Room for Dejan Govc and Wolfgang Mueckenheim - 2013-01-20

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8:23 AM

Good morning Dejan!

You are right and I agree: Every real differs from all possibe FISs - _but not by nodes or digits_. Every real number needs a finite definition! (And there are only countably many.)

The FISs or digits are only tools to express a real number as well as possible. In case that only zeros follow after a certain digit, this expression is even exact, like 0.25 = 1/4. In case that only 3's follow, we can escape to the ternary system and have there 0.1 = 1/3. In case of irrational numbers we have no chance to express them exactly. We can only use as many FISs as we like. Exac…

4 hours later…

12:21 PM

@BelsaZarkin: Hi, Wolfgang,

first of all I have to apologize for showing up so late. I realized I have some things to do today, so I won't have as much time as I thought. Nevertheless, I will try to participate in the discussion as much as I can. (I'll probably have some more time in the evening.)

I agree that "it happens in the infinite" means "it happens never" in this case. But to me, this only means that 0.333... has no finite digite. It does still have a digit at every finite place. The number 0.333... is an infinite series 3/10 + 3/100 + 3/1000 + ... There is an exact expression for…

12:51 PM

Hi Dejan, no reason to apologize. Our discussion is just fun and it shall remain so.

I agree that the sigma or the 1/3 denotes an infinite string of digits 3, but the object denoted cannot be conveyed in mathematical discourse by itself. The question between analysis and set theory remains: Consists the infinite string of all its FISs only? Or is there a number of digits that is larger than all finite numbers together, i.e., larger than all FISs? (That it is larger than *every* FIS is clear in both theories, analysis and set theory and need not be mentioned.)

2 hours later…

2:54 PM

@BelsaZarkin: Hi,

I understand what you mean. Logicians often use phrases like "for all x, P(x)" and "for every x, P(x)". The intended interpretation, however, is *always* "for every x, P(x)". This is an (admittedly) unfortunate terminology. To avoid any confusion, I suggest we *always* read the universal quantifier as saying "for *every* x". (In any case, the words used are just a useful mnemonic, to give us the intuition about what we are talking about (which are always mathematical formulas, which should be as exact *per se* and thus independent of words from natural language)).

Hi Dejan, yes the all-quantifier means "for every". But what I have in mind is not quantifier exchange as you just wrote. Consider Cantor as an example: The diagonal argment proves that the diagonal differs from every list entry. But it is concluded that therefore the diagonal is not in the list. That means it is concluded that for every here means for all. That it obviously correct in the finite realm

but it fails in the infinite domain. Or perhaps could we understand the actual infinity as born by quantifier exchange? In potential infinite we have for every n there is m > n. Actual infinity however says there is m larger than every n. (m being aleph_0)

I agree that we use only reals of the unit interval. And I propose to apply binaries because then the Binary Tree becomes very lucid:

Here is a picture. All nodes can be enumerated in a simple way and have values 0 or 1 only:hs-augsburg.de/~mueckenh/GU/Pruefung%20GU1007.pdf

3:29 PM

Right, but if we work in the realm of the naturals, then there are infinitely many naturals, i.e. for every n there is a natural m > n. There is no natural number m such that m > n for every n. Despite this, the realm itself is actually infinite.

I don't understand the following: if there is a list of numbers a1, a2, a3, ... and a number d that differs from every element in the list, how could it be in the list? If d is in the list, it means d = a_m for some m, by definition of the list. A clarification would be most welcome.

There are infinitely many naturals. But what infinity do you mean? Is there for every n a larger one (potential infinity)? Or are there m = aleph_0 naturals that are more than are in all FISONs? And what is mathematically suitable to distinguish N from all FISONs?

Consider the list 0.0, 0.1, 0.11, 0.111, ... and diagonalize it. You can get 0.111... that differs from every entriy of the list, but it does not differ, at any finite place, from all entries. And differing behind any finite place is impossible. Sorry, I will be back in an hour or so.

(No, problem, I also have to go somewhere. I'll be back in cca. 2 hours.)

