Topology of Liouville Numbers

Asked Jan 16 '20 at 10:56

Active Feb 16 '20 at 05:06

Viewed 93 times

Liouville numbers are defined as the numbers $x\in \mathbb R$ s.t. for ever $i\in\mathbb N$ there exist $n,m\in\mathbb Z$ such that \begin{equation} 0<\left|x-\frac nm\right|<\frac1{m^i}. \end{equation} and they have the property of being well approximated by a sequence of rationals. Is there a (translation invariant) topology on $\mathbb R$ such that the closure of $\mathbb Q$ are the Liouville numbers?

Or maybe an uncountable set included in the Liouville numbers? The problem is always that such topology, if exists, cannot contain any ball.

edited Feb 16 '20 at 05:06

Alex Ravsky

106,166

asked Jan 16 '20 at 10:56

Davide Maran

1,259

I added translation invariant, otherwise one could just take the topology of the balls intersecated with the Liouville numbers – Davide Maran Jan 16 '20 at 11:16
The metric for the balls is translation invariant. – William Elliot Jan 16 '20 at 11:40
That's trivial: let $\mathbb Q$ and the Liouville numbers inherit the usual topology, then define discrete topology on the rest of $\mathbb R$. – Ivan Neretin Jan 16 '20 at 22:43
@IvanNeretin It is not translation invariant. – Paul Frost Jan 17 '20 at 13:51
Darn! Indeed, a shift by $\sqrt2$ will turn some open sets into not-so-open. – Ivan Neretin Jan 17 '20 at 14:42

Topology of Liouville Numbers

0 Answers0