Room for Dejan Govc and Wolfgang Mueckenheim - 2013-01-19

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12:06 AM

This is a continuation of the discussion that began in the comments at math.stackexchange.com/questions/280595/…

12:23 AM

So to summarize: we were discussing infinite strings of digits. One of the main question was: why is the cardinality of the set of finite strings of digits $\aleph_0$, while the cardinality of the set of all strings is $2^{\aleph_0}$? One of the main examples we examined was the infinite string $333\ldots$, which is a union of its finite substrings $3, 33, 333, \ldots$ I believe we so far agree on the following: "Each infinite string of digits is the union of its finite initial segments."

12:33 AM

@WolfgangMueckenheim: for TeX support in chat, see this.

9 hours later…

9:34 AM

Hi Dejan,

Thank you for setting up this chat room. Let me first explain what I understand by a real number. According to Kant we cannot know "the thing itself" but only the impression exerted upon us. Same is valid for numbers _if they exist in a platonic way soemwhere outside of us_. Therefore we can only talk about the _names_ we give them when we think to talk about numbers in mathematical dialogue. (Otherwise, from a non-platonic point of view, the numbers are nothing but their names that live in a system of certain rules that have been set up by us, mainly taken from properties of rea…

1 hour later…

10:48 AM

@BelsaZarkin: Hi Wolfgang,

welcome to chat. I can agree with you that 0.333... is a word consisting of eight digits. But I don't agree that it describes a number. To describe a number, we have to know what exactly *"and so on"* means. Otherwise I might think that 0.333... stands for either 0.33303330333 or perhaps for 0.333666666999999999 or perhaps some other number, which unlike these two, might even have infinitely many digits, whatever that might mean. (Note that the two numbers mentioned are obviously not equal, so using the same symbol for them will get us into trouble.) Instead, to …

11:43 AM

Hi Dejan,

Hello. =)

of course you are right unless "..." is defined as "and continue in a way that is 'obvious'". If someone meant 0.333444555... ans_so_ on, he would try to cheat. Nevertheless this question is not important. We can drop 0.333... if you don't find it clear enough. Let's keep 1/3 or f: N --> {0,1,...,9} for all n f(n) = 3 (Notice, also here we have used "...". What I find important is that we can only use finite strings to describe real numbers in a unique way. Do you agree? Regards, WM

Yes, I agree. We can only use finite strings since we are not infinite beings. In this way, "f: N ---> {0,1,2,3,4,5,6,7,8,9}, f(n) = 3" (which I'd suggest to be our main example, if you agree) is a finite string of symbols. However, it conveys an idea of something infinite. It is just a word that describes some object. The word "rabbit" is also just a string of 6 symbols, but it describes an animal.

11:59 AM

It conveys an idea of something infinite iff we want to represent it by digits. If we use the ternary system, it is rather finite. But if we use digits, we know we would need infinitely many, but we cannot. All that 1/3 as decimal contains is the set of finite initial segments (this term will apear frequently, therefore I propose to use FIS). In the same way every natural number is finite and belongs to a finite initial segment of naturals (FISON). Anything extending beyond

Anything extending beyond could not be treated by means of FISs or FISONs. Do you know the Binary Tree which contains all reals of the unit interval as infinite paths? It nodes are countable. If I construct a paths through every node, then you cannot discern any further path by means of nodes although I have used only countably many paths.

Anything extending beyond could not be treated by means of FISs or FISONs. Do you know the Binary Tree which contains all reals of the unit interval as infinite paths? It nodes are countable. If I construct a paths through every node, then you cannot discern any further path by means of nodes although I have used only countably many paths. - timeout - retry/ cancel

Aren't there uncountably many paths in that infinite binary tree?

As concerns the digits of 1/3, one can prove that the n-th digit is equal to 3 for every natural number n. This is the reason why we represent it as 0.333...; but this imprecise symbol also has a precise meaning. It is the infinite sum $\sum_{n=1}^\infty \frac{3}{10^n}$ of the sequence 3/10^n (I also describe this in this answer of mine).

