I am a first-year undergraduate who stumbled upon natural density. I am working on extending this definition to the subset of rational numbers. While most people would wait until they are older, I am already attempting to solve the problem.

I sent a paper to my Professor showing how I derived my definition of density on the subsets of the rationals. The paper is in this link this link this link. In conclusion I stated,

If $T\subseteq\mathbb{Q}$ and $V(a,b,r,t)=$

$${\left|\left(\bigcup\limits_{\left\{k\in\mathbb{Z}\right\}\cap[0,r]}\bigcup\limits_{\left\{n\in\text{odd}\right\}\cap[0,t]}\left\{\left.\frac{m}{2^k n}\in[a,b]\right|m\in\mathbb{Z}\right\}\right)\right|}$$
then if

$$\underset{-}{D}(T)=\limsup_{(r,t)\to\infty}\limsup_{(a,b)\to\left(-\infty,\infty\right)}\frac{\left|T\cap V(a,b,r,t)\right|}{|V(a,b,r,t)|}$$


$$\overset{-}{D}(T)=\liminf_{(r,t)\to\infty}\liminf_{(a,b)\to\left(-\infty,\infty\right)}\frac{\left|T\cap V(a,b,r,t)\right|}{|V(a,b,r,t)|}$$

then $D(T)$ ,the density of $T$ with respect to $\mathbb{Q}$, exists when


My Professor says if the definition is countably additive and shift invariant then the definition is an invariant mean defined by a Folner Sequence of $\mathbb{Q}$. If the definition is neither, then it is not a density.

He is extremely busy and does not have time to help. He suggested the following research papers:

Density in Arbitrary Semigroups

Density of Invariant Means in Left Amenable Semigroups

Multiplicative Large Sets and Ergodic Ramsey Theory

Density Theorem of Rational Numbers

However, after reading the papers, I am not sure if the density I found belongs to their definition of density.

Yet my definition is an extension of natural density. If sets $T_1$ and $T_2$ are subsets of the rational numbers, with $T_2=\mathbb{N}$ and $T_1\subseteq T_2$, then if $a=0$ it follows that $\lim\limits_{(a,b)\to\infty}\lim\limits_{(r,t)\to\infty}\frac{\left|T_1\cap V(0,b,r,t)\right|}{\left|T_2 \cap V(0,b,r,t)\right|}$ is the same as the definition of natural density.

Is my definition connected to an Invariant mean defined by a Folner Sequence of the Rational Numbers?

  • 1
    $\begingroup$ Isn't $D(\mathbb{N}) = 0$? $\endgroup$ Jul 9, 2017 at 9:46
  • $\begingroup$ @mathworker21 Your are right. Due to my error, I had to change the definitions. If they do not make sense to you, feel free to edit my post. $\endgroup$
    – Arbuja
    Jul 9, 2017 at 16:14
  • $\begingroup$ You say if a=0, then $\lim_{(a,b)\to \infty} ...$ Can you explain what this means? $\endgroup$ Jul 9, 2017 at 17:02
  • $\begingroup$ Also, I really don't think you should be trying to generalize the definition of density for a subset of natural numbers here, because natural numbers should have density 0 in $\mathbb{Q}$. $\endgroup$ Jul 9, 2017 at 17:05
  • 1
    $\begingroup$ @AlexRavsky Thank You! I try so hard to make my posts clear yet I keep making mistakes. Either I do this or I wait several years till I gain enough knowledge. I'm so close, I can't give up now. $\endgroup$
    – Arbuja
    Apr 10, 2020 at 18:05

1 Answer 1


This answer might be long, but I think it will be helpful since there is a lot going on here (I also will try to make it easy to read). First of all, it's very impressive and beneficial that you are starting to work with these ideas now and experimenting. I will first talk about defining a density on $\mathbb{Z}$ and explain how the ideas you mentioned above are applicable.

If you want to define a density on subsets of $\mathbb{Z}$ (which is basically the same as defining one on $\mathbb{N}$), you want to associate, to every subset $E$ of $\mathbb{Z}$, a quantity $\mu(E) \in [0,1]$. Consistent with intuition for density, you want $\mu(\mathbb{Z}) = 1, \mu(E+1) = \mu(E)$, and $\mu(E\cup F) = \mu(E)+\mu(F)$ for each $E,F \in \mathbb{Z}$ that are disjoint.

