A-k-2-n-1-k-2-n-n-N-find-A-

Question Number 205794 by mnjuly1970 last updated on 30/Mar/24

A = {\frac{k}{2^{n}} ∣ 1 ⩽ k ⩽ 2^{n}, n \in N}

f i n d . \bar{A} = ?

Commented by A5T last updated on 30/Mar/24

A=(0,1]? ⇒A^− =R\(0,1] (if A^− means complement of A)

A = (0, 1] ? \Rightarrow \bar{A} = R ∖ (0, 1]

(i f \bar{A} m e a n s c o m p l e m e n t o f A)

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 _□

I will assume k \in N

and \bar{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} := {\frac{k}{2^{n}} ∣ k \in N \land 1 ⩽ k ⩽ 2^{n}}

A = \underset{n \in N}{\cup} A_{n}

We notice that A \subseteq [0, 1]

Let x \in [0, 1 [and n \in N

0 ⩽ x < 1 \Rightarrow 0 ⩽ 2^{n} x < 2^{n}

\Rightarrow 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 \in [0, 1] we have

0 ⩽ 2^{n} x - k_{n} < 1

i . e . 0 ⩽ x - \frac{k_{n}}{2^{n}} < \frac{1}{2^{n}}

In other words \forall x \in [0, 1] \exists (a_{n}) \in A^{N} : a_{n} \to x

which means \bar{A} \supseteq [0, 1]

and of course \dots since we saw that A \subseteq [0, 1]

\bar{A} \subseteq Closure ([0, 1]) = [0, 1]

We can conclude \bar{A} = A_{}

Commented by A5T last updated on 30/Mar/24

Is there any k(1≤k≤2^n ) such that (k/2^n )=0?

I s t h e r e a n y k (1 ⩽ k ⩽ 2^{n}) s u c h t h a t \frac{k}{2^{n}} = 0 ?

Commented by mnjuly1970 last updated on 30/Mar/24

b r a v o s i r . .

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.

A s h o u l d b e (0, 1] i n s t e a d t o s h o w t h e e x c l u s i o n .

Commented by TheHoneyCat last updated on 30/Mar/24

Nope. althoug 0∉A ((1/2^n )) does and that goes to 0 so 0∈A^�

N o p e .

althoug 0 \notin A

(\frac{1}{2^{n}}) does and that goes to 0

so 0 \in \bar{A}

Commented by A5T last updated on 30/Mar/24

I′m talking about A specifically, not A^−

I^{'} m t a l k i n g a b o u t A s p e c i f i c a l l y, n o t \bar{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

Y e a .

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