April 18th, 2017, 12:06 PM  #61 
Senior Member Joined: Aug 2012 Posts: 1,659 Thanks: 427 
ps  I see that your definitions come from your OP, which I never saw because I got lost down your rabbit hole about the dyadics, which as it turns out have nothing at all to do with this. And you have completely redefined $V$ in a nonstandard way. It's like redefining $\mathbb R$ to mean the set of integers. Its only purpose is to confuse people. Better to use $W$ here so $V$ can have its usual meaning. I will stick with my nonnegotiable demand and ask you for a version of your proof that is selfcontained. And I'd really like to see some motivation or explanation for your definition of $k$. Is your use of $0.5$ some sort of adjustment to avoid talking about addition mod 1? ps  Ok fine, in the interests of everyone's sanity (mine's a lost cause but perhaps some reader's) I will stipulate to your redefinition of $V$  which again I say is equivalent to calling $\mathbb R$ the set of integers and expecting to be taken seriously  and everything comes down to this: If you would be kind enough to give some examples, try to motivate this, and explain why you think it gives a uniform distribution on $\mathbb N$, we can get to the bottom of this soon. Last edited by Maschke; April 18th, 2017 at 12:22 PM. 
April 18th, 2017, 02:09 PM  #62 
Senior Member Joined: Aug 2012 Posts: 1,659 Thanks: 427 
Ok I figured this out. Your $0.5$ is yet another red herring, like the dyadics. You are doing exactly what I said you've been doing. You have the set of cosets; you have a collection $V$ of representatives of the cosets (usual $V$ not your redefined $V$). Now you have an enumeration of the rationals in the unit interval $r_1, r_2, \dots$ For any $x \in [0,1)$, $x \in V + r_n$ for exactly one rational $r_n$. This is what you're doing when you take $x  v \in \mathbb Q$ and then map that rational back to $\mathbb N$ via some bijection. And now, you can map that rational back to the natural number $n$. The problem is that each of the translates $V + r_n$ is nonmeasurable. How do we know this? Because $V$ is nonmeasurable and our measure is preserved by translation. It's uniform. So you are only selecting each natural with the "same" probability in the sense that there is no probability at all. You can't say that one nonmeasurable quantity is the "same" or "different" than another, because nonmeasurable sets can't be assigned a sensible measure or probability. This is the exact problem I've pointed out to you many times already. It's true that you have a map that assigns some natural number to each real. The problem is that the inverse image of each natural is a nonmeasurable set. It can't be said to be the same or different than some other nonmeasurable set because it has no measure at all. Last edited by Maschke; April 18th, 2017 at 02:19 PM. 
April 18th, 2017, 03:33 PM  #63  
Senior Member Joined: Jun 2014 From: USA Posts: 316 Thanks: 22  Quote:
For the record, I do not consider an undefined probability as being "no probability at all." If a Vitali set were to be a measurable set as opposed to nonmeasurable, we'd have a uniform distribution over $\mathbb{N}$ in the traditional sense, but since that is not possible given that the sum of the probabilities would be infinite, I suppose this is a backasswards way of showing that Vitali sets are nonmeasurable as you said earlier. My confusion over what you meant by "rational translate" prevented me from understanding you (I thought you meant "shifted copy of the rationals = an equivalence class" when you really just meant to assert that V + r = "another Vitali set"). Ok. The 0.5 thing was to enable me to define a single function $k$ that would map each real to a natural. Without it, I couldn't construct such a function. Now that it's probably pointless, as you were typing this I redid the proof like you asked (as though I was going to try and make publishable) and took your previous suggestions into account. I think my notation gets better and my line of thinking more clear every time I do this, so at least that's something. Here was the latest version: Introduction: It is helpful to consider a simple observation pertaining to finite sets in order to get started. Let $A = \{1, 2\}$ and $B = \{3, 4, 5, 6\}$. Let $f(x) = 1$ if $x$ is odd and $f(x) = 2$ if $x$ is even. Then, function $f$ is a surjection from $B$ onto $A$ that is “uniform” in the sense that selecting an element $x \in B$ uniformly at random will result in the selection of $f(x) \in A$ uniformly at random as well. Note that in order to do this, we have effectively partitioned $B$ into subsets the same size as $A$ so that we could biject those subsets with $A$. Likewise, this work shows how we can partition $[ \,0, 1) \,$ into countable sets that are then bijected with the natural numbers. Consideration is then given to whether selecting an element of $[ \,0, 1) \,$ uniformly at random allows for the selection of a natural number uniformly at random as well. Definitions: Let $V^{( \,0.5, 1) \,}$ be a set containing one and only one element from each Vitali equivalence class on the interval $( \, 0.5, 1 ) \,$ (Vitali equivalence classes are equivalence classes of the real numbers that partition $\mathbb{R}$ under the relation $x \equiv y \iff ( \, \exists q \in \mathbb{Q} ) \, ( \, x  y = q ) \,$). The axiom of choice allows for such a selection. For any real number $r$, let $d(r)$ equal the one and only one element $v \in V^{( \,0.5, 1) \,}$ such that $r  v \in \mathbb{Q}$. Let $h : \mathbb{N} \longmapsto \mathbb{Q} [ \, 0, 1 ) \,$ (here $\longmapsto$ denotes a bijection). Let $x$ be an element of $[ \,0, 1) \,$ selected uniformly at random. Let $k(x) = \begin{cases} h^{1}(x  d(x) + 0.5) && x \geq d(x)  0.5 \\ h^{1}((x + 1)  d(x) + 0.5) && x < d(x)  0.5 \end{cases}$. For each natural number $1, 2, 3, …$, let $V^{1}, V^{2}, V^{3}, …$, respectively, be the sets such that $( \, \forall x \in V^{n} ) \, ( \,k(x) = n ) \,$. We then have $V^{( \,0.5, 1) \,} = V^{h^{1}(0.5)}$, for example. Each $V^{n}$ will be a Vitali set by definition with the collection $\{ V^{n} : n \in \mathbb{N} \}$ forming a partition of $[ \,0, 1) \,$. Comments on Uniformity: A uniform distribution is a concept of translation invariance. For example, if $S$ is a measurable set, we may want the probability of $S$ to be the same as the probability of $\{y : y = z + n, z \in S \}$ for each natural number $n$. In the case of function $k$ over the domain $[ \,0, 1) \,$, however, we end up mapping each element of each nonmeasurable Vitali set $V^{n}$ to a distinct natural number $n$. Where $a, b \in \mathbb{N}$, it is easy to see that the probability of $x$ falling within $V^{a}$ is equal to the probability of $x$ falling in $V^{b}$, but we cannot rely on a Lebesgue measure as a means of establishing probability or creating any sort of cumulative distribution function on $\mathbb{N}$. The probability of selecting any given natural remains undefined.  
