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Here $a_k$ is a sequence of natural numbers.

Several specific examples of this are known to be true, for example the case when $a_k=k$ just tells that the rationals are dense in the reals. The case $a_k=n^k$ is also widely discussed, for example here or here.

However, I couldn't find anything about the general case, so I'm looking for references about it. Also, if this is clearly false and I missed the obvious counterexample I apologize, but seems reasonable to me that one can let $m$ and $k$ grow as large as needed in order to "get close enough" to any real number.

Hanul Jeon
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Zima
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3 Answers3

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Something very similar to this ...

We have to show that $\forall r \in \mathbb{R}$ and $\forall \varepsilon>0, \exists m \in \mathbb{Z}$ and $k\in\mathbb{N}$: $$\left|r-\frac{m}{a_k}\right|<\varepsilon \tag{1}$$

From $$m=\left \lfloor r\cdot a_k \right \rfloor \leq r\cdot a_k < \left \lfloor r\cdot a_k \right \rfloor +1=m+1 \Rightarrow \\ \frac{m}{a_k} \leq r < \frac{m}{a_k}+\frac{1}{a_k}$$ we have $$\left|r-\frac{m}{a_k}\right|<\frac{1}{a_k}\tag{2}$$

Since $a_k\to\infty$ when $k\to\infty$, by "adjusting" $k>0$ we can find one such that $\frac{1}{a_k} < \varepsilon$ and $(1)$ follows.

rtybase
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Given an open interval $O$, choose $a_k$ such that $a_k > \frac{1}{\text{length}(O)}$. This forces that a rational with denominator $a_k$ is in $O$.

MathematicsStudent1122
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    There may be a core of an answer in here, but I think it would be useful to be more explicit about what the conclusion is you want to draw. – Wolfgang Bangerth Jun 10 '24 at 21:21
  • @WolfgangBangerth I think this answer is fine as is. The definition of density is “has a nontrivial intersection with any open set”. – Malady Jun 10 '24 at 22:02
  • @Malady Perhaps, but see whether you can explain this one sentence to a typical 3rd year student learning about density, for example. For one, the second sentence of the answer is only understandable by providing the definition of density. Second, this only shows that for one choice of $O$, the set $m/a_k$ has a nontrivial intersection, but it is at least not obvious that the set is dense (which would require that it has an intersection with all open sets). – Wolfgang Bangerth Jun 11 '24 at 14:57
  • @WolfgangBangerth First, I am a second year student and this answer was perfectly understandable. Also, the second sentence just requires knowing that rationals with denominator $a_k$ are spaced $1/a_k$ far apart, which is obvious. Also yeah, an answer about density used the definition of density. Finally, this argument shows that any open interval intersects the set nontrivially (we took an arbitrary interval, and proved the existence of an element from our set in the interval). Open intervals are a basis, so this shows that in fact any open set intersects it nontrivially. – Malady Jun 11 '24 at 15:30
  • I understand the answer. I suspect some others do as well, including at least one second-year student. But teaching (which this site is devoted to) also implies wanting to reach a larger audience than just the best students, and that often means providing explanations to statements that, while true on their own, are perhaps not as easy to understand for everyone. – Wolfgang Bangerth Aug 13 '24 at 03:46
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One way to show that your set is dense in $\mathbb{R}\ $ is to let $x\in\mathbb{R},\ \varepsilon>0$ and then show that $\ \exists\ K\in\mathbb{N}, M\in\mathbb{Z},\ $ such that $\ \left\lvert x - \frac{M}{a_K} \right \rvert < \varepsilon.$ To this end, let $x\in\mathbb{R},\ \varepsilon>0.$

Since $a_k\to\infty\ $ as $\ k\to\infty,\ \exists\ K\ $ such that $\ a_K > \frac{1}{\varepsilon},\ \implies 0<\frac{1}{a_K} < \varepsilon.\ $ Now define $M$ to be the greatest integer such that $\frac{M}{a_K} < x.\ $ Now, first suppose $\frac{M}{a_K}\leq x-\varepsilon.\ $ Then $\frac{M+1}{a_K} = \frac{M}{a_K} + \frac{1}{a_K} < x - \varepsilon + \varepsilon = x,\ $ contradicting the minimality of $M.$ Therefore we must have $\ x>\frac{M}{a_K}>x-\varepsilon,\ \implies \left\lvert x - \frac{M}{a_K} \right \rvert < \varepsilon.$

Note that this is true if $a_k$ is any sequence of real numbers tending to infinity, not necessarily a sequence of natural numbers tending to infinity.

Adam Rubinson
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  • Thank you, I was wandering if it was true even for real sequences, and you answered that as well. – Zima Jun 11 '24 at 06:54