But more importantly, can you explain why you think there are only countably many infinite paths in the infinite binary tree?

12:59 PM

Sorry, our server was down.

Welcome back.

I can explain that even with infinitely many levels it is impossible to distinguish more than countably many path by nodes. So if you believe that there are more, they must be distinguished in another way.

What exactly do you mean by "distinguish [them] by nodes"?

I prove my assertion in the following way: In the Binary Tree I construct a FIS of path to every node. Then I append an infinite tail of my choice (like 000... or 111... or 010101... or the bit sequence of pi) but I don't let you know what I have appended. As you are not able to discern any further path, it is clear that further paths cannot be defined by nodes. On the other hand, everything in a Cantor-list is proved by nodes or digit at finite places.

If I tell you that I use only tails 000..., then you can say that the path for 1/3, namely 0.010101... is missing in the tree, although you cannot find a node of this path that is missing. This means, you cannot find out by considering the covered nodes whether 1/3 is in the Binary Tree (since I used it as tails) or not. That I mean by distinguishing by nodes.

Sorry for another inconvenience, I have to leave for some hours due to professional obligations. I think I'll be back in the evening. Regards, WM

If you agree, we may think in terms of a decimal tree. So, for example, for the sequence 333... (by which I again do not mean this literally, but instead the function f: N ---> {0,1,2,3,4,5,6,7,8,9}, f(n)=3) we would always choose the node 3. This defines an appropriate path. If I choose any node in this path, from that node on, I still always choose 3. So, the path for 333... is not in fact missing. (If we are indeed talking about infinite paths.)

Ok, see you in the evening.

3 hours later…

4:39 PM

Hi Dejan,
yes you have defined the path f: N ---> {0,1,2,3,4,5,6,7,8,9} with f(n)=3, or 0.333... or 1/3 or 10/30. But you did it with a _finite_ definition. In fact you cannot find out from the covered nodes of the decimal tree that I present you, whether I used the path of 1/3 or not, can you? I do not deny that 1/3 exists, but all its nodes at finite levels of the tree can be covered by paths which differ from 1/3, namely by all its FISs. 0.3000..., 0.33000..., ... You can say: 1/3 is the path that has no zeros whereas all FISs have zeros. In _every_ FIS I can find an index with digit 0 …

Hi, welcome back.

Do you have an alarm clock for this room?

No, I was just reading some articles on the internet, and I noticed your message appearing.

(A (1) appears next to the title of the page, if there is a new message.)

A, I see. Thank you.

Hmmm ... Interesting. But the string 333... can indeed be distinguished from any other string: suppose I have some other string g: N ---> {0,1,2,3,4,5,6,7,8,9}. Then the string g differs from the string f: N ---> {0,1,2,3,4,5,6,7,8,9}, f(n) = 3 if and only if there is a natural number n0 such that f(n0) is different from g(n0). I.e. two strings are different if and only if we can distinguish them at some finite level.

Right, so if I understand correctly, what bothers you here is the leap from this g to all g?

So, the basic claim is that there is a profound difference between "for every x, P(x) is true" and "P(x) is true for all x"?

5:00 PM

It bothers me that in fact every number is given by a finite definition. (Even a finite string like 0.333 requires the implicit understanding that only zeros will follow.) Now, there are only countably many definitions because the definitions (necessarily finite, if applicable) are a subset of the set of finite words.

Further I can cover the decimal tree in that way that I append to every FIS of a path not only one infinite tail like 010101... but all infinite tails which can be finitely defined. Then I cover the tree so densely that you cannot even talk about a further path. But I have used only countably many.

Yes, the basic claim is that there is a profound difference between "for every x, P(x) is true" and "P(x) is true for all x"? Consider a Cantor list containing all FISs. For every FIS the diagonal will deviate at a finite place. For all FISs the diagonal cannot deviate at a finite place because all FISs are there. It is like the decimal tree. It contains all FISs.

