Question Number 67057 by Tony Lin last updated on 22/Aug/19
$$\left(\mathrm{1}\right){find}\:\cap_{{n}=\mathrm{1}} ^{\infty} \left[\mathrm{0},\:\frac{\mathrm{1}}{{n}}\right) \\ $$$$\left(\mathrm{2}\right){find}\:\cup_{{n}=\mathrm{2}} ^{\infty} \left[\frac{\mathrm{1}}{{n}},\:\mathrm{1}−\frac{\mathrm{1}}{{n}}\right] \\ $$
Answered by Kunal12588 last updated on 22/Aug/19
$$\left(\mathrm{1}\right)=\left[\mathrm{0},\mathrm{1}\right]\cap\left[\mathrm{0},\frac{\mathrm{1}}{\mathrm{2}}\right)\cap\left[\mathrm{0},\frac{\mathrm{1}}{\mathrm{3}}\right)\cap…\infty\:{terms}\:=\:\left\{\mathrm{0}\right\} \\ $$$$\left(\mathrm{2}\right)=\left[\frac{\mathrm{1}}{\mathrm{2}},\frac{\mathrm{1}}{\mathrm{2}}\right]\cup\left[\frac{\mathrm{1}}{\mathrm{3}},\frac{\mathrm{2}}{\mathrm{3}}\right]\cup\left[\frac{\mathrm{1}}{\mathrm{4}},\frac{\mathrm{3}}{\mathrm{4}}\right]\cup…\cup\left[\mathrm{0},\mathrm{1}\right]=\left[\mathrm{0},\mathrm{1}\right] \\ $$$${are}\:{they}\:{correct} \\ $$$${i}\:{don}'{t}\:{have}\:{any}\:{experience}\:{of}\:{solving}\:{these} \\ $$$${type}\:{of}\:{questions}.\:{don}'{t}\:{even}\:{know}\:{meaning} \\ $$$${of}\:\cup_{{n}=\mathrm{1}} ^{\infty} \\ $$
Commented by Tony Lin last updated on 22/Aug/19
$${yeah},\:{your}\:{answer}\:{is}\:{correct} \\ $$