Robert Shaw wrote:
>
>>> Generally, composing just two functions will give an uncountable
>> >
>> Countable :-)
>>
>Uncountable, if you allow infinite sequences.
>If you've got functions f and g you can do the
>diagonal argument on any list of sequences.
>
But I was arguing against your presumed assertion that
"composing two functions a finite number of times..."
because this was the a priori rule that generated the
countable set that must get as close as possible to the
target set:
>>> If you start with 3 you can only generate a countable infinity
>>> of numbers. Take the factorial n times, then the square root
>>> m times. Because of the sparsity of the factorials I'd suspect
>>> this set isn't dense.
>>>
>>> Generally, composing just two functions will give an uncountable
>>> infinity of numbers which I think means the set of numbers
>>> must be dense over a nonzero interval.
>
>I'm not clear what you mean here.
>Can you give a uncountable set of numbers whose
>closure doesn't include any connected sets?
>
Ok. I interpreted your assertion above [>>>] as:
Every infinite set restricted to a finite interval is dense
in the interval
When it should be:
Every infinite set restricted to a finite interval accumulates
around at least one point
Now, for your second query:
>Can you give a uncountable set of numbers whose
>closure doesn't include any connected sets?
Hell, yes! Cantor's set, that is closed, uncountable, and doesn't
include any interval.
[for those that are still listening: Cantor set is generated by
removing from [0, 1] the middle part (1/3, 2/3), then removing
from each remaining part the middle part, and so on. Or,
equivalently, the numbers that can be represented in base 3
by 0.abcdef... where all digits are 0 or 2]
Alberto Monteiro
