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I will give some examples of universal quantification where the "for all" aspect holds or is violated in the limit. In some examples a failure is preserved, in others it is not.

(1) $\forall$ $n \in \mathbb{N}$ the sum $\Sigma_{k=1}^n\frac{9}{10^k}$ is less than 1. The sum over all these very terms $\Sigma_{n \in \mathbb{N}}\frac{9}{10^n}$ is 1. Here it does not matter in the limit that all finite terms fail.

(2) $\forall$ $n \in \mathbb{N}$ the first $n$ digits of the antidiagonal in Cantor's list differ from the first $n$ entries of the list. And in the limit the digits of the antidiagonal differ from all entries of the Cantor list too. Here we can conclude from "all fail" to the "failure of all". Here it does matter in the limit that all finite terms fail.

(3) The $n$th level in the binary tree has $N(n) = 2^n$ nodes. The limit of $N(n)$ is infinite. Here again, like in (1) it does not matter in the limit that all finite terms fail.

(4) $\forall$ $n \in \mathbb{N}$ the number of paths in the binary tree that can be distinguished at level $n$ is finite, namely $P(n) = 2^n$. In the limit the number of paths that can be distinguished however is uncountable. Also here it does not matter that all finite terms fail to distinguish infinitely many paths. Moreover, $\forall$ $n \in \mathbb{N}$ it does not only not matter that $P(n)$ is finite and equal to $N(n)$ but the limit of $P(n)$ is much larger than that of $N(n)$.

My question is how to give a logical and consistent account justifying the differences in the respective results.

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    $\begingroup$ Are you by any chance the author of this question? matheducators.stackexchange.com/questions/12574/… $\endgroup$
    – Steven Gubkin
    Oct 26, 2017 at 18:39
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    $\begingroup$ Sure he is. Two of the twelve current incarnations of Wolfgang Mückenheim. $\endgroup$
    – Uwe
    Oct 26, 2017 at 19:41
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    $\begingroup$ I'm voting to close this question as off-topic because the user is abusing an education site to try and convince others of his theories about infinity. $\endgroup$
    – Steven Gubkin
    Oct 27, 2017 at 12:10
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    $\begingroup$ @user37237 If the question was asked honestly, that is if you were actually open to thinking about and digesting the answer, then it would still be a better fit for math.stackexchange instead of here, since it is a question about the mathematics itself, not how to teach it. However, I do not recommend posting there because you have shown no indication that you will try to comprehend the answers. In fact you seem convinced that Cantor was wrong, and that everyone else is making the same mistake. It is sad, but I think there is very little that can be done to help you. $\endgroup$
    – Steven Gubkin
    Oct 27, 2017 at 17:06
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    $\begingroup$ @user37237 You are right that a discussion is not possible in such an environment. The stackexchange websites are not designed for discussions. They are designed for questions with definite answers. These answers have been given all ready. I would certainly be willing to have a chat with you sometime about this. It seems you have not properly digested the definition of a real number, and this is at the root of your difficulties. $\endgroup$
    – Steven Gubkin
    Oct 27, 2017 at 18:23

2 Answers 2

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The logical and consistent account is to actually use logic. The limit has nothing to do with the behaviour 'before' that. When you go to the mall, you are "on the way" all the way until you reach, at which point you are no longer "on the way". $ \def\nn{\mathbb{N}} $

When $f(n) \to c$ as natural $n \to \infty$, it means nothing more or less than:

For every neighbourhood $S$ of $c$, there is some natural $m$ such that, for every natural $n > m$ we have that $f(n)$ is within $S$.

This $c$ could very well satisfy lots of properties that $f(n)$ do not satisfy for any natural $n$. This explains (1) and (3).

(4) is rather different, because the use of the term "limit" is different. You do not have $P(n)$ tending to an uncountable cardinal as $n \to \infty$. You are looking at the cardinality of the 'limit' object generated by some process, not the limit of the cardinality of the sequence of objects generated by the process. And here the 'limit' object is not quite a limit object in the earlier sense but rather a union.

Furthermore your statement of (2) is simply wrong. Ensuring different digits does not guarantee the decimal is different. $0.999\cdots = 1.000\cdots$. This shows clearly that logic is crucial to everything. We cannot reasonably talk about whether some property holds or is violated in the 'limit' without fully understanding the logical structure of that property.

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  • $\begingroup$ The statement of (2) is somewhat sloppy. What Cantor did is exactly described by: "The digit sequence $d_1,d_2,...,d_n$ differs from the correcponding digit sequences of the first $n$ entries." Since every diagonal digit $d_n$ belongs to a finite sequence we cannot say more. The conclusion to the infinite case requires the same logic as the conclusion in the other cases. $\endgroup$
    – user37237
    Oct 26, 2017 at 20:03
  • $\begingroup$ (4) is not different at all. $P(n)$ is equal to $N(n)$ for every $n$. Analysis proves that the limits are equal. Of course we have to take the analytical limit. This restricts the distinguishable paths to a countable set. $\endgroup$
    – user37237
    Oct 26, 2017 at 20:06
  • $\begingroup$ @user37237: Stop talking nonsense. I already gave a counter-example to your statement of (2) in my post. If you don't understand it, don't criticize. If you don't even want to admit that mistake, forget about talking about (4). $\endgroup$
    – user21820
    Oct 27, 2017 at 6:15
  • $\begingroup$ The diagonal argument is based on digits only and covers merely all finite places. The conclusion to the infinite limit requires same logic as in (1). Can you understand that? Your counter example would show that the diagonal argument fails. $\endgroup$
    – user37237
    Oct 27, 2017 at 9:02
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    $\begingroup$ @user37237: My counter-example only shows that you fail in understanding the correct diagonal argument. Enough. Please take an introductory course in real analysis from a proper university, or at least work through a proper textbook such as Spivak's. I won't respond any further. $\endgroup$
    – user21820
    Oct 27, 2017 at 9:49
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There are no different results of quantification. To see this we have to be clear about real numbers first.

