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Suppose $G$ is an open subset of the real number that is not upper bounded. Is there a real number $x > 0$ such that the set of all integer multiples of $x$ intersects $G$ at infinitely many points?

That is, is it true that $\exists x \in \mathbb{R}$ such that $\{mx\mid m\in \mathbb{Z}\}\bigcap G$ is infinite?

My intuition tells me yes, since the fact that $G$ is not upper bounded seems to be a major factor here, but I can’t seem to prove it.

AdditIdent
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IUissopretty
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  • Is $G$ a subset of $\mathbb R$? – 5xum Nov 16 '18 at 08:03
  • Could you explain more? – Darman Nov 16 '18 at 08:04
  • Yes, I made an edit to the question I asked. It should be slightly clearer now. – IUissopretty Nov 16 '18 at 08:06
  • i think its, $G \subset \mathbb{R}$ does $\exists~ (x \in \mathbb{R}) > 0 ~s.t~ A = {ax | a \in \mathbb{Z}}$ with $A \cap G$ being infinitely large. – Vaas Nov 16 '18 at 08:16
  • What is your proof – Darman Nov 16 '18 at 08:17
  • If x was rational in the question, answer was no. – Darman Nov 16 '18 at 08:19
  • just spitballing here, but if such an x exists, and since the Reals are a well ordered chain, couldnt you just use open balls around x to make a countable ordering? – Vaas Nov 16 '18 at 08:19
  • Where is $G$ being open used here? – YiFan Tey Nov 16 '18 at 08:54
  • It's false if $G$ isn't open. Let $G={c^k, k\in \mathbb{N}}$ with $c > 1$ and $c$ transcendental.

    Let $x \in \mathbb{R}, p>q, m>n \in \mathbb{N}$ such that $px = c^m$ and $qx = c^n$. Then $\frac{p}{q} = c^{n-m}$ which is a contradiction.

     – Rchn Nov 16 '18 at 16:38
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    $G$ is open means $G$ is a countable union of open intervals. If $(a,\infty ) \subset G$ or $G$ admits a subsequence of open interval with bounded below diameter, things are easy. – AdditIdent Nov 16 '18 at 17:30

1 Answers1

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Yes. Suppose $\nexists x$ such that $\{nx:n\in\mathbb{Z}\}\cap G$ is infinite, then for any $\epsilon>0$, $\{n\epsilon:n\in\mathbb{Z}\}\cap G$ is finite.

Let $n_0 = \max \{n \in\mathbb{Z}:n\epsilon \in G\}$.

Take $a\in G$ such that $a > n_0\epsilon$. $[\because G$ is unbounded above$]$.

Let $S = \{x \in G: (a,x) \subset G\}.$ $S\ne \phi$ as $G$ is open.

Let $b=\sup S$.

So $(a, b) \subset G$. Then $|b-a|<\epsilon$ also implies $b=a$. We have arrived at a contradiction.

Offlaw
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