Let

be such that x ≠ y.
Which of the following statements is true for every

?

Correct option is A

​Given:

• $x,y\in [0,1]$ and $x\ne y$.
• We need to determine which statement is true for every $ϵ>0$.

Analysis of Options

1. Option (a):
   There exists a positive integer NNN such that $|x-y|<2$^$n$ϵ for every integer $n\ge N$.

   • $|x-y|$ is a fixed positive number because $x\ne y$.
   • Since 2nϵ2^n \epsilon2ⁿ 
     ϵ grows exponentially with $n$, there will always exist a large enough $N$ such that ∣x−y∣<2nϵ|x - y| < 2^n \epsilon∣x−y∣<2ⁿ
      ϵ for all $n\ge N$.
   • This statement is true.

2. Option (b):
   There exists a positive integer $N$ such that 2nϵ<∣x−y∣2^n \epsilon < |x - y|2ⁿ  ϵ<∣x−y∣ for every integer $n\ge N$.
   

   • For large $n$, 2nϵ2^n \epsilon2ⁿ 
     ϵ grows exponentially and will eventually surpass any fixed $|x-y|$.
   • This statement is false.

3. Option (c):
   There exists a positive integer $N$ such that ∣x−y∣<2−nϵ|x - y| < 2^{-n} \epsilon∣x−y∣<2^-n 
    ϵ for every integer $n\ge N$.

   • As 2−nϵ2^{-n} \epsilon2^-n  
     ϵ becomes arbitrarily small for large $n$, it is not guaranteed that ∣x−y∣<2−nϵ|x - y| < 2^{-n} \epsilon∣x−y∣<2^-n ϵ for large $n$.
   • This statement is false.

4. Option (d):
   For every positive integer $N$, ∣x−y∣<2−nϵ|x - y| < 2^{-n} \epsilon∣x−y∣<2−nϵ for some integer $n\ge N$.

   • Similar to (c), this statement is false because $|x-y|$ is fixed and cannot always satisfy $|x-y|<2$^-n  ϵ.
     
     

Final Answer: (a) There exists a positive integer NNN such that $|x-y|<2$^$n$ ϵ for every integer $n\ge N$.

Another method:

$$x, y \in [0, 1] \, \& \, x \neq y \implies |x - y| > 0 \, \& \, \epsilon > 0 \\\text{By Archimedean principle, there exists a positive integer } N \text{ such that } N \cdot \epsilon > |x - y| \\\implies n \cdot \epsilon > |x - y|; \, \forall n \geq N \\\implies 2^n > n \cdot \epsilon > |x - y| \\\implies 2^n \cdot \epsilon > |x - y|; \, \forall n \geq N$$​