I've found some duplicate questions on the website about this exact proof but I can't really understand them. The only definition that I know for denseness is that if $M \subset \mathbb{R}$ then $M$ is dense in $\mathbb{R}$ if for any interval $(a,b)$ there's at least one element from $M$ in $(a,b)$. None of the proofs used this fact or at least they didn't use it in a way that was obvious to me. Please help!!
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5Can you elaborate which part you have trouble understanding? The proof goes essentially like this: $1.)$ $G:={ m\alpha +n \ \vert \ m,n \in \mathbb{Z}$ is a subgroup of $\mathbb{R}$. It is enough to show that $\inf { \vert g \vert \ \vert \ g\in G \setminus {0} } =0$. Indeed, this means that we can find for every $\varepsilon>0$ some $g_\varepsilon\in G\setminus {0}$ such that $\vert g_\varepsilon \vert\leq \varepsilon$. However, then $\mathbb{Z} g_\varepsilon \subseteq G$ and every point in $\mathbb{R}$ has distance at most $\varepsilon$ to $\mathbb{Z} g_\varepsilon$. – Severin Schraven Oct 16 '23 at 20:37
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As the differences of elements in $G$ are again in $G$ it is enough to find any two distinct elements in $G$ which are close. In order to do so one would like to use the pigeonhole principle. Let $H=\mathbb{Z}\alpha$. Consider the quotient map $P:H \rightarrow \mathbb{R}/\mathbb{Z}$ (note it is injective). By the pigeonhole principle we get that the image of $H$ must accumulate at some point. However, this means, that for every $\varepsilon>0$ there exists $g_1, g_2\in H$ (with $P(g_1)\neq P(g_2)$) and some $m\in \mathbb{Z} \subseteq G$ such that $0\neq \vert g_1-g_2-m \vert \leq \varepsilon$. – Severin Schraven Oct 16 '23 at 20:47
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3The proofs at the question you link do skip over some justification! Severin mentions the intended argument. I'll spell it out a bit more: if you show that for any $\epsilon > 0$, you can find $m_\epsilon, n_\epsilon$ with $0 < m_\epsilon \alpha + n_\epsilon < \epsilon$, it follows that your set is dense in $\Bbb R$. To show it intersects the interval $(a, b)$, take $\epsilon = b - a$. The set of integer multiples of $m_\epsilon \alpha + n_\epsilon$ is contained in your set. But it must intersect $(a, b)$, since adjacent multiples have distance less than $b - a$. So it's dense! Does that help? – Izaak van Dongen Oct 16 '23 at 21:42
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@IzaakvanDongen So I can take the interval $(a,b)$ and split it into $k$ subintervals then take some $x_1, \dots, x_{k+1}$ and from there'll know that at least 2 of them are in the same interval hence less than $b-a$ apart. I hope I'm thinking in the right direction. What should I be taking for the $x_i$ values? – Marin Oct 17 '23 at 05:49
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I'm not sure that made any sense... – Marin Oct 17 '23 at 05:52
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@SeverinSchraven Why is it enough to show that $\inf { \vert g \vert \ \vert \ g\in G \setminus {0} } =0$. I haven't studied group theory yet so I don't know how you reach that conclusion, sorry. – Marin Oct 17 '23 at 12:20
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Hi moxy. That's not quite my argument - the "splitting into subintervals" part of the proof is just used to give us the black box "for any $\epsilon > 0$, there is some $m, n$ with $0 < m\alpha + n < \epsilon$. Are you happy that you can follow the proof of that fact? The rest of the proof depends entirely on the following lemma: "if $(a, b)$ is an interval and $0 < \delta < b - a$, then the set ${\dotsc, -2\delta, -\delta, 0, \delta, 2\delta, \dotsc}$ contains some element that lies in $(a, b)$". Are you happy with this fact? (Try thinking about the case $\delta = 1$, and $b = a + 1.00001$) – Izaak van Dongen Oct 17 '23 at 13:14
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@IzaakvanDongen I don't understand the proof of the $0 < m\alpha + n < \epsilon$ part. The lemma does hold when I tried the example but I have no idea how to prove it. – Marin Oct 17 '23 at 14:20
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OK. The idea of the argument for the $\epsilon$ part is "if we look at the sequence $m \alpha\bmod 1$, then at some point there must be two terms that are close together. The difference between these terms is small, and the difference can be written as $m\alpha+n$. This argument is given at your link and by Severin. Can you say any more about what you don't understand there? For the lemma: my example is basically the whole lemma. You can always divide by $\delta$, and then it becomes "$\Bbb Z$ always intersects $(a,b)$ if $b-a>1$". To prove it: take $\lfloor \tfrac 12(a+b+1)\rfloor$. – Izaak van Dongen Oct 17 '23 at 14:35
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@IzaakvanDongen Ok I managed to convince myself the lemma is true. I am assuming $m\alpha \mod 1$ means "the fractional part of $m\alpha$." How do we know that at some point two terms must be close together? Oh I think I got it. We take the $[0, 1]$ interval and split it into an arbitrary $n$ subintervals each of length $1/n$. Then if we take the values $0, \alpha \mod 1, 2\alpha \mod 1, \dots, n\alpha \mod 1$ Then at least two, say $i\alpha \mod 1$ and $j\alpha \mod 1$ with $i < j$. So that means, $j\alpha - i\alpha < 1/n$ and since $n$ was arbitrary it means... something. – Marin Oct 17 '23 at 14:51
