Surjective function g and the floor function

[mahler1]

Homework Statement .
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.

 

[haruspex]

I don't understand how that defines g.

 

[mahler1]

I don't understand how that defines g.

Sorry, I've corrected notation. There is no $f$, it's always $g$.

 

[pasmith]

Homework Statement .
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 $\lfloor \frac {n} {x}\rfloor <\lfloor \frac {n+1} {x}\rfloor$

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