Kevin Wampler wrote:
> Mueen Nawaz wrote:
>>         In point of fact, it's the Cantor set (i.e. what is the
>> probability that a number you pick will be from the Cantor set?):
> 
> I think that it's unfortunately a touch more involved than that.  For
> instance:
> 
> 0.200000....
> 0.122222....

	That's only a technicality that I was hoping no one would notice.
Strictly speaking, the Cantor set is the set of all numbers that _can_
be written without a 1 in the base 3 expansion (as in your case) - as
opposed to just numbers that *must* be written without a 1. The
important point is that there are only a countably infinite number of
such "exceptions", and so those exceptions are "tiny" compared to the
whole Cantor set, because the Cantor set has the cardinality of the reals.

	Put another way, I showed that the even if I allow some sequences with
a single 1 to sneak in, the probability is still 0.


> I think represent the same number on the real line, but one would be in
> the set and one wouldn't.  Nevertheless I do think that the Cantor set

	One would be in the set of sequences we're interested in, and the other
wouldn't. In terms of the Cantor set, they're both in it as they are the
same number.

>> So the probability of getting any infinite sequence that doesn't
>> have one of the states is zero. I'm willing to bet that this generalizes
>> to any base.
> 
> 
> It does, but the proof is probably easier if you compute the limit as
> the sequence length goes to infinity rather than visualize it
> geometrically:
> 
>     lim_{n->inf} a^n = 0 if 0 <= a < 1

	Well, we all know *that*.<G> I was pointing out a different way of
looking at it.

	(Incidentally, I'm not sure I understand your limit. What is 'a' and
why is it fixed?

> 1 = sum_{0:inf} p
> 
> Since no real number has this property, p cannot exist.  Thus there is
> no probability (zero or otherwise) that a random sequence is in S.  Said
> in more standard terminology: S has no measure.

	I didn't read the whole thing - xors make my head hurt. But I got the
general idea, having seen similar arguments before. When I was learning
integration theory using measures, I noticed that integrals were always
defined over _measurable_ sets.

	Moral of the story: Don't apply probability to problems constructed
using the axiom of choice.<G>

-- 
Guitar for sale. Very cheap. No strings attached.


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