Right, I believe that's one of the consequences of Cantor's diagonalization: not every real number can be described by words. This also seems to agree with my intuitive understanding of numbers: every number can be seen as an infinite string of digits. However, it is conceivable that there might be no pattern in such a string. In that case, a string would still exist (in a Platonic sense) but there would be no finite way of describing it.

In order to exclude the idea of quantifier exchange better say: "for every x, P(x) is true" and "for all x, P(x) is true"

About the Cantor list of FISs: yes we can enumerate all possible FISs. But Cantor diagonalization will yield a new real number, which is not a FIS.

(Since it differs from every FIS.)

(You can try this explicitly on a piece of paper.)

Here I disagree: Cantor's argument concerns only differences d_nn =/= a_nn at finite places. We cannot save it by accepting undefinable reals. Cantor's argument is in direct contradiction with the decimal tree as far as paths are defined by nodes. All finite paths belong to a countable set. And no diagonalization within the tree is possible. All we can do is to escape to paths which are longer than all FISs.

But 0.333... is not an undefinable real. It is defined by f: N ---> {0,1,2,3,4,5,6,7,8,9}, f(n) = 3. And yet it differs from every FIS, so it might as well be the number generated by the diagonal argument.

5:14 PM

I know that the diagonal differs from every FIS, but it cannot differ from all FIS - by definition of all. On the other hand it is Cantor's argument that the diagonal always differs at a finite place. Elsewhere it is impossible.

How do you define "all"?

0.333... is definable - but not by nodes. And it does not differ from all its FISs. Therefore it cannot be generated by the diagonal argument. Remember: For every n there is a FIS of 1/3 in the list. What is 1/3 more than all its FISs? Yet a digit behind all finitely indexed digits?

I define "all" FISs of 1/3 by the sequence 0.3, 0.33, 0.333, ... I define all FISs of all numbers by all functions from (1, 2, ..., n) --> {0, 1, ..., 9}, for all n in N.

I believe 0.333... differs from all its FISs. I justify this claim as follows: a FIS is a g: N ---> {0,1,2,3,4,5,6,7,8,9} for which there is a natural number n0, such that g(m) = 0 for all m >= n0. In particular g(n0) = 0. On the other hand, f(n0) = 3, for our string f. Therefore the string defined by f(n) = 3 differs from all possible FISs. Where exactly do you disagree?

(And: would you agree with the definition of FIS I just presented?)

5:29 PM

If you say 1/3 differs from all possible FISs, then you use in fact a proof for every possible FIS. All FISs of 1/3 contain all digits 3 that reside at finite indices. And more is not possible. Remember the decimal tree: You cannot distinguish whether it has been constructed using 1/3 or using only its FISs. If 1/3 would differ from all its FISs, shouldn't you be able to answer the question whether or not 1/3 has been used for construction?

Yes I would agree with your definition of FIS of 1/3.

Ok, in that case, I believe i still do not understand your definition of "all". What exactly does the statement "for all x, P(x) is true" mean to you?

In current logic for all is the same as for every. Your argument correctly states that the string defined by f(n) = 3 differs from every possible FISs because every possible FIS has some zeros - infinitely many, by the way. I look at the digits of 1/3 and see that each one is at a finite place. Hence each one is within a FIS. And I state: For every digit 3 of 1/3 there is a FIS of 1/3 that contains this digit (and ten times more, if you like).

(In particular, I am interested, what would consistute a proof of such a statement?)

Therefore if these FIS are used to construct the path of 1/3 in the decimal tree then nobody can distinguish that path from the path of 1/3.

My proof lies in the fact that nobody can find out from the completely constructed decimal tree whether or not 1/3 has been used to construct it. I think this "proof by ignorance" is more convincing than most.

Actually it can be determined: just put all the FISs together and you will get precisely 1/3. Isn't it so?

5:45 PM

All (or at least infinitely many) FIS of 1/3 will yield a result that cannot be distinguished from 1/3. Yes.

