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Q: uniform distribution over a countably infinite set ( No Answer,   8 Comments )
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 Subject: uniform distribution over a countably infinite set Category: Science > Math Asked by: qarl-ga List Price: \$150.00 Posted: 05 Nov 2005 07:29 PST Expires: 30 Nov 2005 18:30 PST Question ID: 589400
 ```yes, yes, i know: there is no such thing as a uniform distribution over a countably infinite set. what i'd like is a mathematical argument that says "when you uniformly choose a natural number, the expected value is infinite." surely one exists.``` Request for Question Clarification by leapinglizard-ga on 05 Nov 2005 08:16 PST ```Your statement cannot be proven, for it is not true. I can refute it with a proof by contradiction. Would you accept that as an answer? leapinglizard``` Clarification of Question by qarl-ga on 05 Nov 2005 11:45 PST `that's disappointing. really? yes, a counter proof is an acceptable answer.` Clarification of Question by qarl-ga on 05 Nov 2005 14:27 PST ```let me clairify, again - without using inconsistent jargon. i'd like an argument that says, roughly "when you select a natural number at rando (and all numbers have the same chance of being selected) you can expect that number to be extremely large." a mathematician once advised me that such an argument is possible using a "Jeffries Noninformative Prior". if that helps. K.``` Clarification of Question by qarl-ga on 13 Nov 2005 13:32 PST ```hey all - this question keeps getting locked - presumably because someone is starting to answer it - but never finishes. the lock lasts for 8 hours, which makes it impossible to continue the discussion below. so - if you're planning to answer, make sure you do answer, and don't let us hang...``` Clarification of Question by qarl-ga on 14 Nov 2005 13:21 PST `question now worth \$150 - for proof/disproof.`
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 The following answer was rejected by the asker (they received a refund for the question). Subject: Re: uniform distribution over a countably infinite set Answered By: leapinglizard-ga on 15 Nov 2005 13:19 PST
 ```Dear qarl, First of all, we should agree that infinity is not a number but a property. For example, we can say that a set contains an infinity of members, but it makes no sense to say that any member of this set is itself infinity. By definition, an infinite set has more members than any finite set. The set containing all natural numbers is one example of an infinite set. Its membership is greater than that of any finite set of natural numbers. In particular, it contains every possible natural number. If we pick a natural number at random, it will surely not be infinity, for it is always possible to produce a greater number. If we pick a number n, there is always the greater value n+1, and countless more besides. This is true for any n. However, we shall see that the expected value of a random selection from the set of all natural numbers is infinite in a different sense. In the remainder of our discussion, we use "random" to mean that every member of the set has an equal chance of being chosen, which is to say that we are selecting under a uniform distribution. You propose the experiment of making a random selection from the set of all natural numbers, and wish to establish that the result will be very large. The expected value of such an experiment is not infinity, as I explained above, but it does have the following interesting property. Claim The expected value of a random selection from the set of all natural numbers is greater than the expected value of a random selection from any finite set of numbers. So no matter how many big numbers you collect into a finite set, the expected value of picking a number from this set is smaller than the expected value of a selection among all natural numbers. If we restrict ourselves to a set containing a single number n, then we must pick n. And no matter how big this n may be, the expected value of a randomly chosen natural number is greater than n. Thus, our Claim implies the following Corollary. Corollary The expected value of a random selection from the set of all natural numbers is greater than any given natural number n. In that precise and highly practical sense, the expected value of a randomly chosen natural number is infinite. Since the Claim leads directly to the Corollary, we need only prove the Claim. My proof, which does not rely on any particular ordering of the natural numbers, is as follows. It is a proof by contradiction. --------------------------------------------------------------------- Proof Suppose the Claim is false. This implies that there is a finite set S such that the expected value of a random selection from S is at least as great as the expected value of a random selection from N, the set of all natural