Re: Cantor set = irrationals

On 15 Nov 2006 14:07:38 -0800, "eugene" <jane1806@xxxxxxxxxx> wrote:

Is it possible to construct fat Cantor set which consists only of
irrational numbers ?

Certainly. In fact fatness has nothing to do with it, you
can make the set as thin as you like.

You construct the middle-thirds set by starting with [0,1]
and succesively removing middle thirds, right? Instead,
do this:

Start with [a,b], where a and b are irrational. Enumerate
the rationals as q_1, q_2, ... . Now at the n-th stage of
the construction, instead of remving the middle third
of each interval that exists at that stage, remove an
open subinterval with a few properties: (i) the endpoints
of every interval are irrational (ii) each interval
you remove has length >= half the length of the interval
you're removing it from, (iii) _if_ it happens that q_n
is in the union of the intervals that exist at the
n-th stage, make certain that q_n is an element of
one of the intervals removed at that stage.

(Part (iii) might not work so well if q_n is an
endpoint of one of the current intervals. But that
can't happen, because of (i). Note that modifying
condition (ii) allows you to get various sorts
of "thinness".)

Thanks


************************

David C. Ullrich
.