The distance requirement is a distraction, because for all number \(n\), if it is found that, in an instance (t), that any one of the runners are at least \(\cfrac{1}{n}\) in front and back from other runners, then it is also true that in the distance from \(\cfrac{1}{n+1}\) to \(\cfrac{1}{n}\) taking reference from the same runner, that another runner be place there (either side of this reference runner) with any speed.
Then the distance condition for \((n+1)\) runners is satisfied. In fact for all \(m\ge n\) this snapshot is true.
However this \(n+1\) runner must be placed with speed \(v_{n+1}\) such that all runners has a common starting point in time and space (started together (same time) at the same place, with arbitrary speed).
Can speed \(v_{n+1}\) be arbitrarily any speed, over a placement allowance of \(d_{\epsilon}=\cfrac{1}{n}-\cfrac{1}{n+1}\)?
No, \(v_{n+1}\) can only varies in around \(\pm d_\epsilon+m_{n+1}\) circles. Not the other way round. \(v_{n+1}\) is restricted but can be infinitely many.
Let turn back the clock anyway, given time \(t\), the duration that returns every runner to the starting point.
Consider, is it possible that,
\(v_{n+1}.t=m_{n+1}-\cfrac{1}{n+1}-\epsilon\)
where \(m_{n+1}\) is any integer greater than \(1\) (completing \(m_{n+1}\) full circles) less \(\cfrac{1}{n+1}+\epsilon\), \(\epsilon\) is up to the length \(d_{\epsilon}\) above.
\(m_{n+1}\) is unbounded and the expression \(\epsilon\) can be made proportionally infinitely, small but exact.
And we place the \(n+1\) runner at a distance \(\epsilon\) from the \(\cfrac{1}{n+1}\) boundary away from the \(n\) runner.
With this placement, \(n\) lonely runners implies \(n+1\) lonely runners and with an \(n=n_o\) to start (done computationally or be found in any discussion of this question), we would have proven the Lonely Runner Conjecture.
Unfortunately, runner \(n+1\) added does not have arbitrary speed! There's no proof.
