Monday, 12 April 2010

set theory - Questions about ordering of reals and irrationals

(3) Here's how to construct an example. We can assume the segment in question was from 0 to 1 non-inclusively. Also, I will write source numbers in binary and target numbers in base 4.



Consider first the number 1/2, in binary 0.1000... Let's map it to 0.1[0102010...] — it doesn't matter what's inside [...] as long as it's irrational. Now we decide to map all numbers of the form 0.0... to 0.0... and 0.1... to 0.2... . Clearly, so far we didn't break anything.



Now, let's take a different rational number, e.g. 1/4 = 0.010... Similarly, we decide to map it to one of irrationals of the form 0.01[10011010...] and the segment [1/4, 1/2] is ready to go to 0.02...



Select the next rational number, e.g. 1/3 = 0.0101(01). It's breaking in half the segment destined to go to 0.02... No problem, again we select some irrational 0.021[010012...] for 1/3 and move left and right subsegments to 0.020... and 0.022...



Now, so far I was using xxx0, xxx1 and xxx2. But let's sometimes move segments to xxx1, xxx2 and xxx3. Let's do it whenever I'm on a level which is a square of natural number.



We're still increasing, yahoo!



Repeat this process for all rationals ordered by denominator. For any rational we have selected an irrational number by definition. For any irrational, it's the limit of segments broken down into parts. Each breakdown reveals exactly one digit of the result — so we reconstruct it digit-by-digit in ternary. It has infinite number of digits. Moreover, these digits never become periodical thanks to the fact that each $n^2$-digit was shifted by 1. So the result is irrational too.



Since every two irrationals are separated by rational, this function is always increasing. Qued erat construirum.

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