April 18th, 2017, 04:29 PM  #64  
Senior Member Joined: Jun 2014 From: USA Posts: 316 Thanks: 22  Quote:
 
April 18th, 2017, 04:29 PM  #65  
Senior Member Joined: Aug 2012 Posts: 1,659 Thanks: 427  Quote:
However you are mathematically wrong when you talk about the equality or inequality between two undefined quantities. Such a thing is not allowed in mathematics. It's meaningless. You are of course entitled to have your own personal mathematics in which you may compare two undefined quantities. But in so doing, you depart from standard mathematics; and certainly from anything I can talk to you about. I don't know anything about the mathematics in your head. I know a bit about the mathematics outside your head. And we have made progress this morning so I am heartened. You have to choose whether you are going to keep talking to yourself about your own personal mathematics; or whether you would like to talk to others about standard mathematics. Quote:
And as I recall, this all started a few years ago when you got this idea for a uniform distribution on $\mathbb N$, then a few weeks ago when I told you about $\mathbb R / \mathbb Q$. But you know the idea doesn't work because the rational translates of a choice set for $\mathbb R / \mathbb Q$ must be nonmeasurable. Now we know what that coolsounding phrase means. Quote:
You have to make a choice here. Standard math or your own personal math. Which are you interested in? Quote:
So what most people do is they accept this fact, put it in their back pocket, and go ahead and learn some more measure and integration theory. The REAL point of all this is to get Lebesgue integration off the ground, and then functional analysis, and then quantum physics. The goal is to get to the $L^p$ spaces, not slog around fighting with logic. SOME people prefer to just have a personal fight with mathematics. This I could never understand. And yet I'm personally drawn to try to explain math to these people. It's strange, I've always been this way. Quote:
But we can think of it this way. We start with the reals; we mod out the rationals; then we translate a choice set by the rationals and we get back the reals. We go from having an uncountable union of countable sets; and we end up with a countable union of uncountable sets. Quote:
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I'll be happy to make any constructive suggestions I can, stylewise.  
April 18th, 2017, 05:06 PM  #66 
Senior Member Joined: Aug 2012 Posts: 1,659 Thanks: 427  
April 18th, 2017, 09:11 PM  #67 
Senior Member Joined: Aug 2012 Posts: 1,659 Thanks: 427 
I really like what you wrote. I could quibble at this and that (and I will when I get around to it. For now I'm quibbled out) but if I saw your post on a forum I'd say that's a very nice piece of mathematical exposition. I really like your analogy of finite uniform surjections. That's a perfect way to get people to understand the big (and strange) picture of locating a random real in one of the rational translates of the Vitali set and mapping it back to a natural number. I'm beginning to come around to your philosophical point of view. Even though we don't have a measure, we do at least have a mapping. Once we choose a bijection or enumeration the rest is straightforward. And no path through the process is privileged in any way. Every natural has an "equal chance" of being picked. So when we prove that "equal chance" can not be sensibly defined in this situation, the mind rebels. We want to say $\frac{1}{\infty}$. But we can't. Once again we are at the mystery of the infinitesimal. Infinitesimals are banned from contemporary math but not from our intuition. If you don't mind my saying, for someone untrained in higher math you have great insight into this proof and good facility with abstractions in general. The nonmeasurable set is generally taught to first year grad students and it doesn't go down easily! It's a proof with a lot of moving parts. It's a challenge to visualize the translations and you saw them instinctively. That's a little uncanny in fact. You are to be commended for all this. Last edited by Maschke; April 18th, 2017 at 09:23 PM. 
April 19th, 2017, 04:23 PM  #68  
Senior Member Joined: Jun 2014 From: USA Posts: 316 Thanks: 22  Quote:
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