## Re: Proportions of Infinity

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• Unless he denies the Continuum Hypothesis, kevinsworkingemail was already saying that there is a 1-1 function between the sets, when he said they were both
Message 1 of 53 , Sep 1, 2003
Unless he denies the Continuum Hypothesis, "kevinsworkingemail" was already
saying that there is a 1-1 function between the sets, when he said they
were both uncountable.

His point, it seems to me, is only that cardinality doesn't equate
unerringly to measure (size), which mathematicians use to handle
probabilities in this context. That's true.

Devising and then analyzing any particular procedure for choosing a number,
to get to the probabilities in this question, is at least hard. For
example,kevinsworkingemail's procedure doesn't terminate, so it doesn't
terminate in a choice. Not sure how serious an objection that is. But,
also, as K. Kahn seems to be suggesting, a reasonable but different
procedure can give a different result. I think he's saying this: if we
choose a "preliminary number" using the original coin flip procedure, and
then [if we can imagine a "then" after a non-terminating procedure] take
its reciprocal, it looks like this slightly different procedure exactly
reverses the probabilities by the same analysis. So finding and analyzing
a choice-procedure to compute this kind of probability is subtle or
impossible. I don't know which.

But the probabilities in this question would not usually be computed like
that. People use measure. Say, Lebesgue measure. (It's fairly standard.)
The ratio

(measure of (0,1)/(measure of (1,infinity))

is 1/infinity

= 0, more or less.

That's the usual reason math people, I think, evaluate this particular
probability as zero.

On the other hand, some sets of real numbers are not measurable in the
Lebesgue theory, so there is room for another approach. But my guess is,
people would just look for a more expansive theory of measure to handle
those sets.

Barry

>Sorry about the 3 month delay but I am confused about the following.
>
>--- In Fabric-of-Reality@yahoogroups.com, PaintedDevil@a... wrote:
>> In a message dated 5/6/2003 12:46:47 PM GMT Daylight Time,
>> kevinsworkingemail@y... writes:
>>
>>
>> For example, the irrational numbers between 0 and 1 and between 1
>and
>> infinity both form uncountably infinite sets, yet if you choose an
>irrational
>> number at random (e.g. with a (countably?) infinite series of coin
>tosses to
>> generate the digits before the decimal point (in binary) and
>another infinite
>> series to get the digits after the decimal point) your chances of
>getting a
>> value between 0 and 1 is zero compared to your chances of getting
>one between
>> 1 and infinity.
>>
>
>This example makes sense until I think about the fact that the
>reciprocal of all those irrational number between 1 and infinity are
>between 0 and 1. There is a one-to-one function between the two
>sets. So it must depend upon how you generate the irrational number
>as to whether you're likely to find it between 0 and 1 or greater
>than 1. Why is one way of generating numbers better than another?
>
>-ken kahn
>
>
>
>
>
>
>
>Your use of Yahoo! Groups is subject to http://docs.yahoo.com/info/terms/
• Unless he denies the Continuum Hypothesis, kevinsworkingemail was already saying that there is a 1-1 function between the sets, when he said they were both
Message 53 of 53 , Sep 1, 2003
Unless he denies the Continuum Hypothesis, "kevinsworkingemail" was already
saying that there is a 1-1 function between the sets, when he said they
were both uncountable.

His point, it seems to me, is only that cardinality doesn't equate
unerringly to measure (size), which mathematicians use to handle
probabilities in this context. That's true.

Devising and then analyzing any particular procedure for choosing a number,
to get to the probabilities in this question, is at least hard. For
example,kevinsworkingemail's procedure doesn't terminate, so it doesn't
terminate in a choice. Not sure how serious an objection that is. But,
also, as K. Kahn seems to be suggesting, a reasonable but different
procedure can give a different result. I think he's saying this: if we
choose a "preliminary number" using the original coin flip procedure, and
then [if we can imagine a "then" after a non-terminating procedure] take
its reciprocal, it looks like this slightly different procedure exactly
reverses the probabilities by the same analysis. So finding and analyzing
a choice-procedure to compute this kind of probability is subtle or
impossible. I don't know which.

But the probabilities in this question would not usually be computed like
that. People use measure. Say, Lebesgue measure. (It's fairly standard.)
The ratio

(measure of (0,1)/(measure of (1,infinity))

is 1/infinity

= 0, more or less.

That's the usual reason math people, I think, evaluate this particular
probability as zero.

On the other hand, some sets of real numbers are not measurable in the
Lebesgue theory, so there is room for another approach. But my guess is,
people would just look for a more expansive theory of measure to handle
those sets.

Barry

>Sorry about the 3 month delay but I am confused about the following.
>
>--- In Fabric-of-Reality@yahoogroups.com, PaintedDevil@a... wrote:
>> In a message dated 5/6/2003 12:46:47 PM GMT Daylight Time,
>> kevinsworkingemail@y... writes:
>>
>>
>> For example, the irrational numbers between 0 and 1 and between 1
>and
>> infinity both form uncountably infinite sets, yet if you choose an
>irrational
>> number at random (e.g. with a (countably?) infinite series of coin
>tosses to
>> generate the digits before the decimal point (in binary) and
>another infinite
>> series to get the digits after the decimal point) your chances of
>getting a
>> value between 0 and 1 is zero compared to your chances of getting
>one between
>> 1 and infinity.
>>
>
>This example makes sense until I think about the fact that the
>reciprocal of all those irrational number between 1 and infinity are
>between 0 and 1. There is a one-to-one function between the two
>sets. So it must depend upon how you generate the irrational number
>as to whether you're likely to find it between 0 and 1 or greater
>than 1. Why is one way of generating numbers better than another?
>
>-ken kahn
>
>
>
>
>
>
>
>Your use of Yahoo! Groups is subject to http://docs.yahoo.com/info/terms/
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