Either a real number can be represented by an actually infinite sequence of digits (i.e., of a series of partial sums of fractions) or this is not the case. If so, then all real numbers of the unit interval $[0, 1)$ can also be represented as actually infinite and continuous paths in the Binary Tree. If not, then no real number can be represented in a Cantor-list either.

Now let's consider the four cases.

(1) $\forall n \in \mathbb{N}: \sum_{k=1}^n\frac{9}{10^k} < 1$ implies $\sum_{n \in \mathbb{N}}\frac{9}{10^k} < 1.$

How else could logic work? If we gather a set of green balls, we will not get a red cube. If we take a set of failures, we will not get a success.

We can even define: Sum all fractions $9/10^n$ that fail to yield 1. Like in every infinite sequence that means we have to take $\aleph_0$ fractions, namely one for every natural index $n \in \mathbb{N}$ that fails.

Correct is $\lim_{n \rightarrow \infty} \sum_{k=1}^n\frac{9}{10^k}= \sum_{k=1}^{\infty}\frac{9}{10^k}=1.$

(2) With this interpretation the diagonal argument holds. The infinite sequence of failures, i.e., of diagonal digits $d_n$ represents a real number that is not in the Cantor-list.

(3) With respect to the number of nodes $N(n)$ as function of the level number $n$ in the Binary Tree we apply the same logic again. Like in (1) the terms of an increasing sequence do not contain its limit. Since every $N(n) < \aleph_0$ there is no level with $\aleph_0$ nodes. Only the limit of $N(n)$ has $\aleph_0$ nodes. And since there are infinitely many levels with finite number of nodes, the complete, actually infinite Binary Tree has $\aleph_0$ nodes too.

(4) It is easy to see that at level $n$ we can distinguish precisely $N(n) = 2^n$ paths, (It is irrelevant how many may there be "in principle". We consider in mathematics what we can distinguish.) This implies that we cannot distinguish more than countably many paths in the whole Binary Tree. An upper estimate is obtained by putting all nodes of all levels of the Binary Tree on one common level, say at $\omega$.

More is not possible to distinguish. But remember that the interpretation of infinite representations by digit sequences or paths has been introduced to circumvent the lack of finite definitions. This approach has been proved to fail.

What is the answer to this seeming contradiction of uncountably many real numbers according to the diagonal argument and only countably many according to the Binary Tree? There is no actual infinity at all. There is no complete Cantor-list that could not absorb further entries, and hence there cannot be a complete diagonal number. And of course there is no complete Binary Tree either. It has only served to prove its nonexistence like Cantor's diagonal argument only has served to prove the nonenxistence of completed infinity.

Remark: This lack of completeness makes it clear that the sum over "all" natural numbers in (1) cannot yield the sum 1. We need the limit like in all such cases.

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    $\begingroup$ I have not read this whole answer, but you make a distinction between $\displaystyle \sum_1^\infty \frac{9}{10^n}$ and $\displaystyle \lim_{n \to \infty} \sum_{k=1}^n \frac{9}{10^n}$. There is no distinction between these. The limit is in fact the definition of the infinite sum. Look at any calculus book or real analysis book. So it is very hard to begin a discussion when you seem to use different unspecified definitions for even very basic concepts. $\endgroup$
    – Steven Gubkin
    Oct 29, 2017 at 14:38
  • $\begingroup$ @Steven Gubkin: Most calculus books are confusing the two limits: all terms of the sequence and the numerical limit of the sequence. If you really want to understand then try to find out how many partial sums are less than 1. The answer is: There are $\aleph_0$ terms, one for every natural index. That implies that the sum over as many terms is less than the numerical limit. Note further that $\infty$ is not a natural number. $\endgroup$
    – user37237
    Oct 29, 2017 at 21:46
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    $\begingroup$ @Heinrich No. The definition of an infinite sum is as I have said. You cannot really argue with a definition. $\endgroup$
    – Steven Gubkin
    Oct 29, 2017 at 23:56
  • $\begingroup$ @Steven Gubkin: The definition of the infinite sum differs from the sum over all terms that fail to yield the sum. What logic would it be to define: The sum over all terms that fail to yield the limit yields the limit? Often textbooks confuse both. But that does not make it correct. It shows only that mathematics is in a very bad state. Ask yourself: How many terms fail to yield the sum? instead of believing in bad textbooks! $\endgroup$
    – user37237
    Oct 30, 2017 at 6:53
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    $\begingroup$ @Heinrich We have a definition for a sum of finitely many numbers. Passing to summing infinitely many numbers requires some care. The standard definition is the limit definition I give above. Summing infinitely many numbers certainly does require a definition, because the operation $+$ is binary, and on the face of it only allows you to sum two numbers at a time. So you need to give me a definition in order to continue this conversation (which we shouldn't technically be having on this site). If you do not think you need a definition, I would say that is at the root of your difficulties. $\endgroup$
    – Steven Gubkin
    Oct 31, 2017 at 12:21

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