Isn't this precisely the fact that lim_{n->infty} 0.333...3 = 1/3? (Where 0.333...3 is understood to have n 3's?)

Yet the limit of a sequence differs from all its terms.

Pardon me: every term.

And we can easily see here that 1/3 was used to construct this sequence. Since otherwise the limit would be something else.

I do not understand. In the decimal tree you claim that you can distinguish whether or not I have used 333... as tails appending the finite paths? That is impossible, because even without tails all digits of 1/3 would be covered by the finite paths.

Yes, indeed. But they would be covered by an infinite number of finite paths. And infinitely many paths taken together then determine 1/3 uniquely.

(It is the same as listing every single digit of 1/3.)

At least I see it that way. If you disagree at any point, you are most welcome to clarify.

If 1/3 is nothing but its FISs, then it cannot be distinguished in a Cantor-list from all its FISs.

This gets dramatic if we apply this knowledge to the irrational numbers.

I think I understand what you mean.

6:00 PM

And remember, all that happens in a Cantor-list, happens withing FISs. If all are there, then the diagonal argument shows that the countable set of FISs is uncountable by equating for all and for every.

1/3 is nothing but its FISs. (This is just a variant of the definition of a real number by Cauchy sequences.)

But beware: if 1/3 cannot be distinguished from all its FISs, this does not in any way imply that it is itself a FIS.

No, 1/3 is not a FIS. But it cannot be distinguished by nodes or digits in a Cantor-list from all its FIS. To distinguish 1/3 from all its FISs, we need a finite definition of 1/3. An infinite sequence of nodes is not suitable. But there are only countably many finite definitions.

But we do not need to distinguish 1/3 from all its FISs. 1/3 is all its FISs.

And the diagonal argument does not show that the countable set of FISs is uncountable, because "all" and "every" are used in a different sense in logic than you use them.

Nice to see that we agree. 1/3 is all its FISs. So it does not increase the cardinality of the set of numbers 0.3, 0.33, 0.333, ... (A set is different from its elements, but this does not express itself in the mathematics of real numbers and paths. 1/3 does not constitute another path than all its FISs.) So limits do not increase cardinality. Now apply this argument to irrational numbers.

The diagonal argument proves a diagonal path deviating from every path of the decimal tree at a finite place. This is impossible.

Up to every level all possible FISs are present. This means the diagonal argument fails to produce a new path within the realm of finite levels. But there is no "behind".

The diagonal argument proves that there is a diagonal path deviating from every FIS. This does not mean it deviates from all FISs.

6:14 PM

If the diagonal does not deviate from all FISs, then it cannot increase the cardinal number of the set of all FISs.

Right

but the set of all possible sequences of FISs is already uncountable.

Why? All possible FISs contain all possible sequences of FISs because every FIS is the sequence of its predecessors. Again, consider the decimal tree: The set of all paths that can be distingusihed by nodes is countable. This holds even for all infinite paths (that I construct). Further sequences can only be defined by finite definitions, not by nodes. But all finite definitions belong to a countable set.

Let me see if I understand you ...

Dear Dejan, it was a pleasure to discuss with you, but now I have some other responsibilities. I will come back tomorrow with certainty. Thanks a lot! And have a good night! Regards, WM

A number is determined by (or one might say "equal to") its sequence of FISs. In other words: a number is equal to all its FISs. Now, I don't understand what you mean by all possible FISs?

It was a pleasure for me, too. Have a nice evening, see you.

(I'll still pose my question here, so you can think about it when you have time.)

6:35 PM

So, what bothers me is the following: all possible FISs (as a whole) do not determine any particular real number. Therefore it is very easy to produce a real number that differs from all possible FISs: take any real number you want. It will differ from all possible FISs because all possible FISs do not determine a real number. (I might of course be misunderstanding what you mean by "all possible FISs".) What do you think?

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Jan '1319

$Mathematics$ Room for Dejan Govc and Wolfgang Muec…

Continuation of a discussion about cardinality from math.SE

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