numbers. We express this situation with the inequality mean(S) >= mean(N) . [1] Now, the expected value of a random selection from S is bounded from above by the greatest value in S, which we shall call m. mean(S) <= m [2] This is because no matter what number we pick from S, it will not be greater than m, the greatest member of S. Since no member of S is greater than m, it follows that the expected value cannot be greater than m either. Now, the set of all natural numbers N includes the subset T = { 1, 2, 3, ..., 2m-2, 2m-1, 2m } [3] no matter how it is ordered. Note that T has exactly 2m members. The expected value of a random selection from T is m+0.5, because the mean of T is mean(T) = (1 + 2 + 3 + ... + 2m-2 + 2m-1 + 2m) / 2m [4] = m * (2m + 1) / 2m [5] = (2m + 1) / 2 [6] = m + 1/2 . [7] But in addition to the numbers appearing in T, the set of all natural numbers N contains many more numbers, all of which are greater than 2m. Thus, the expected value of a random selection from N must be greater than m+0.5 . mean(T) < mean(N) [8] m+0.5 < mean(N) [9] But inequality [2] above tells us that m+0.5 is still greater than mean(S), the expected value of a random selection from S. Hence, mean(S) must be even smaller than the expected value of a random selection from N. Putting together inequalities [2] and [9], we obtain the following. mean(S) <= m < m+0.5 < mean(N) [10] mean(S) < m+0.5 < mean(N) [11] mean(S) < mean(N) [12] This contradicts inequality [1], which is the contrary of the claim. Therefore, the Claim cannot be false. It must be true. --------------------------------------------------------------------- If you don't see how to compute mean(T) = m+0.5, let me bring to your attention the following article on summing an arithmetic series. MathWorld: Arithmetic Series http://mathworld.wolfram.com/ArithmeticSeries.html Note in particular equation (10), which in our case yields sum = 1/2 * 2m * (1 + 2m) = m * (2m + 1) and that is exactly the numerator in line [5] above. Regards, leapinglizard``` Request for Answer Clarification by qarl-ga on 16 Nov 2005 12:51 PST ```hey leapinglizard - i feel like such a party-pooper. everyone brings me a proof - and i always find some hole in it. i'm beginning to feel very bad about all this... but i think i've found a hole in yours, too. but perhaps not, let's talk about it. my concern is your underlying (unmentioned) assumption that "the expected value of a randomly selected natural number" has a specific value - especially given that your proof is by contradiction. you show that since it doesn't have one value, it must have the other. but if it has no value at all, then it's not an either-or situation. consider the infinite sum (-1)^i for all natural numbers i. (which longhand is (-1 +1 -1 +1 -1 +1....)) the sum doesn't converge, so we say it has no value. if we began by assuming it had a value, we would quickly get ourselves into all kinds of trouble, and show it to be equal to anything we wanted. or, consider this question: what is the probability of choosing an even number from the set of natural numbers. as shown in the comments, this probability has no value. but if we start by assuming it does, we could then argue (by contradiction) that it can't be less than 1/2, and it can't be greater than 1/2, so it must be 1/2. but in fact it can be shown to be not 1/2 either - because it has no value. ... so what do you think? K.``` Clarification of Answer by leapinglizard-ga on 16 Nov 2005 13:56 PST ```My proof does not mention a specific value for the result of the random selection. It shows that the expected value of randomly selecting among the natural numbers is greater than any given natural number. In this sense, the expected value is infinite. You cannot produce a natural number that is as great or greater than the expected value of making a random selection among all natural numbers. This can be proven, as I have done, by elementary algebraic manipulation. leapinglizard``` Request for Answer Clarification by qarl-ga on 16 Nov 2005 16:14 PST ```erm - i think you do assume a value. when you say "a < b" leads to a contradiction, so "a >= b" - the underlying assumption is that "a" and "b" both have values (and have value in a completely ordered set.) in this case, as in the case with infinite sums, there's the possiblilty that "a = NaN" (in IEEE parlance) and so the negation of "a < b" is not "a >= b", but instead "(a >= b") or (a = NaN) or (b = NaN)". so to complete your proof, you'll need to show that "a != NaN".