Indeed. Define a 'dyadic interval' to be an interval with endpoints n/2^k and (n+1)/2^k) for some integers n, k. Then your function has the property that every open dyadic interval is mapped to a dyadic interval of the same length. Also, if alpha is a non-dyadic-rational, it can be written as the intersection of a nested family of open dyadic intervals of length tending to zero. So any sequence converging to alpha eventually lands within an arbitrarily small dyadic interval, which itself maps to an arbitrarily small dyadic interval containing f(alpha). Best wishes, Adam P. Goucher

Sent: Thursday, November 24, 2016 at 3:29 AM From: "Bill Gosper" <billgosper@gmail.com> To: math-fun@mailman.xmission.com Subject: [math-fun] The base 2-i, digits 0,i^0..3 spacefill

x xor <non dyadic-rational>: Discontinuous at the dyadic rationals. Continuous everywhere else? --rwg

Date: 2016-11-23 02:51 From: Joerg Arndt <arndt@jjj.de> To: math-fun <math-fun@mailman.xmission.com> * Bill Gosper <billgosper@gmail.com> [Nov 23. 2016 10:21]:

...

[...] He says it's really just prf dressed up. And he makes the remarkable observation that the frac4 dragon gosper.org/4flopfour.png is the bar graph of x xor 2/3 ! gosper.org/xor667fade.png It appears that x xor r makes a fractile for every rational r. --rwg

Correct: write 1 in binary as 0.11111..., then observe that (x xor r) + ((1-x) xor r) == 1.

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