Non Decreasing Subsequence in [0,1]

[Bachelier]

Not a homework problem.

I was reading this analysis book by Korner and in it there was a question about Bolzano-Weierstrass property in $$\mathbb{R}$$. it states

$$\text{Find a sequence} \ x_n \in [0,1], \text{such that given any x} \in [0,1], \text{we can find} \ n_1 < n_2 < ... s.t. \ x_n_j \rightarrow x \ as j \rightarrow \infty$$

 

[micromass]

Try to take a sequence that enumerates all the rational numbers. I believe that could do it...

 

[Bachelier]

you mean something like the Farey sequence.

 

[disregardthat]

Basically it is sufficient that your sequence as a set is dense in [0,1] (this is not immediately obvious, but apparent if you think of it, since any open interval around x will contain an infinite amount of elements). A sequence ordering the rationals will certainly do, but a simple explicit formula could be $$y(n,k) = k/2^n$$ where k ranges from 0 to 2^n for each n, and letting $$x_{n} = x(\lfloor \log_2n \rfloor,n-2^{\lfloor \log_2n \rfloor})$$. What has been done is just forming y(n,k) into a sequence.

This will generate the set of dyadic numbers in [0,1] which is a dense subset.

 

[Bachelier]

Clear...Thank you guys.