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April 18th, 2017, 11:06 AM   #61
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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 non-negotiable demand and ask you for a version of your proof that is self-contained. 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:

Quote:
Originally Posted by AplanisTophet View Post

If $x < v - 0.5$, then $k(x) = h^{-1} ( \,(x + 1) - v + 0.5) \,$.
If $x \geq v - 0.5$, then $k(x) = h^{-1} ( \,x - v + 0.5) \,$.
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.
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Last edited by Maschke; April 18th, 2017 at 11:22 AM.
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April 18th, 2017, 01:09 PM   #62
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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 non-measurable. How do we know this? Because $V$ is non-measurable 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.
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Last edited by Maschke; April 18th, 2017 at 01:19 PM.
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April 18th, 2017, 02:33 PM   #63
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Originally Posted by Maschke View Post
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 non-measurable. How do we know this? Because $V$ is non-measurable 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.
Yes. That is why I provided the layman's definition of "select an element of an infinite set uniformly at random" in the OP. The distribution is uniform in the sense that yes, each natural does have an equal probability of being selected, however, that probability is undefined. You'll recall that this all started because I said that $\frac{1}{\infty}$ is undefined in the other thread...

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 non-measurable, 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 back-asswards way of showing that Vitali sets are non-measurable 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 non-measurable 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.
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April 18th, 2017, 03:29 PM   #64
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Quote:
Originally Posted by AplanisTophet View Post
Also, given:
...
3) a definition of "select an element of an infinite set uniformly at random" that simply means, in layman's terms, that one and only one element of an infinite set will be selected using a process where all elements of the set have an equal chance of being selected, then it is possible to select a natural number uniformly at random...
Quote:
Originally Posted by romsek View Post
of course you are. I will gladly pay you on Tuesday.
So um, yeah, it’s Tuesday. Given my above post (undefined, sure, but equal), can I get paid now or what?!
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April 18th, 2017, 03:29 PM   #65
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Quote:
Originally Posted by AplanisTophet View Post
Yes. That is why I provided the layman's definition of "select an element of an infinite set uniformly at random" in the OP. The distribution is uniform in the sense that yes, each natural does have an equal probability of being selected, however, that probability is undefined.
I'm glad we are on the same page now.

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:
Originally Posted by AplanisTophet View Post
You'll recall that this all started because I said that $\frac{1}{\infty}$ is undefined in the other thread...
I have no such recollection, because it is my habit to ignore all these undefined expression threads on the Internet. The topic doesn't interest me.

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 cool-sounding phrase means.

Quote:
Originally Posted by AplanisTophet View Post
For the record, I do not consider an undefined probability as being "no probability at all."
If you depart from standard math and decide to talk about your own personal math, that is your right, but I would not be able to come along for that ride. My value here is as someone who knows the standard catechism. I use that word deliberately. I don't hold that standard math is right. Only that it's standard math. You want to talk standard math, I'm here. You want to make stuff up, I wouldn't be able to participate.

You have to make a choice here. Standard math or your own personal math. Which are you interested in?

Quote:
Originally Posted by AplanisTophet View Post
If a Vitali set were to be a measurable set as opposed to non-measurable, 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 back-asswards way of showing that Vitali sets are non-measurable as you said earlier.
Well the moral of the story is usually expressed like this: There is no uniform measure on the real numbers that is countably additive and that normalizes the unit interval. That last phrase just means that it assigns the value $1$ to the unit interval.

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:
Originally Posted by AplanisTophet View Post
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.
Here I will confess to throwing some jargon at you. It took me a long time to grok the rational translates of $V$. There's nothing wrong with finding all these concepts slippery and hard to get one's mind around.

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:
Originally Posted by AplanisTophet View Post
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
Instead of using a bijection, the standard proof uses an enumeration of the rationals to produce an enumeration of the translates. It's really a matter of style, your idea of a bijection is perfectly ok. Perhaps enumerations are easer to visualize than $h$ and $h^{-1}$ but there's really nothing wrong with your idea of using a bijection.

Quote:
Originally Posted by AplanisTophet View Post
, 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:
I'll take a look at this later. At this point we agree that according to standard math, we can't compare undefined quantities, hence we have not found the mythical (and impossible) uniform distribution on $\mathbb N$.

I'll be happy to make any constructive suggestions I can, style-wise.
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April 18th, 2017, 04:06 PM   #66
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Originally Posted by AplanisTophet View Post
The probability of selecting any given natural remains undefined.
You brought joy to my heart.

Your exposition is much, much better. More comments later.
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April 18th, 2017, 08:11 PM   #67
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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.
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Last edited by Maschke; April 18th, 2017 at 08:23 PM.
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April 19th, 2017, 03:23 PM   #68
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Quote:
Originally Posted by Maschke View Post
You brought joy to my heart.

Your exposition is much, much better. More comments later.
Quote:
Originally Posted by Maschke View Post
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.
That truly brings joy to my heart as well.
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