``` Clarification of Answer by leapinglizard-ga on 16 Nov 2005 20:20 PST ```I bring to your attention the Law of the Excluded Middle. MathWorld: Law of the Excluded Middle http://mathworld.wolfram.com/LawoftheExcludedMiddle.html If the Claim is not false, then it must be true. In my proof, I posit that it is false, and I arrive at a contradiction. Therefore it is true. leapinglizard``` Request for Answer Clarification by qarl-ga on 16 Nov 2005 20:58 PST ```yes. and i'm telling you that when you posit that the statement is false, you incorrectly take its negation. the original statement is "The expected value of a random selection from the set of all natural numbers is greater than the expected value of a random selection from any finite set of numbers." the correct negation of this statement is "The expected value of a random selection from the set of all natural number is less than or equal to the expected value of a random selection from any finite set of numbers, OR the expected value of a random selection from the set of natural numbers has no value OR the expected value of a random selection from any finite set of numbers has no value." the last term of the statement is easily shown to be false - that leaves the second term. perhaps, if you gave a mathematical definition of what you mean by "the expected value of a random selection from the set of natural numbers" we could avoid the "no value" situation. but as long as we leave it undefined, the problem persists.``` Clarification of Answer by leapinglizard-ga on 16 Nov 2005 22:09 PST ```Well, the expected value of drawing a number from any set under the uniform distribution is the same as the arithmetic mean of the numbers in the set. So the mathematical definition you are looking for is that of the arithmetic mean, also known as simply the mean or the average. See equation (1) on the following page for the formula. MathWorld: Arithmetic Mean http://mathworld.wolfram.com/ArithmeticMean.html As N tends toward infinity, so does the mean of the set {x_i}. Or if X is a random variable with a uniform distribution, we say that the expected value of X tends toward infinity as the domain of X tends toward the set of all natural numbers. leapinglizard``` Request for Answer Clarification by qarl-ga on 17 Nov 2005 00:29 PST ```so, we're agreed that "mean" has no definition for an infinite set? the 1/N definition i saw in your Wolfram reference gets into trouble for N=inf. and by your last comment, it sounds as if you're relying on a limit argument, which i think suffers from the same trouble as in the comments.``` Clarification of Answer by leapinglizard-ga on 17 Nov 2005 00:58 PST ```No, we are not agreed. Every set, whether finite or not, has a mean. For example, the mean of the infinite set {..., -2, -1, 0, 1, 2, ...} is 0. The mean of the infinite set {1, 2, 3, ...} is infinitely large. In other words, it is larger than any given natural number. leapinglizard``` Request for Answer Clarification by qarl-ga on 17 Nov 2005 02:08 PST ```really? are you sure? can you provide a reference? (your wolfram reference defined means only for finite sets.)``` Clarification of Answer by leapinglizard-ga on 17 Nov 2005 03:08 PST ```Yes, I'm sure. Every set has a sum, and every set has a cardinality. The mean is the sum divided by the cardinality. leapinglizard``` Clarification of Answer by leapinglizard-ga on 17 Nov 2005 03:10 PST ```There is one exception: the empty set. Since the empty set has cardinality zero, its mean is undefined. But every non-empty set has a mean. leapinglizard``` Request for Answer Clarification by qarl-ga on 17 Nov 2005 11:37 PST ```aw crap. no, not every set has a sum. http://mathworld.wolfram.com/RiemannSeriesTheorem.html``` Clarification of Answer by leapinglizard-ga on 17 Nov 2005 11:59 PST ```Yes, every set has a sum. The summing operation is well-defined for finite as well as infinite sets. The Riemann Series has nothing to do with the sum of a set. leapinglizard``` Request for Answer Clarification by qarl-ga on 17 Nov 2005 12:10 PST ```i'm getting the feeling you don't have a firm grip on these topics. so what's the sum of {...-1/8, -1/6, -1/4, -1/2, 0, 1/3, 1/5, 1/7....}? can you provide a reference for the definition of "mean" over an infinite set (other than your own?)``` Clarification of Answer by leapinglizard-ga on 17 Nov 2005 16:01 PST ```I hold the degree of Master of Mathematics from an eminent university. I know what I'm talking about. leapinglizard``` Request for Answer Clarification by qarl-ga on 17 Nov 2005 16:34 PST ```that's very nice for you - and i have lain waste to more than one phd thesis from another eminent university. so perhaps with both our big brains, we can resolve this issue. you have repeatedly stated that summation is well defined for countably infinite sets. can you please specify this definition? and, what value does this definition provide for the sum of the alternating harmonic series? given than Riemann showed it converges to any value you'd care to make it - it doesn't seem promising that it has a "single well defined value". seriously - if you have a valid point here - please explain it.``` Request for Answer Clarification by qarl-ga on 17 Nov 2005 16:45 PST ```or, more simply - what is the sum of the set of integers, I = {..., -3, -2, -1, 0, 1, 2, 3, ...