I believed all the mechanics of Cantor when studying in grad school, probably because the faculty and textbooks support each other. However, I am less certain now, particularly since much of the theory relies on de facto arguments as to Lebesgue measure or cardinality.

I may be missing the obvious, but I present the following.

Let $r_1 = \frac{1}{4}$ and $r_2 = \frac{1}{2}$. Then $2 (r_1 - r_2)$ has twice the measure precisely because $2r_1$ and $2r_2$ are twice as far apart as $r_1$ is from $r_2$.

We accept the bijection as preserving the cardinality between the two intervals, thinking for any real number $X$ such that $2r_1 < X < 2r_2$, it must be true that there exists $2r = X$ such that $r_1 < r < r_2$.

If the interval is a continuum of points between $r_1$ and $r_2$, then there exists $r_3$ such that $r_1$ and $r_3$ are 'right next to each other. We can never identify $r_3$ distinctly any more than we can distinctly identify all the points mapped from $[r_1, r_2]$ to $[2r_1, 2r_2]$.

However, it must then be true that $2r_3 - 2r_1 = 0$. Otherwise, exists an $X$ such that $2r_1 < X < 2r_3$ and $r_1 < \frac{X}{2} < r_3$, contradicting the conditions of a continuum of points.

Zeno? This can be repeated for every point between $r_1$ and $r_2$, so that there is no distance between $r_1$ and $r_2$.

Status quo is an uncountable number of points (i.e. an uncountable number of zeros) can add up to a positive Lebesgue measure. This is because there is a Lebesgue measure for $r_2 - r_1$ and the measure of countably infinite zeros are argued to be zero, so de facto the answer for adding up an uncountable number of zeros in this case is the value of $r_2 - r_1$. I no longer find this type of reasoning sufficient.

Upon closer examination, I don't accept there is any level of infinity that accumulates a positive measure when the elements accumulated are of measure zero. All the points would fit upon themselves. The Greek 'infinitesimals' make more sense to me. There must be some distance between 'the center' of any two points, no matter how small, else the aforementioned paradox results.

Perhaps the problem is the tendency to project what is perceived physically into math proofs of difficult to grasp concepts. For example, a rubber ball can be squeezed, in theory to a point, an argument abstracted in topology. However, there is a shocking amount of space between any two molecular or atomic components that form the underlying structure of the rubber ball. If the components were pressed so tightly as to remove all the excess space and form a ball, assuming it would be stable, there is no squeezing a ball down to point or anything close to that.

Moreover, the limit as $n$ approaches infinity of $n(1/n)$ is $1$, but $(\lim n)(\lim 1/n)$ is zero. If $\lim 1/n$ is an infinitesimal, so that any finite number is smaller than any epsilon greater than zero, then the two representations above are equal.

If I follow this uneasy line of inquiry, I then have $1 / (\lim 1/n)$ equal to $\aleph_0$ and some similar representation exists with respect to $\aleph_1$.

I might mention this started when I realized the Archimedean property implied a rational number existed between every real number in a Vitali set, thereby bounding the cardinality of the Vitali set. However, the status quo is $\aleph_0$, each with $\aleph_0$ members, form a bijection with Cantor's "dynamic triangle"--that is his diagonal forms a triangle with the number of elements of the base approaching $\aleph_0$. I find that the number of elements in this dynamic triangle increase too quickly for a bijection with the natural numbers. This alleviated the conundrum introduced by the Archimedean property when applied to elements of a Vitali set, but opened the door for the issues presented here.

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"Relies on de facto arguments as to Lebesgue measure or cardinality". I honestly have no idea what that could possibly mean. –  Arturo Magidin Mar 27 '12 at 16:22
Your first serioues error lies in the alleged implication: "If the interval is a continuum of points between $r_1$ and $r_2$, then there exists $r_3$ such that $r_1$ and $r_3$ are 'right next to each other'." How does the premise allegedly imply the consequent? What is it about "continuum" that implies a well-ordering? And the consequent is, in fact, incorrect: between any two distinct real numbers there exist a real number distinct from both. (There are minor errors before, such as confusing numbers with measures; $r_1-r_2$ is a number, not something that has a measure). –  Arturo Magidin Mar 27 '12 at 16:25
Just out of curiosity, what did you study in the university? (Mathematics? Engineering?) What level? (Undergrad? grad? you have a PhD?) What mathematical content did you focus on? (Set theory? Analysis? Applied mathematics?) the answers to this might very well help me shape an answer to your question. –  Asaf Karagila Mar 27 '12 at 18:36
The Greek philosopher’s name was Ζήνων, with initial zeta, so the English is Zeno, not Xeno. –  Brian M. Scott Mar 28 '12 at 2:26
@Brian: Maybe Zeno had a child with Xena? –  anon Mar 28 '12 at 2:30

