Question Number 205794 by mnjuly1970 last updated on 30/Mar/24
$$ \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\mathrm{A}\:=\:\left\{\:\frac{{k}}{\mathrm{2}^{{n}} }\:\mid\:\mathrm{1}\leqslant\:{k}\:\leqslant\:\mathrm{2}^{{n}} \:,\:{n}\in\mathbb{N}\:\right\} \\ $$$$\:\:\:\:\:{find}\:.\:\:\overset{\:−} {\mathrm{A}}\:=\:? \\ $$$$ \\ $$
Commented by A5T last updated on 30/Mar/24
![A=(0,1]? ⇒A^− =R\(0,1] (if A^− means complement of A)](https://www.tinkutara.com/question/Q205795.png)
$${A}=\left(\mathrm{0},\mathrm{1}\right]?\:\:\:\Rightarrow\overset{−} {{A}}=\mathbb{R}\backslash\left(\mathrm{0},\mathrm{1}\right] \\ $$$$\left({if}\:\overset{−} {{A}}\:{means}\:{complement}\:{of}\:{A}\right) \\ $$
Answered by TheHoneyCat last updated on 30/Mar/24
![I will assume k∈N and A^ to be the closure of A. Simply because other meanings make the question too trivial. But please, next time, define your things Let A_n :={(k/2^n )∣k∈N ∧1≤k≤2^n } A=∪_(n∈N) A_n We notice that A⊆[0,1] Let x∈[0,1[ and n∈N 0≤x<1 ⇒0≤2^n x<2^n ⇒0≤⌊2^n x⌋≤2^n x<⌊2^n x⌋+1≤2^n with ⌊•⌋ denoting the integer part. so taking k_n :=⌊2^n x⌋ for any x∈[0,1] we have 0≤2^n x−k_n <1 i.e. 0≤x−(k_n /2^n )<(1/2^n ) In other words ∀x∈[0,1] ∃(a_n )∈A^N : a_n →x which means A^ ⊇[0,1] and of course... since we saw that A⊆[0,1] A^ ⊆Closure([0,1])=[0,1] We can conclude A^ =A _□](https://www.tinkutara.com/question/Q205796.png)
$$\mathrm{I}\:\mathrm{will}\:\mathrm{assume}\:{k}\in\mathbb{N} \\ $$$$\mathrm{and}\:\bar {{A}}\:\mathrm{to}\:\mathrm{be}\:\mathrm{the}\:\mathrm{closure}\:\mathrm{of}\:{A}. \\ $$$$\mathrm{Simply}\:\mathrm{because}\:\mathrm{other}\:\mathrm{meanings}\:\mathrm{make}\:\mathrm{the}\:\mathrm{question} \\ $$$$\mathrm{too}\:\mathrm{trivial}. \\ $$$$\mathrm{But}\:\mathrm{please},\:\mathrm{next}\:\mathrm{time},\:\mathrm{define}\:\mathrm{your}\:\mathrm{things} \\ $$$$ \\ $$$$ \\ $$$$\mathrm{Let}\:{A}_{{n}} :=\left\{\frac{{k}}{\mathrm{2}^{{n}} }\mid{k}\in\mathbb{N}\:\wedge\mathrm{1}\leqslant{k}\leqslant\mathrm{2}^{{n}} \right\} \\ $$$${A}=\underset{{n}\in\mathbb{N}} {\cup}{A}_{{n}} \\ $$$$\mathrm{We}\:\mathrm{notice}\:\mathrm{that}\:{A}\subseteq\left[\mathrm{0},\mathrm{1}\right] \\ $$$$\mathrm{Let}\:{x}\in\left[\mathrm{0},\mathrm{1}\left[\:\mathrm{and}\:{n}\in\mathbb{N}\right.\right. \\ $$$$\mathrm{0}\leqslant{x}<\mathrm{1}\:\Rightarrow\mathrm{0}\leqslant\mathrm{2}^{{n}} {x}<\mathrm{2}^{{n}} \\ $$$$\Rightarrow\mathrm{0}\leqslant\lfloor\mathrm{2}^{{n}} {x}\rfloor\leqslant\mathrm{2}^{{n}} {x}<\lfloor\mathrm{2}^{{n}} {x}\rfloor+\mathrm{1}\leqslant\mathrm{2}^{{n}} \\ $$$$\mathrm{with}\:\lfloor\bullet\rfloor\:\mathrm{denoting}\:\mathrm{the}\:\mathrm{integer}\:\mathrm{part}. \\ $$$$ \\ $$$$\mathrm{so}\:\mathrm{taking}\:{k}_{{n}} :=\lfloor\mathrm{2}^{{n}} {x}\rfloor\:\mathrm{for}\:\mathrm{any}\:{x}\in\left[\mathrm{0},\mathrm{1}\right]\:\mathrm{we}\:\mathrm{have} \\ $$$$\mathrm{0}\leqslant\mathrm{2}^{{n}} {x}−{k}_{{n}} <\mathrm{1} \\ $$$${i}.