Proving that a sequence arising by a one-to-one indexing of $\{\sqrt[3]n-\sqrt[3]m\mid n,m\in\mathbb{N}\}$ is dense in $\mathbb R$ 
Let $(a_k)$ be a sequence arising by an arbitrary one-to-one indexing of the elements of the matrix $\{ \sqrt[3]{n} - \sqrt[3]{m} \mid n,m \in \mathbb{N}\}$. Show that every real number is a limit point of this sequence.

Try
I guess to simplify matters, itd be easy if we write $a_k$ as
$$a_{n,m} = \sqrt[3]{n} - \sqrt[3]{m},$$
we need to prove that for any $ x \in \mathbb{R}$ we are able to construct a subsequence of $a_{m,n}$ say $a_{m_k, n_l}$ such that it converges to $x$.
Maybe can we take $n = [2x]^3 $ and $m = [x]^3$, then $a_{m,n} = [2x]-[x] \to x \in \mathbb{R}$.
Is this correct?
 A: $\def\N{\mathbb{N}}$Denote $b_{n, m} = \sqrt[3]{n} - \sqrt[3]{m}$ for $n, m \in \N$. For a fixed $x \in \mathbb{R}$ and any $n \in \N$, define $c_n = n - [3xn^{\frac{2}{3}}]$. If $x = 0$, then clearly $\lim\limits_{n → ∞} b_{n, c_n} = 0$. Otherwise as $n → ∞$,$$
c_n \sim n,\quad n - c_n = [3xn^{\frac{2}{3}}] \sim 3xn^{\frac{2}{3}},
$$
and$$
\lim_{n → ∞} b_{n, c_n} = \lim_{n → ∞} \frac{n - c_n}{\sqrt[3]{n^2} + \sqrt[3]{nc_n} + \sqrt[3]{c_n^2}} = \lim_{n → ∞} \frac{3xn^{\frac{2}{3}}}{3n^{\frac{2}{3}}} = x.
$$
Now suppose $\{a_k\}_{k \geqslant 0}$ is defined by $a_k = b_{φ(k)}$ where $φ: \N → \N^2$ is a bijection. Inductively define\begin{gather*}
k_0 = φ^{-1}(0, c_0),\quad k_{l + 1} = \min\{φ^{-1}(n, c_n) \mid n \in \N,\ φ^{-1}(n, c_n) > k_l\}.
\end{gather*}
The sequence $\{k_l\}_{l \geqslant 0}$ is well-defined since $\{φ^{-1}(n, c_n) \mid n \in \N\}$ has infinitely many elements. Therefore,$$
\lim_{l → ∞} a_{k_l} = \lim_{l → ∞} b_{φ(k_l)} = x.
$$
A: Let $D = \{n^{1/3}-m^{1/3}: m,n\in \mathbb N\}.$ Note that if $d\in D,$ then $-d\in D.$ So it's enough to show every $x>0$ is the limit of a sequence in $D.$
Now as others have noticed, if $a>0,$ then
$$\tag 1 (a+1)^{1/3}-a^{1/3} < \frac{2}{3a^{2/3}}.$$
This follows from the mean value theorem.
Now suppose $x>0$ is given. Let $0<r<x.$ We'll be done if we show there exists $d\in D$ such that $x-r\le d \le x.$ Start by choosing $N\in \mathbb N$ such that $1/N^{2/3}<r.$ The idea is that $d= (N+k)^{1/3}-N^{1/3}$ will work for some $k\in \mathbb N.$
First note that $(N+k)^{1/3}-N^{1/3}$ increases to $\infty$ as $k\to \infty.$ It follows that there is a maximum $k_0$ such that $ (N+k_0)^{1/3}-N^{1/3} \le x.$ Suppose $(N+k_0)^{1/3}+N^{1/3}< x-r.$ Then by $(1),$
$$(N+k_0+1)^{1/3}-N^{1/3}$$ $$ =(N+k_0+1)^{1/3}-(N+k_0)^{1/3}+(N+k_0)^{1/3}-N^{1/3} $$ $$< r+x-r=x.$$
This is a contradiction, since $k_0$ is the largest $k$ with this property. It follows that $x-r\le (N+k_0)^{1/3}-N^{1/3}\le x,$ and this is what we wanted to show.
A: What you propose doesn't work, the term $[2x]-[x]$ is constant in $m$ and $n$. But I think this is how it can work: look at the map $$f : x \longmapsto \sqrt[3]{x} - \lfloor \sqrt[3]{x}\rfloor.$$What this map basically does is, it takes the third root and just looks at the part right of the decimal point. What I want to show is that with this map we can approximate every number in $(0,1)$ when we restrict it to $\mathbb N$, because then we can approximate every number in $\mathbb R$ with the expressions $\sqrt[3] n - \sqrt[3] m$.
So assume there is an open set $U \subset (0,1)$ such that $f(n) \not\in U$ for every $n \in \mathbb N$, for simplicity say $U = (a,b)$ for some $a,b \in (0,1)$. For every $n \in \mathbb N$ there is a number $k \in \big[n^3 , (n+1)^3 \big] \cap \mathbb N$ with the property that $f(k) < a$ and $f(k+1) > b$. That is because there is a number $k$ with $f(k)<a$ because $f(n)=0$ and our assumption $f(\mathbb N ) \cap U = \emptyset$ then yields $f (k+1) >b$. Now we look at the derivative of $f$, which exists in every interval of the form $\big(n^3 , (n+1)^3 \big) \subset \mathbb R$. We have $$f' ( x ) = \frac{1}{3} \frac{1}{\sqrt[3]{x^2}},$$which is monotonously decreasing. This yields $$f(k+1) - f(k)\leq \frac{1}3 \frac{1}{\sqrt[3]{k^2}},$$ hence a contradiction for big enough $n$ and therefore $k$.
