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Surjective function g and the floor function

  1. Nov 16, 2013 #1
    The problem statement, all variables and given/known data.
    Let ##A## be the set of sequences ##\{a_n\}_{n \in \mathbb N}##:
    1) ##a_n \in \mathbb N##
    2) ##a_n<a_{n+1}##
    3) ##\lim_{n \to \infty} \frac {\sharp\{j: a_j \leq n\}} {n}## exists.


    Call that limit ##\delta (a_n)## and define the distance (I've already proved this is a distance) ##d(a,b)=|\delta (a) -\delta (b)|+k^{-1}## where ##k=\{min j : a_j≠b_j\}## and ##d(a,a)=0## for any two sequences ##a## and ##b##. Prove that the function ##g:(A,d) \to [0,1]## defined as ##g(\{a_n\})=\delta(\{a_n\})## is surjective and continuous.


    The attempt at a solution.
    I didn't have problems to prove that this function is continuous (in fact, it's uniformly continuous), but I am totally lost at the surjectivity part. The only elements in the codomain to which I could associate two members of the domain where ##0## and ##1##. Someone suggested me to consider for each ##x \in (0,1)##, the sequence ##{x_n}## with ##x_n=\lfloor \frac {n} {x}\rfloor##. It's immediate this sequence is of natural numbers, but how can I prove that ##\lfloor \frac {n} {x}\rfloor <\lfloor \frac {n+1} {x}\rfloor ## and that ##\delta (x_n)=x##?

    If someone has a better suggestion/idea of how could I prove surjectivity, you are welcome to tell me.
     
    Last edited: Nov 17, 2013
  2. jcsd
  3. Nov 17, 2013 #2

    haruspex

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    I don't understand how that defines g.
     
  4. Nov 17, 2013 #3
    Sorry, I've corrected notation. There is no ##f##, it's always ##g##.
     
  5. Nov 17, 2013 #4

    pasmith

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    The floor function is non-decreasing, so immediately [itex]x_n \leq x_{n+1}[/itex]. To exclude equality, consider the fact that [itex]\frac{n+1}x - \frac nx = \frac 1x[/itex] and use the condition that [itex]0 < x < 1[/itex] to show that there must exist an integer between [itex]\frac{n}{x}[/itex] and [itex]\frac{n + 1}{x}[/itex].
     
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