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.
Best Answer
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.