{e}.\:\mathrm{0}\leqslant{x}−\frac{{k}_{{n}} }{\mathrm{2}^{{n}} }<\frac{\mathrm{1}}{\mathrm{2}^{{n}} } \\ $$$$\mathrm{In}\:\mathrm{other}\:\mathrm{words}\:\forall{x}\in\left[\mathrm{0},\mathrm{1}\right]\:\exists\left({a}_{{n}} \right)\in{A}^{\mathbb{N}} :\:{a}_{{n}} \rightarrow{x} \\ $$$$\mathrm{which}\:\mathrm{means}\:\bar {{A}}\supseteq\left[\mathrm{0},\mathrm{1}\right] \\ $$$$\mathrm{and}\:\mathrm{of}\:\mathrm{course}…\:\mathrm{since}\:\mathrm{we}\:\mathrm{saw}\:\mathrm{that}\:{A}\subseteq\left[\mathrm{0},\mathrm{1}\right] \\ $$$$\bar {{A}}\subseteq\mathrm{Closure}\left(\left[\mathrm{0},\mathrm{1}\right]\right)=\left[\mathrm{0},\mathrm{1}\right] \\ $$$$\mathrm{We}\:\mathrm{can}\:\mathrm{conclude}\:\bar {{A}}={A}\:\:_{\Box} \\ $$
Commented by A5T last updated on 30/Mar/24
$${Is}\:{there}\:{any}\:{k}\left(\mathrm{1}\leqslant{k}\leqslant\mathrm{2}^{{n}} \right)\:{such}\:{that}\:\frac{{k}}{\mathrm{2}^{{n}} }=\mathrm{0}? \\ $$
Commented by mnjuly1970 last updated on 30/Mar/24
$${bravo}\:{sir}\:.. \\ $$
Commented by TheHoneyCat last updated on 30/Mar/24
no, k/x=0 => k=0x=0 and 0 is smaller than 1, so that's excuded...
Commented by TheHoneyCat last updated on 30/Mar/24
thanks
Commented by A5T last updated on 30/Mar/24
![A should be (0,1] instead to show the exclusion.](https://www.tinkutara.com/question/Q205802.png)
$${A}\:{should}\:{be}\:\left(\mathrm{0},\mathrm{1}\right]\:{instead}\:{to}\:{show}\:{the}\:{exclusion}. \\ $$
Commented by TheHoneyCat last updated on 30/Mar/24
$${Nope}. \\ $$$$\mathrm{althoug}\:\mathrm{0}\notin{A} \\ $$$$\left(\frac{\mathrm{1}}{\mathrm{2}^{{n}} }\right)\:\:\mathrm{does}\:\mathrm{and}\:\mathrm{that}\:\mathrm{goes}\:\mathrm{to}\:\mathrm{0} \\ $$$$\mathrm{so}\:\mathrm{0}\in\bar {{A}} \\ $$
Commented by A5T last updated on 30/Mar/24
$${I}'{m}\:{talking}\:{about}\:{A}\:{specifically},\:{not}\:\overset{−} {{A}} \\ $$
Commented by TheHoneyCat last updated on 30/Mar/24
One way to see A is as the set of numbers in ]0,1] that have finite digits in basis 2. Saying Closure(A)=[0,1] means saying all numbers have an expression in basis two provided you allow infinite digits.
Note that if you change k/(2^n) to k/(10^n) you get the same question but in basis 10.
actually k/(b^n) gives you that in any basis b...
Commented by A5T last updated on 30/Mar/24
$${Yea}. \\ $$