Resolving problem in proof of Theorem 1 in V. Sós "On the distribution mod 1 of the sequence $n\alpha$"

Question

I am reading V. Sós' "On the distribution mod 1 of the sequence $n\alpha$" which can be found on page 127 of its journal issue. This is one of the early proofs of the "three gap theorem" that there are at most three gaps between consecutive points in the points $n \alpha \mod 1$ for $0 \leq n \leq N$. In her proof of Theorem 1 at the bottom of p. 130 she is dealing with Case B where $N - k_1 < k_{\ell} < k_N$ and $k_{\ell+1} = k_{\ell} + k_1 - k_n$. She says that $n + k_N - k_{\ell} > N$, which, using $k_N \geq k_{\ell}$ gives $n >N$. I can verify both premises but this chain of logic does not work. I suspect that there is a correct argument for concluding that $n > N$ as Sós' proof is well accepted in the literature but I have not found it yet. I thought perhaps we need to use $N-k_1 < k_{\ell}$ which is the only restriction in Case B that has not been used yet, but so far I have not figured out a way to use it.

Answer 1

I have patched Sos' argument by reading Świerczkowski, S. "On successive settings of an arc on the circumference of a circle." Fund. Math. 46 (1959), pp187–189. Replacing the "Case B" argument on p.130 of Sos' article with the following completes the proof

Note that if $$k_{N-1}\alpha-, i\alpha-, 0-$$ then either $i= k_N$ or $i > N$. Thus Lemma 1 implies that $k_{N-1}=k_{N}-k_1$. If $n \geq k_{l} + k_1 - k_N$ then remark 8 gives $$ (k_N-k_1)\alpha-, (n-k_l+k_N-k_1)\alpha-, 0 $$ which the same as $$ k_{N-1}\alpha-, (n-k_l+k_N-k_1)\alpha-, 0- $$ So either

• $n-k_l+k_N-k_1 = k_N$ which implies $n > N$ because in Case B, $N-k_1 < k_l$; or
• $n-k_l+k_N-k_1 > N$ which implies $n > N$ because $n \geq k_{l} + k_1 - k_N$.

If $n < k_{l} + k_1 - k_N$ we use remark 8 to obtain $$ (k_l-n)\alpha-, 0-, (k_l-k_1-k_N-n)\alpha- $$ And since the entire interval has the width of $$ k_{N-1}\alpha-, 0-, $$ we get $$ k_{N-1}\alpha-, (k_l-n)\alpha-, 0-, (k_l-k_1-k_N-n)\alpha- $$ so either

• $k_l-n = k_N$ which implies $k_N < k_l$ for a contradiction with the Case B condition $k_l < k_N$; or
• $k_l-n > N$ which implies that $k_l > N$ for another contradiction.