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Nov 18, 2014 at 19:13 comment added Ilya Bogdanov @Douglas Zare: Yes, it seems that this modification works; the proof I can imagine looks similar to what's above.
Nov 18, 2014 at 18:56 comment added Douglas Zare @Yaakov Baruch: I looked at that, too. A modification I haven't checked, but which seems to work is to use ternary digits, and use a map $i_6 i_5 i_4 i_3 i_2 i_1 i_0 \mapsto (0.(-i_0)(-i_2)(-i_4)(-i_6), 0.(-i_1)(-i_3)(-i_5))$. That is, choose a fixed encoding of the digits so that $.02222$ is not next to $.10000$.
Nov 18, 2014 at 15:03 vote accept Vesselin Dimitrov
Nov 18, 2014 at 14:44 comment added Yaakov Baruch Ilya: thank you for the prompt reply. It's a very clever example! (By the way, how to you typeset "@"?)
Nov 18, 2014 at 14:28 comment added Ilya Bogdanov @Yaakov: No; it is close to my earlier (unsuccessful) attempt. Check what happens with the numbers $i=\overline{011\dots100}$ and $j=\overline{100\dots011}$ which differ by 7. My (hopefully working) recent approach is different --- see my first comment.
Nov 18, 2014 at 14:18 comment added Yaakov Baruch Ilya: I happened to be investigating the properties of sending index $i=\overline{ \dots i_{6} i_5 i_4 i_3 i_2 i_1 i_0}$ to the point $(0.i_0 i_2 i_4\dots, 0.i_1 i_3 i_5\dots)$, in binay notation. It seems somewhat related to your idea. Could it work too?
Nov 18, 2014 at 13:31 comment added Ilya Bogdanov In fact, in the second part we construct the sequence of integer points $(m(i),n(i))$ such that for all $i>j$ either $0<|m(i)-m(j)|<4\sqrt{i-j}$ or $0<|n(i)-n(j)|<4\sqrt{i-j}$.
Nov 18, 2014 at 12:55 history answered Ilya Bogdanov CC BY-SA 3.0