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Problem: Describe $\bigcap_{n=1}^{\infty}[-1,1-\frac{1}{n}]$ and $\bigcup_{n=1}^{\infty}[-1,1-\frac{1}{n}]$. And prove that their results are true.

Let $A_{n}=\left [-1,1-\frac{1}{n} \right ]$. We see that \begin{equation*} A_{1}=\left [-1,0\right ]\qquad A_{2}=\left [-1,\frac{3}{4}\right ]\qquad \dots\qquad A_{\infty}=\left [-1,1\right ] \end{equation*} so we have \begin{equation*} \bigcap_{n=1}^{\infty}A_{n}=A_{1}\cap A_{2}\cap A_{3}\cap \dots\cap A_{\infty}=\left [ -1,0 \right ] \end{equation*} and \begin{equation*} \bigcup_{n=1}^{\infty}A_{n}=A_{1}\cup A_{2}\cup A_{3}\cup \dots\cup A_{\infty}=\left [ -1,1\right] \end{equation*} If I try to google something about it, the second one should have been $\bigcup_{n=1}^{\infty}A_{n}=[-1,1[$. It doesn't really make sense why. If it's true, then $A_{\infty}$ must be wrong. Could you please elaborate?

And how do I prove them more precisely by showing two different ways: $\subseteq $ and $\supseteq $? I've tried to look for some inspirations in See page 73, Example 3b but it doesn't really give me some ideas how to prove on my own.

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    $\begingroup$ Question : Is $+1$ in any of the $A_n$? If not, then how can it be in the union? $\endgroup$ – Prahlad Vaidyanathan Oct 22 '13 at 16:20
  • $\begingroup$ There is no such thing as $A_{\infty}$. $\endgroup$ – fleablood Aug 31 '16 at 13:15
  • $\begingroup$ Arrgh. Why the heck do people edit these zombie posts?????? $\endgroup$ – fleablood Aug 31 '16 at 13:17
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You interpret the notation $\bigcup_{n=1}^{\infty}[-1,1-\frac{1}{n}]$ a bit too literally. Unlike finite unions, this infinite union does not contain a term corresponding to the "upper limit" $\infty$. To spell out this notation correctly, one does not write $A_{1}\cup A_{2}\cup A_{3}\cup \dots\cup A_{\infty}$ but rather $A_{1}\cup A_{2}\cup A_{3}\cup \dots$ (without a last term). Over the hyperreals, one can have such infinite terminating unions with a last term having an infinite index $H$ (better notation than $\infty$), but even then this last term will not contain the number 1. This is because $1/H$ is not zero; it is a nonzero infinitesimal.

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There is no $A_\infty$: the notation is traditional and unfortunate. If you write $$ \bigcap_{n\ge1}A_{n} $$ you'll probably will figure out better what's the set you want to compute. The variable $n$ is considered to take on natural number values and $\infty$ is not a natural number.

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I see this problem as a classic homework in Basic Set Theory. Remember,

Definition of $\displaystyle\bigcup_{i=1}^{\infty}$ : $ x\in \bigcup_{i=1}^{\infty} \left[-1,1-\frac{1}{n} ,\right] \Longleftrightarrow \exists \;n \in\mathbb{N}-\{0\}, \mbox{ such that } x\in \left[-1,+1-\frac{1}{n} \right] $

Note that, for all $x\in[-1,1)$ there is $n_x\in\mathbb{N}-\{0\}$ such that $ -1<x<1-\frac{1}{n_x}<1. $ Then $x\in \left[-1,1-\frac{1}{n_x}\right]$. By cause $\left[-1,1-\frac{1}{n_x}\right]\subset \bigcup_{i=1}^{\infty} \left[-1,1-\frac{1}{n} ,\right]$ we have $x\in \bigcup_{i=1}^{\infty} \left[-1,1-\frac{1}{n} ,\right]$. Therefore, $\left[-1,1\right)\subset \bigcup_{i=1}^{\infty} \left[-1,1-\frac{1}{n}\right]$. For outher hand, $ \left[-1,1-\frac{1}{n}\right]\subset \left[-1,1,\right) $ for all $n\in\mathbb{N}-\{0\}$ implies $\bigcup_{i=1}^{\infty} \left[-1,1-\frac{1}{n} ,\right]\subset \left[-1,1\right)$. For the characterization of equality of sets via subset and superset can conclude, $$ \bigcup_{i=1}^{\infty} \left[-1,1-\frac{1}{n} ,\right]= \left[-1,1\right). $$

The other part of the exercise is done in a similar way using the definition below and conveniently exchanging in demonstrating the existential quantifier by universal quantifier. But the difference now is that $ \bigcap_{i=1}^{\infty} \left[-1,1-\frac{1}{n} ,\right]= \left[-1,0\right). $

Definition of $\displaystyle\bigcap_{i=1}^{\infty}$: $\quad x\in \bigcap_{i=1}^{\infty} \left[-1,1-\frac{1}{n}\right] \Longleftrightarrow \forall \;n \in\mathbb{N}, \mbox{ hold } x\in \left[-1,+1-\frac{1}{n} \right] $

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