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Mueen Nawaz wrote:
> 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.
Oh yeah, I certainly wasn't debating the accuracy of the result. It
appears that you already considered this detail anyway, so my
nit-picking was entirely unnecessary.
I did like the Cantor set analogy by the way. I didn't forsee it when I
was reading your post and was delighted when you pointed it out.
>
>> 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.
Yeah, that's what I meant.
>> lim_{n->inf} a^n = 0 if 0 <= a < 1
>
> (Incidentally, I'm not sure I understand your limit. What is 'a' and
> why is it fixed?
For base m a would be (m-1)/m -- the probability that a single given
digit of the random sequence doesn't contain a 1 in base m.
>> 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.
In that case I'll spare you the trouble of being tempted to read it
later. I just adapted the construction of a Vitali set from the real
line to infinite binary sequences by replacing addition with xor. A few
small things changed as a result, but nothing major.
I'm enveious you had an integration theory class that got into measure
theory. I don't recall such a course being available back when I was
taking math classes, but it would have been pretty fun had there been.
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