As others have noted, it’s not clear what your question actually is. It is clear, however, that it rests on some profound misunderstandings. I’ll deal with as many as I can.

Let $r_1=\frac14$ and $r_2=\frac12$. Then $2(r_1−r_2)$ has twice the measure precisely because $2r_1$ and $2r_2$ are twice as far apart as $r_1$ is from $r_2$.

If, as appears to be the case, by measure you mean Lebesgue measure, your statement makes no sense. (Some) subsets of $\Bbb R$ are measurable and have an associated measure. However, $2(r_1-r_2)$ is simply the number $-\frac12$; it is not a subset of $\Bbb R$, so you can no more talk about its measure than you can about its color. You can talk about the measure of the set $\{-1/2\}$, but that’s clearly not what you have in mind, since it’s $0$. Perhaps you mean the magnitude of $2(r_1-r_2)$, i.e., $|2(r_1-r_2)|=1/2$, which is indeed twice the magnitude of $r_1-r_2$.

If the interval is a continuum of points between $r_1$ and $r_2$, then there exists $r_3$ such that $r_1$ and $r_3$ are 'right next to each other.

Utterly false: the existence of such points would flatly contradict the assertion that the interval in question was a continuum. There are several mathematical notions of continuum, but the most relevant here is that of a linear continuum, which by definition is densely ordered, meaning precisely that it does not have two adjacent elements. This of course completely vitiates your attempt to construct a Zeno-like ‘paradox’. Before venturing any further into this territory, you really ought to get a better grasp of the order structure of the real line. It has its own complexities, but I consider it substantially simpler than Lebesgue measure.

Upon closer examination, I don't accept there is any level of infinity that accumulates a positive measure when the elements accumulated are of measure zero.

What you choose to accept is entirely up to you. However, your failure to accept that Lebesgue measure works the way it does merely shows that you don’t understand it. There’s nothing wrong with this: it’s not a particularly simple construct. It is, however, a bit foolish to suppose that everyone is out of step but you.

Moreover, the limit as $n$ approaches infinity of $n(1/n)$ is $1$, but $(\lim n)(\lim 1/n)$ is zero.

The first statement is true; the second is not. Since $\lim_{n\to\infty}n=\infty$ and $\lim_{n\to\infty}1/n=0$, your second expression is undefined, not $0$. Nor is there anything wrong with this $-$ to anyone who has actually gone through and understood the mathematics of limits.

If $\lim 1/n$ is an infinitesimal, so that any finite number is smaller than any epsilon greater than zero, then the two representations above are equal.

There are in fact rigorous ways to extend the real numbers with infinitesimals, but in those extended systems it remains true that $\lim_{n\to\infty}1/n=0$.

I might mention this started when I realized the Archimedean property implied a rational number existed between every real number in a Vitali set, thereby bounding the cardinality of the Vitali set.

This is true, but almost certainly not in the way that you believe. The actual result is that if $X$ is any linearly ordered set, and $Y$ and $D$ are subsets of $X$ such that (1) $|D|=\omega$, and (2) between any two members of $Y$ there is a point of $D$, then $|Y|\le 2^\omega$. In other words, the only bound imposed on the cardinality of your Vitali set is one that you already had trivially by virtue of the fact that it’s a subset of $\Bbb R$: its cardinality is at most $2^\omega$.

However, the status quo is $\aleph_0$, each with $\aleph_0$ members, form a bijection with Cantor's "dynamic triangle"--that is his diagonal forms a triangle with the number of elements of the base approaching $\aleph_0$.

