Consider the collection of intervals of real numbers $A_n=\left[1-\frac{1}{n}, 2-\frac{1}{n}\right]$, where $n \in \mathbb N$
(a) State the intervals corresponding to $A_1$, $A_2$, $A_3$
(b) State the intersection of the whole collection of intervals $\bigcap_{n=1}^\infty A_n$
(c) State the union of the whole collection of intervals $\bigcap_{n=1}^\infty A_n$
For (a) I got:
$$A_1=\left[0,1\right]=\{0,1\}$$
$$A_2=\left[\frac{1}{2},\frac{3}{2}\right]=\left\{\frac{1}{2},1,\frac{3}{2}\right\}$$
$$A_3=\left[\frac{2}{3},\frac{5}{3}\right]=\left\{\frac{2}{3},1,\frac{4}{3},\frac{5}{3}\right\}$$
Before going further I just wanted to make sure this is right.
If it is correct, then for b would the intersection be those values explicitly in $A_1$, $A_2$, and $A_3$? Meaning the answer to b is b $=\{1\}$?
$\endgroup$ 81 Answer
$\begingroup$Note that $\forall n\in \mathbb{N}$,
$$0 \leq \color{red}{1-\frac{1}{n}}< \color{blue}{1-\frac{1}{n+1}} <1 \leq \color{red}{2-\frac{1}{n}}< \color{blue}{2-\frac{1}{n+1}}<2$$
So $$\bigcap_{k=1}^{n} A_{k}=\left[ 1-\frac{1}{n},1 \right]$$ and
$$\bigcup_{k=1}^{n} A_{k}=\left[ 0, 2-\frac{1}{n} \right]$$
Hence $$\bigcap_{k=1}^{\infty} A_{k}=\{ 1 \}$$ and
$$\bigcup_{k=1}^{\infty} A_{k}=[0, 2)$$
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