the author assume that $m=\infty$.

The author assumes, for the sake of argument, that $m$ is an element of the set $$\bigcap_{n=1}^\infty A_n$$

Then, the author proves that this is a contradiction.

In other words, the author proves the statement

$$\forall m\in\mathbb N: m\notin \bigcap_{n=1}^\infty A_n$$


That's all. There is not saying "$m$ is infinity". In the argument, we use the definition

$$x\in \bigcap_{n=1}^\infty A_n \iff \forall n\in\mathbb N: x\in A_n.$$

That's just a definition that allows us to simply construct the set we will call $\bigcap_{n=1}^\infty A_n$ from now on. Using this definition, we can prove that for our particular selection of sets $A_n$, $$\bigcap_{n=1}^\infty A_n=\emptyset.$$

The argument goes (with the reason how we know each point is true written in italics):

  1. Let $m\in \bigcap_{n=1}^\infty A_n$ (assumption)
  2. Therefore, for all $n\in\mathbb N: m\in A_n$ (follows from point $1$ and the definition of $\bigcap_{n=1}^\infty A_n$)
  3. Therefore, $m\in A_{m+1}$ (by setting $n=m+1$ in point 2)
  4. However, we also know that $m\notin A_{m+1}$ (by the definition of $A_{m+1}$)
  5. Points $3$ and $4$ form a contradiction. (*because point $3$ is a negation of point $4$, and "$p$ and not $p$" is a contradiction).
  6. Therefore, the assumptions we used must be incorrect, and there is no element of $\bigcap_{n=1}^\infty A_n$. (reductio ad absurdum)
  7. Therefore, $\bigcap_{n=1}^\infty A_n=\emptyset$ (by the definition of equality of sets)

Every statement I wrote follows logically from the previous statements, and the only time I mention $\infty$ is in writing $\bigcap_{n=1}^\infty A_n$. In $\bigcap_{n=1}^\infty A_n$, all the $\infty$ sign means is captured in the definition of $\bigcap_{n=1}^\infty A_n$.

5xum
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The argument presented in the text seems to have given 'infinity' a double meaning. First, it allows it to be a 'solid state number' where m could be substituted to be it. Then it allows m+1 to let the argument run through, this would have to assume that 'infinity' is mutable and not a fixed number. Hence creating two different definitions for it.

What makes you think the text is talking about "infinity" at all? It's not! It's talking about ordinary, finite, natural numbers. The variable $m$ is not representing "infinity" or anything like that: it's just a natural number, which is assumed to be an element of the set $\bigcap_{n=1}^\infty A_n$. Since $m$ is a natural number, so is $m+1$. None of this really has anything to do with "infinity"; the only appearance of "infinity" is that our set is an intersection of infinitely many other sets.

Eric Wofsey
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  • Oh I see now thanks! I thought wrongly because it was tempting to think to disprove something you'd go to extremes first and plugging in an m that's somewhat equivalent to infinity was what did me in. – Kenshin Feb 20 '18 at 16:46