Any function $\mu : \mathcal{P}(\mathbb{Z}) \to [0,1]$ such that $\mu(\mathbb{Z}) = 1$ and $\mu(E\cup F) = \mu(E)+\mu(F)$ whenever $E,F$ are disjoint is called a mean. The second condition is called finite additivity, since it clearly implies $\mu(E_1\cup\dots\cup E_n) = \sum_{i=1}^n \mu(E_i)$ whenever $E_1,\dots,E_n$ are disjoint. Your professor should not have said you want countable additivity since then $1 = \mu(\mathbb{Z}) = \sum_{k=-\infty}^\infty \mu(\{k\}) = \sum_{k=-\infty}^\infty 0 = 0$, impossible. If $\mu(E+1) = \mu(E)$ for all $E \subseteq \mathbb{Z}$, then $\mu$ is called shift invariant. After some thought, it is natural to just define a density to be any shift-invariant mean.

The issue of finding a shift-invariant mean, i.e. a density, comes from the fact that we need to define $\mu$ on every subset of $\mathbb{Z}$. This is hard. If we try $\mu(E) = \lim_{n \to \infty} \frac{|E\cap [-n,n]|}{2n+1}$ (where $|A|$ represents the size of $A$), then the limit might not exist. Here's where Følner sets come into play. The idea is, we define a sequence of invariant means that are not shift invariant, but they get closer and closer to being shift invariant. If we take the limit of them (whatever that means), then we should get something that is still an invariant mean but now is shift invariant. (Come back to read this idea again after reading further)

Let's now execute this idea. For each $n \ge 1$ and $E \subseteq \mathbb{Z}$, define $\mu_n(E) = \frac{|E\cap [-n,n]|}{2n+1}$. I leave it to you to check that $\mu_n$ is an invariant mean (this is not too hard). Note $|\mu_n(E+1)-\mu_n(E)| = |\frac{|(E+1)\cap[-n,n]|-|E\cap[-n,n]|}{2n+1}| = |\frac{|E\cap[-n-1,n-1]|-|E\cap[-n,n]|}{2n+1} \le \frac{1}{2n+1}$, so $\mu_n$ is close to shift invariant. Here's the part where analysis comes into play - it allows us to take some kind of limit of the $\mu_n$'s. The set of all means is compact in the topology of pointwise convergence. This means there is some subsequence $(\mu_{n_k})_k$ of the $\mu_n$'s and some mean $\mu$ so that for each $E \subseteq \mathbb{Z}$, $\lim_{k \to \infty} \mu_{n_k}(E) = \mu(E)$. The proof is that $[0,1]^{\mathbb{Z}}$ is compact by Tychonoff's Theorem and the set of all means is closed. To learn more about this, try reading some introductory notes or textbook to point set topology. But for now, just trust me that we can take such a subsequence $(\mu_{n_k})_k$.

The good thing is that for each $E \subseteq \mathbb{Z}$, $|\mu(E+1)-\mu(E)| = \lim_{k \to \infty} |\mu_k(E+1)-\mu_k(E)| \le \lim_{k \to \infty} \frac{1}{2k+1} = 0$, that is, $\mu(E+1) = \mu(E)$. So now $\mu$ is shift-invariant and defined on every subset of $\mathbb{Z}$.

The critical thing we used is that the $\mu_n$'s were almost shift invariant. This stemmed from the fact that $-1+[-n,n]$ is basically the same as $[-n,n]$, compared to the size of $[-n,n]$. This is what a Følner sequence is in an arbitrary group. Precisely, a Følner sequence in a group $G$ is a sequence of finite subsets of $G$ so that $\frac{|g.F_n \Delta F_n|}{|F_n|} \to 0$ for each $g \in G$, where $g.F_n = \{gf : f \in F_n\}$. One can do a similar procedure to the one I did on $\mathbb{Z}$ with any Følner sequence to end up with a mean that is "shift invariant", which for an arbitrary group means $\mu(g.F) = \mu(F)$ for each $F \subseteq G$. Note since $1$ generates $\mathbb{Z}$, having $\mu(E+1) = \mu(E)$ for each $E \subseteq \mathbb{Z}$ is equivalent to $\mu(E+k) = \mu(E)$ for all $k \in \mathbb{Z}, E \subseteq \mathbb{Z}$.

Now to directly answer your question. Your definition of density does not necessarily make sense for each subset of $\mathbb{Q}$ since the limits involved might not exist. But your definition is good in the sense that it is intuitive and finitely additive when the limits do exist. There's not much else I could say in addition to what I said above. The main point is, it's hard to define a density on every subset, so we use the help of Følner sets.

  • 1
    $\begingroup$ This is brilliant! Thank You! $\endgroup$
    – Arbuja
    Jul 9, 2017 at 18:19

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