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The choice of $n$ is arbitrary so the point was to show that within the interval $[0, 1]$ we can have two of our $m\alpha \mod 1$ values be as close as we want. I am kind of stuck at that point now. How can their difference be written as $m\alpha + n$? – Marin Oct 17 '23 at 14:52
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Say $\lvert{m_1\alpha}-{m_2\alpha}\rvert<\epsilon$, where ${\cdot}$ denotes fractional part. You can always write ${\beta}=\beta+n$ whenever $\beta\in\Bbb R$, for some $n\in\Bbb Z$ (take $n=-\lfloor\beta\rfloor$). Hence we can write ${m_i\alpha}=m_i \alpha+n_i$ for some $n_i\in\Bbb Z$, for $i=1,2$. Therefore the difference is $(m_1-m_2)\alpha+(n_1-n_2)$. This is of the desired form! These facts become more obvious if you know some group theory. This comment chain is getting long! If you think you understand the proof, feel free to write it up in an answer and I'll have a look. – Izaak van Dongen Oct 17 '23 at 15:06
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@IzaakvanDongen I don't know any group theory yet (I've just started my analysis course 2-3 weeks ago). I will attempt to write a proof and post it. Thank you for helping me thus far! – Marin Oct 17 '23 at 15:10
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@IzaakvanDongen I've written down my proof so far. – Marin Oct 17 '23 at 16:20
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@moxy The argument why it is enough is spelled out right afterwards, starting after "Indeed..." – Severin Schraven Oct 17 '23 at 16:37
1 Answers
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This is what I've come up with so far:
Let $S = \{ m\alpha + n \mid m,n\in \mathbb{Z}\}$ for a fixed irrational number $\alpha$ and $\{x\} = x - \lfloor x\rfloor, x \in \mathbb{R}$. To prove that $S$ is dense in $\mathbb{R}$ we need to find $m_\epsilon$ and $n_\epsilon$ s.t. $0 \lt m_\epsilon\alpha + n_\epsilon \lt \epsilon, \forall \epsilon\gt0$.
Let $k \gt \frac{1}{\epsilon}$ and consider the sequence of $k+1$ values: $\{0\alpha\}, \{1\alpha\}, \{2\alpha\}, \dots, \{k\alpha\}$. Each element of this sequence is in the interval $[0, 1]$. If we split the interval $[0,1]$ into $k$ subintervals, each of length $\frac{1}{k}$ then by the pigeonhole principle we know that at least two of the values in the sequence are in the same subinterval $[\frac{t}{k}, \frac{t+1}{k}]$, for $0 \le t \lt k-1$. Let's call those values $\{i\alpha\}$ and $\{j\alpha\}$ where $\{i\alpha\}\lt\{j\alpha\}$. Then, $\{j\alpha\}-\{i\alpha\} = (j\alpha - \lfloor j\alpha\rfloor) - (i\alpha - \lfloor i\alpha\rfloor) = (j-i)\alpha - (\lfloor i\alpha\rfloor - \lfloor j\alpha\rfloor)\in S$, since it's of the form of $S$'s elements. Therefore, $\mid \{j\alpha\} - \{i\alpha\}\mid \lt \frac{1}{k} \lt \epsilon \implies S$ is dense in $\mathbb{R}$.
I need help with the final step I'm not exactly sure it makes much sense.
Marin
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Nice work. I would write it as follows: "... Let's call those values ${i\alpha}$ and ${j\alpha}$ where ${i\alpha} < {j\alpha}$. Then $(j - i)\alpha + (\lfloor i\alpha \rfloor - \lfloor j\alpha \rfloor) = {j\alpha} - {i\alpha}$ is of the desired form, and since ${i\alpha}$ and ${j\alpha}$ are in the same interval, we have $0 < {j\alpha} - {i\alpha} \le \tfrac 1k$ (to justify this further, use the fact that ${j\alpha} \le \tfrac{t + 1}k$ and ${i\alpha} \ge \tfrac tk$)." PS: do you see why we can assume ${i\alpha} \ne {j\alpha}$? – Izaak van Dongen Oct 17 '23 at 16:31
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So I don't need the $x \in [\frac{t}{k}, \frac{t+1}{k}]$ part? Also ${i\alpha} \neq {j\alpha}$ since $i \neq j$, right? – Marin Oct 17 '23 at 16:49
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Indeed not. Two real numbers being in the interval $[c, d]$ automatically implies their difference is at most $d - c$, for the reason I gave in parentheses. In your answer now, it's not true that $\lvert x - {(j - i)\alpha}\rvert < \tfrac 1k$ in general - $x$ could be quite big while $ {(j - i)\alpha}$ is quite small. – Izaak van Dongen Oct 17 '23 at 16:53
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You should assume ${i\alpha} < {j\alpha}$, rather than $i < j$ (otherwise the difference might accidentally be negative). Aside from that, I think you could do with a sentence justifying why ${i\alpha} \ne {j\alpha}$. What do you mean by "the final step"? Proving that this version of density implies your version of density? – Izaak van Dongen Oct 17 '23 at 17:03
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@IzaakvanDongen By final step I meant the part after the "Therefore, ..." Also how should I go about writing the justification of ${i\alpha} \ne {j\alpha}$? Is it even possible for them to ever be equal for $i \neq j$? – Marin Oct 17 '23 at 17:12