}. by one ordering: 0 + 1 - 1 + 2 - 2 +... = 0 + 0 + 0 + 0 +... i get 0. by another: 0 + 1 + (2-1) + (3-2) + (4-3) +... = 0 + 1 + 1 + 1 + 1 + ... i get inf.``` Clarification of Answer by leapinglizard-ga on 17 Nov 2005 17:14 PST ```You are correct, there are sums that cannot be evaluated. This does not mean that we cannot otherwise express them or reason about them. leapinglizard``` Request for Answer Clarification by qarl-ga on 17 Nov 2005 19:54 PST ```so - let's be clear - we're agreed that not all sums can be evaluated (i.e. put in a 1-1 "natural" correspondence with the reals.) so we're not dealing with simple numbers, presumably. so what set are we talking about here? when you say one sum is "less than" another sum, what are you talking about? can you please, PLEASE, provide me with some external reference for this "less than" relation - and proof that it is a total ordering? if you don't start answering my specific questions, i'm going to ask that we end this discussion, and you allow someone else to answer my question.``` Clarification of Answer by leapinglizard-ga on 18 Nov 2005 01:52 PST ```Very well. leapinglizard``` Clarification of Answer by leapinglizard-ga on 18 Nov 2005 02:15 PST ```I mean to say, I don't know what you mean. A total order is defined within a set, not between sets. It occurs to me that I should not have presented the proof as a reductio ad absurdum, since I don't make use of the contradicted premise anyway. It works perfectly well as a direct proof. See below. leapinglizard Claim The expected value of a random selection from the set of all natural numbers is greater than the expected value of a random selection from any finite set of numbers. Corollary The expected value of a random selection from the set of all natural numbers is greater than any given natural number n. Proof The expected value of a random selection from a finite set S is bounded from above by the greatest value in S, which we shall call m. mean(S) <= m [1] This is because no matter what number we pick from S, it will not be greater than m, the greatest member of S. Since no member of S is greater than m, it follows that the expected value cannot be greater than m either. Now, the set of all natural numbers N includes the subset T = { 1, 2, 3, ..., 2m-2, 2m-1, 2m } [2] no matter how it is ordered. Note that T has exactly 2m members. The expected value of a random selection from T is m+0.5, because the mean of T is mean(T) = (1 + 2 + 3 + ... + 2m-2 + 2m-1 + 2m) / 2m [3] = m * (2m + 1) / 2m [4] = (2m + 1) / 2 [5] = m + 1/2 . [6] But in addition to the numbers appearing in T, the set of all natural numbers N contains many more numbers, all of which are greater than 2m. Thus, the expected value of a random selection from N must be greater than m+0.5 . mean(T) < mean(N) [7] m+0.5 < mean(N) [8] But inequality [2] above tells us that m+0.5 is still greater than mean(S), the expected value of a random selection from S. Hence, mean(S) must be even smaller than the expected value of a random selection from N. By combining inequalities [1] and [8], we obtain the following. mean(S) <= m < m+0.5 < mean(N) [9] mean(S) < m+0.5 < mean(N) [10] mean(S) < mean(N) [11] Therefore, the expected value of a random selection from N is greater than the expected value of a random selection from S.``` Request for Answer Clarification by qarl-ga on 18 Nov 2005 11:51 PST ```> It works perfectly well as a direct proof. yeah, i had noticed that, too. but it still doesn't get around my central grouse, which i can pinpoint to the following line: > But in addition to the numbers appearing in T, the set of all > natural numbers N contains many more numbers, all of which are > greater than 2m. Thus, the expected value of a random selection > from N must be greater than m+0.5 . if the expected value of a random selection from N has no value (as the sum of the integers has no value) then you can't say it must be greater than m+0.5. they may be "unrelated" - as happens with non-total orderings. when you say "greater than" i can only assume you're talking about (a) the "normal" ordering on numbers or (b) some other total ordering. but we've already seen that these objects don't map nicely to numbers, so perhaps you mean "greater than" defined on some algebra of sums. but here i see no nice way to define such a total ordering - so if you do have one, you'll need to tell me what it is. or you could argue that the expected value of a random selection from N does has some well-defined (numeric) value. but as of yet, you have not provided me with a good definition, other than insisting one exists. a simple external reference to such a definition would quickly resolve this matter - but as you repeated refuse to give me one - i am left to wonder whether you're making it up. so, to summarize. i believe you have proven the following statement: The expected value of a random selection from the set of all natural numbers, if it exists, is greater than the expected value of a random selection from any finite set of numbers. but that's not the whole enchilada. in fact, it misses the most important part. this sort of problem occurs quite frequently when dealing with infinite series. the fact that you're not familiar with it is what worries me about your mathematical experience.