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:




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 $=\{1\}$?

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    $\begingroup$ In fact $[0,1]\not=\{0,1\}.$ Why do you think so? $\endgroup$ – awllower Jun 3 '16 at 1:05
  • $\begingroup$ @awllower is it because there are an infinite amount of numbers between $0$ and $1$? $\endgroup$ – hax0r_n_code Jun 3 '16 at 1:09
  • $\begingroup$ Yes, $[0,1]$ is the set of all real numbers between $0$ and $1.$ And it is in fact in bijection with the whole $\mathbb R.$ $\endgroup$ – awllower Jun 3 '16 at 1:18
  • $\begingroup$ @awllower so that means $A_1 = \mathbb R$? $\endgroup$ – hax0r_n_code Jun 3 '16 at 1:32
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    $\begingroup$ free_mind you seem to have the notation problem [a,b] (the so-called closed interval) denotes the set of ALL (in this case) real numbers between a, b, including a, b, -- whereas the open interval (a,b) is the same thing, without the endpoints a, b. So set you mind on the fact that these sets contain infinite (and continuous) set of numbers, not a finite set, and then you go from there. When they say "state the intervals" they do not mean you to recount for every single number, rather to use this interval notation to get an explicit feel for these intervals for various n... $\endgroup$ – Rado Jun 3 '16 at 1:51

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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