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Here user Mario Carneiro formulated statement which I find really interesting and useful and I proved it and I would like to know is my proof correct?

If sequence $\{a_n\}_{n=1}^{\infty}$ implies the following conditions:

1) $a_1<a_2<\dots<a_{n}<\dots$

2) $a_n\to \infty$ as $n\to \infty$

3) $a_{n+1}-a_{n}\to 0$ as $n\to \infty$

Then $\{a_n+m: (n,m)\in \mathbb{N}\times \mathbb{Z}\}$ is dense in $\mathbb{R}^1$.

Proof: Taking $x\in \mathbb{R}^1$ and let $\epsilon>0$ be given. Then $\exists k\in \mathbb{N}$ such that $a_k<x<a_{k+1}$ (assuming that $x>a_1$). Then $\exists N:=N_{\epsilon}$ such that $n\geqslant N$ implies $a_{n+1}-a_n<\epsilon$.

1) If $k\geqslant N$ then $|x-a_k|<\epsilon.$

2) If $k<N$ then $\exists M\in \mathbb{N}$ such that $x+M\in(a_l, a_{l+1})$ where $l\geqslant N$ then $|x+M-a_l|<\epsilon.$

If $x\leqslant a_1$ then $\exists K\in \mathbb{N}$ such that $x+K>a_1$ and we can apply previous reasoning for $x+K$. Q.E.D.

Remark: If $A$ is dense subset of $\mathbb{R}^1$ then: 1) for any $r>0$ the set $rA$ also dense subset of $\mathbb{R}$. 2) for any $c\in \mathbb{R}$ the set $c+A$ also dense.

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RFZ
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2 Answers2

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There are just two very small corrections that need to be made. First, there is a $k\in\Bbb N$ such that $a_k\color{red}{\le}x<a_{k+1}$. And secondly, in $(2)$ there is an $M\in\Bbb N$ such that $x+M\in\color{red}[a_\ell,a_{\ell+1})$ for some $\ell\ge N$.

Brian M. Scott
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  • Everything else is in OK? – RFZ Nov 14 '15 at 20:56
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    @RFZ: It would be better English to say ‘If the sequence ${a_n}_{n=1}^\infty$ satisfies the following conditions’, but I didn’t see any other problems with the mathematics. – Brian M. Scott Nov 14 '15 at 20:57
  • Yes you're right. I am trying to make my English better :) Thank you very much for your attention! +1 – RFZ Nov 14 '15 at 20:59
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    @RFZ: You’re welcome! (And your English is already quite good.) – Brian M. Scott Nov 14 '15 at 21:00
  • Thanks a lot for your permanent help! Nice to hear it from you :) – RFZ Nov 14 '15 at 21:01
  • Dear Brian! Sorry that bother you but day ago I proved that problem from Rudin which was really hard to me since I thought on it more that two days. Can you check my solution please? I would be really greatful. – RFZ Nov 17 '15 at 06:58
  • @RFZ: Do you mean this one? I’m going to bed now, but I’ll take a look tomorrow. (I warn you, though, it’s been 30 years since I last looked at Riemann-Stieltjes integrals, so I may be a bit slow!) – Brian M. Scott Nov 17 '15 at 10:14
  • Thanks a lot, dear Brian! It would be great! :) – RFZ Nov 17 '15 at 15:50
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Since the original proof I had in mind differs a little, I'll write it down here. (Also it turns out that the monotonicity assumption $a_1\le a_2\le\dots$ is not necessary.)

Take $\epsilon>0$. By assumption (2), there is some $N$ such that $a_{n+1}-a_n\le\epsilon$ for all $n\ge N$.

Lemma: The sequence $a_n$ is $\epsilon$-dense in $[a_N,\infty)$. Proof: Take $x\ge a_N$ and let $k\ge N$ be the least integer such that $a_k\ge x$. Existence is guaranteed by assumption (3). Then $a_k-\epsilon\le a_{k-1}<x$ since $a_{k+1}-a_k\le\epsilon$ and $\lnot(a_{k-1}\ge x)$, so $a_k\in[x,x+\epsilon)$ and thus $a_k$ is $\epsilon$-close to $x$.

Returning to the main theorem, given $x\in\Bbb R$ we can find an $m\in\Bbb Z$ such that $x-a_N-m\in[0,1)$ (explicitly, $m=\lfloor x-a_n\rfloor$). Then by the lemma there is an $a_k$ which is $\epsilon$-close to $x-m$, and so $a_k+m$ is $\epsilon$-close to $x$. Thus $\{a_n\}_{n=1}^\infty+\Bbb Z$ is $\epsilon$-dense in $\Bbb R$ for all $\epsilon$, which is equivalent to it being dense in $\Bbb R$.

Mario Carneiro
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