``` Request for Answer Clarification by qarl-ga on 18 Nov 2005 12:57 PST ```gurgle... it has been brought to my attention that the question i'm asking is really very deep and not well understood: http://www.missouri.edu/~klinechair/on-line%20papers/Standard%20Decision%20Theory%20Corrected.doc as such - it's not a very appropriate question for google answers. (although, it would have been nice if someone here had sent me that link.)``` Clarification of Answer by leapinglizard-ga on 18 Nov 2005 13:22 PST ```You conceived of the experiment of picking a natural number, any natural number, under the uniform distribution. Of course this is not possible in any practical sense, but let us consider it in the abstract. I claim that the expected value n of this experiment is infinitely large. To prove this, I challenge you to name a value m that is greater than n. No matter what value you name for n, I point out that the set of all natural numbers includes the subset {1, 2, ..., 2m}, the mean of which is m+1/2. But if we add 2m+1 to this set, its mean can only increase. As you go on adding 2m+2, 2m+3, and so forth, the mean keeps on increasing. Hence, the expected value must be infinite if we can conceive of the experiment at all. This fully answers your original question. leapinglizard``` Request for Answer Clarification by qarl-ga on 18 Nov 2005 13:54 PST ```given that you have again ignored my specific questions - i ask that you disengage from this question and allow someone else to answer it. thank you.``` Clarification of Answer by leapinglizard-ga on 18 Nov 2005 14:20 PST ```I am willing to address any concerns you may have about my answer to your question. I cannot go farther afield. To minimize the scope for confusion, I succinctly rephrased the proof in my previous Clarification. I see no flaws in it. leapinglizard``` Clarification of Answer by leapinglizard-ga on 18 Nov 2005 14:24 PST ```Actually, I do see a typographical error that does not affect the argument. Where I wrote "No matter what value you name for n", please read m for n. I append an amended version for your reference. leapinglizard You conceived of the experiment of picking a natural number, any natural number, under the uniform distribution. Of course this is not possible in any practical sense, but let us consider it in the abstract. I claim that the expected value n of this experiment is infinitely large. To prove this, I challenge you to name a value m that is greater than n. No matter what value you name for m, I point out that the set of all natural numbers includes the subset {1, 2, ..., 2m}, the mean of which is m+1/2. But if we add 2m+1 to this set, its mean can only increase. As you go on adding 2m+2, 2m+3, and so forth, the mean keeps on increasing. Hence, the expected value must be infinite if we can conceive of the experiment at all.``` Request for Answer Clarification by qarl-ga on 18 Nov 2005 14:27 PST `no. sorry. please retract your answer or i will begin the refund process.` Clarification of Answer by leapinglizard-ga on 18 Nov 2005 14:33 PST ```I regret that you are not pleased with my efforts. Of course you are free to request a refund at any time. leapinglizard```
 Reason this answer was rejected by qarl-ga: ```i believe my researcher was wildly under-qualified to answer my question. it's hard to know for sure, as he quickly began changing the subject when specific questions were posed. regardless - my question requires an existence proof - which my researcher could not give me - and seemed confused as to what i was asking. as such, the answer was incomplete (and largely worthless, as it was the existence proof that i needed, not the rather simple "goes to infinity" argument which he gave. once existence is shown, any of the arguments in the comments are sufficient to show "goes to infinity".)```

 Subject: Re: uniform distribution over a countably infinite set From: ipeirotis-ga on 05 Nov 2005 18:22 PST
 ```Assume that you draw x from the set of numbers 1..N and there is a uniform probability of picking any of the numbers. Therefore Pr{x=i} = 1/N The mean is: Sum_{i=1}^N Pr{x=i} * i = Sum_{i=1}^N (1/N) * i = (1/N) * Sum_{i=1}^N i = (1/N) * N(N+1)/2 = (N+1)/2 ==> mean of the distribution = (N+1)/2 Since in your case, your set contains all natural numbers N->+oo ==> Correct, the expected value is infinity.```
 Subject: Re: uniform distribution over a countably infinite set From: qarl-ga on 06 Nov 2005 09:35 PST
