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Let $G$ be an open dense subset of $\mathbb{R}$ with the usual metric. Prove that for $x$ in $\mathbb{R}$ there exists $a$ and $b$ belonging to $G$ such that $x=a-b$.

Eric Stucky
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UNM
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  • Can you add where you're stuck? – egreg Jul 26 '13 at 22:40
  • Interesting trivia: this is true if $G$ is a dense set with nonzero Lebesgue measure (it doesn't have to be open) or a dense, nonmeager set with Baire property. This is a consequence of Steinhaus theorem. This also generalizes to locally compact groups with Haar measure. But all of this is a bit too involved for such a basic question. :) – tomasz Jul 26 '13 at 22:44
  • Sir, I haven't yet reached upto compact sets. Just started studying Metric spaces! – UNM Jul 26 '13 at 22:49
  • @MAB: That's what I thought, that's why I just post it as a trivia comment, not an answer. :) – tomasz Jul 26 '13 at 22:50
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    If this was false, you could find $x$ such that $x+b\not\in G$ for every $b\in G$. That is $G\subseteq G^c-x$. Take the closure to reach a contradiction. – Julien Jul 26 '13 at 23:02

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Hint:

  1. Find a pont $b_0\in G$ and an open interval $(b_0-\epsilon,b_0+\epsilon)\subseteq G$ around it.
  2. Given $x$, find a point $a\in G$ with $|a-(b_0+x)|<\epsilon$
  3. Let $b=a-x$

Why do these three steps work?

Hagen von Eitzen
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