This is incomprehensible, I’m afraid. My best guess is that you’ve somehow conflated the usual argument that the reals are uncountable with one of the proofs that the set of rationals (or perhaps $\Bbb N\times \Bbb N$) is countable, but it’s not at all clear what you mean by ‘Cantor’s “dynamic triangle”’.

I find that the number of elements in this dynamic triangle increase too quickly for a bijection with the natural numbers.

Since I don’t yet know what this ‘dynamic triangle’ is, I can’t say whether your finding is correct or not. I will point out, however, that personal incredulity weighs nothing in the balance against mathematical proof.

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+1, I was going to write this sort of answer, but I really have nothing to add on yours! I can contribute with two possibly useful links for the last part about $\aleph$ numbers and calculus: My answers to this question and to that question. –  Asaf Karagila Mar 28 '12 at 10:47
The linear continuum reference does not preclude that a rational number exits (Archimedean Property) between any two members. Therefore, the answer avoids rather addresses the cardinality of the rationals as implied by a Vitali set. Cantor does not enumerate a matrix of countably infinite rows and columns that represents the rational numbers, he enumerates a triangle within the matrix. It's dynamic because the number of elements in the triangle increase with each additional diagonal. –  Pat Mar 28 '12 at 14:52
@Pat: Your implication is invalid. For example, a rational exists between any two real numbers (or elements of any subset of $\mathbb{R}$, really), but $\mathbb{R}$ is still uncountable; there are still (uncountably many!) real numbers between any two reals, too. Each row of the (lower) "triangle" of rationals is finite, so a listing of $\mathbb{Q}$ is easy: enumerate the 1st row, then the 2nd, then the 3rd, etc. and every rational will appear on this list, how "long" it takes is moot. Finally, try not to use such terms without explanation (unless you want to alienate your readers). –  anon Mar 28 '12 at 16:38
@Pat: Your original post talked about a bound on the cardinality of a Vitali set imposed by the rationals, so I addressed that. Now you talk about the reverse, namely, what the existence of a Vitali implies about the cardinality of the rationals. Beyond implying that there are infinitely many rationals, which we certainly knew anyway, it implies nothing. As for the enumeration of the rationals, you’re simply mistaken: the entire matrix is enumerated by the diagonal enumeration. And there is nothing dynamic about the enumeration; it can be given explicitly by equation. –  Brian M. Scott Mar 28 '12 at 20:08

Yes, there is something like Zeno's paradox for measure of infinite sets. The measure of a set containing a single point {$x_\alpha$} is 0, yet the union of uncountable many such sets can make up the interval from 0 to 1, $$\bigcup_{x_\alpha\in (0,1)}\{x_\alpha\}=(0,1),$$ which has measure 1. The measure of any finite subset of these sets is 0, but add them all together and the measure is 1.

This is unavoidable, just like Zeno's paradox with limits. Measures of unions of sets only behave under countable unions, and not uncountable ones.

Your statement about limits is incorrect. $$\lim_{n\rightarrow \infty} n\frac{1}{n} = 1$$ However ,$$(\lim_{n\rightarrow \infty} n )(\lim_{n\rightarrow \infty} \frac{1}{n} )=(\infty)(0),$$ which is simply not well defined, and is not 0. It is a similar issue as above, only finite limits behave nicely.

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I appreciate the answers, but no one has addressed the question. There is no magical arm waving that answers such conundrums any more than one can assert the excluded middle with respect to the the existence or non-existence of God, conditional in the current information set. –  Pat Mar 27 '12 at 22:40
So, Pat, what is the question? I searched for a question mark in your exposition (it was too long for me to actually read it) and only found one, and the "sentence" with that question mark reads, in full, "Xeno?" Maybe you could summarize your "question" with an actual question, so we could better address it (and maybe you could deal with Arturo's comments as well). –  Gerry Myerson Mar 27 '12 at 22:57
I guess I am not sure what your question is. –  user16124 Mar 28 '12 at 1:43
@user16124 You're taking the first limit w.r.t $x$, it should be w.r.t $n$ instead. Trivial correction, but I thought I'd point it out. –  ThisIsNotAnId Mar 28 '12 at 3:31