 ```i'm not so sure about this proof - similar proofs have shown that the probability of selecting an even (or odd) natural number is 1/2, AND 1, AND 0 (and anything you'd like to make it) done by "choosing" different orderings on the natural numbers. now your proof doesn't suffer these contradictions - but i'll need some additional information explaining why the technique "works" for "largeness" but not for "evenness/oddness".```
 Subject: Re: uniform distribution over a countably infinite set From: ipeirotis-ga on 06 Nov 2005 21:33 PST
 ```Can you provide the proofs that show that the probability of picking an odd or even number can be 0 or 1?```
 Subject: Re: uniform distribution over a countably infinite set From: qarl-ga on 06 Nov 2005 23:24 PST
 ```consider the sequence: {1,2,4,3,6,8,10,5,12,14,16,18,20,7...} where a single odd number is followed by an exponentially increasing number of even numbers. all natural numbers are represented, and only once. in the limit, the ratio of even numbers to odd numbers is 1, and in the limit all natural numbers are covered.```
 Subject: Re: uniform distribution over a countably infinite set From: mathprof-ga on 07 Nov 2005 14:02 PST
 ```I think I see a way for you to accept ipeirotis-ga's argument. Given an ordering of the positive integers, not necessarily the usual ordering, consider a set S_n which is the first n in the ordering. You might be asking for the average value of that set S_n; i.e., you might be satisfied if we could prove that the liminf of the average value goes to infinity as n->infinity, for every ordering. (I use liminf in case we're not apriori sure that the limit exists.) But the average value of the elements of S_n is always bounded from below by (n+1)/2, because it is always bounded from below by the average using the usual ordering. So, regardless of the choice of ordering, the avg value of the set S_n always is (n+1)/2 or greater, hence tends to infinity as n->infinity. Does this work?```
 Subject: Re: uniform distribution over a countably infinite set From: qarl-ga on 08 Nov 2005 11:58 PST
 ```> Does this work? heh. the very fact that you're asking means i think it doesn't. presumably the proof should be self-evidently true. i do hear what you're saying, tho. your proof shows that ipeirotis's argument doesn't succumb to the same attack that the even/odd arguments do. but that's a far cry from saying it doesn't succumb to any attack. granted - i'm not exactly asking for a proof here - as i'm not even using valid definitions. but i'm still hoping for better insight into the problem. as such, i'm doubling the bounty. pass this along to your math-grad-student friends.```
 Subject: Re: uniform distribution over a countably infinite set From: manuka-ga on 08 Nov 2005 23:12 PST
 ```It might help to rephrase mathprof's proof - which actually does strike me as being a self-evident one to follow - without explicitly mentioning orderings, which are fundamental to this sort of thing but also make it harder to think about sometimes. 8-) So, phrase it differently: Pick an arbitrary set of n natural numbers. Then the minimum value for their sum is 0 + 1 + ... + (n-1) = n(n-1)/2, so their average is at least (n-1)/2 [in contrast to some of the other contributors I'm allowing 0 as a natural number]. As n tends to infinity, the average of *any* set of n natural numbers therefore also tends to infinity; this statement is very easy to show in an epsilon-style proof by essentially the argument just given. (i.e. Let N > 0. Let M = 2N + 1. Then for all n > M, by the reasoning above any arbitrary set of n natural numbers has average at least (n-1)/2 > (2N+1-1)/2 = N. Thus as n -> oo, the average -> oo also [recall that the definition of "as x -> oo, y -> oo" is "for any N > 0, there exists an M > 0 such that whenever x > M, y > N"].) Of course, to pick an arbitrary set of n natural numbers implies choosing the first n elements of an ordering on the natural numbers, and to let n tend to infinity implies choosing a specific ordering on the whole of the natural numbers. But it's easier to think about it in a non-technical way if you think of it in terms of random selection - which was how the original question was proposed, after all.```
 Subject: Re: uniform distribution over a countably infinite set From: qarl-ga on 09 Nov 2005 12:52 PST
 ```here's my concern - which is at the limit of my math abilities - so i'll have to trust the "true" mathematicians. are these arguments sufficient to transfer a property of finite subsets onto the complete infinite superset? i'm worried of falling into the following trap: proof - the set of natural numbers is finite. pick an arbitrary set of n natural numbers. then this set is finite. as n tends to infinity, *any* set of of n natural numbers remains finite. at the limit, the set is the set of natural numbers, which is then finite. (and - if i haven't said so before, i am much oblidged to everyone who's helped. thank you.)```