3:52 PM

Yes, the number 0.111... does not differ at any finite place from all entries. That is correct. But if you look at all of its digits together, the sequence of digits, taken together, does differ. You can also say that the graph of the function f: N ---> {0,1,2,3,4,5,6,7,8,9}, f(n) = 1 differs from the graphs of functions g_m: N ---> {0,1,2,3,4,5,6,7,8,9}, g_m(n) = 1 if n <= m, g_m(n) = 0 if n > m. And the graph of f differs from all graphs of g_m. The point is that the difference does not come from any finite place. It comes from all the finite places taken together.

1 hour later…

5:00 PM

I can only see that the graph of f differs from _every_ graph of g_m. If we agree that f contains nothing than all g_m, how can it differ from all g_m, i.e., from itself?

I agree with your example 123. "To be able to discern it from all elements of the least, you have to examine at least two digits *together*." Yes. But in my example we can examine as many digits we like (and as many digits as exist at finite places). We do not find a difference, since all FISs together do not leave a free space. They cover all digits 1 that reside at finite places. And we had agreed that there are no fur…

Should read: I know that this is very unfamiliar.

2 hours later…

7:06 PM

@BelsaZarkin: The point was that we can examine any finite number of digits in 0.1111111... and not find a difference. When we examine *infinitely many* digits, the difference is there.

This fact corresponds to the fact that in the binary tree a path is not necessarily determined by a single node. You are trying to say: OK, every path has a *final* node and this node determines the path. But an infinite path does not have a final node. After every single node there is another node. (And by examining every single node we can discern 0.333... from every FIS.) Only the finite paths have final…

Welcome back, Dejan. You say: "by examining every single node we can discern 0.333... from every FIS." I do not deny that. You say: "every path has a final node and this node determines the path." But the union of all FISs has no final node. So I ask again: How can you distinguish the infinite path 0.333... from the union of all its FISs?

Hi. That is exactly my point. Only finite paths have final nodes. Therefore an infinite path is not determined by a single node, but rather by a sequence of nodes.

You say: "There are however, also uncountably many infinite paths, which are not determined by final notes, since they have no end". In fact they are not defined by nodes at all, as you can see when I ask you to discern by nodes one path that is not in the Binary Tree (or decimal tree) that I have constructed with countably many infinite paths. Do you agree to this point?

I believe we can discern that path by nodes. Just not by a single node.

7:22 PM

Then, please, look at the picture of the Binary Tree that I recently gave you by link. This Binary Tree is infinite and complete. No node is lacking, no edge is lacking. Nevertheless it has been constructed by countably many infinite paths. If you can discern a path that I did not use, please let me know. I pay everyone who can do so $1000.

Right ... As I was saying: an infinite path is not determined by a single node. It is determined by a sequence of nodes (where it travels). This uniquely determines such a path. The point is that there are uncountably many sequences of nodes; and countably many nodes.

(In other words: I disagree that there are countably many paths. Why exactly do you think there are countably many of them?)

You ask: "Why exactly do you think there are countably many of them?" I think that there are not more paths that can be determined by nodes because you and everybody else is unable to earn the $1000. You cannot find out by nodes any further path because all ipossible nfinite sequences of nodes are already there in the infinite tree.

Exactly. And there are uncountably many such sequences. Or am I wrong?

(I.e. uncountably many paths determined by sequences of nodes.)

Please decide this question by yourself. Can you discern by nodes any path that I have not used to construct the complete infinite tree? I have used countably many, but I don't tell you which I have used, because you claim to be able to find uncountably many, definable by nodes.

I'll explain my thoughts on this in a moment, then we can see where the source of the disagreement lies.

7:35 PM

Let you time. I have to leave but will be back in one or two hours.

Ok, I'll think about it in the meantime.

7:51 PM

This depends a lot on what you mean by "definable by nodes". Do you mean finite paths, determined by their final nodes? If that is so, than I can easily exhibit the path that is missing from your tree. It is the path / ---> 3 ---> 33 ---> 333 ---> 3333 ---> ...; This path is defined by the sequence of its nodes, i.e. the sequence a_n = (10^n - 1)/3. This sequence has a finite definition: its formula.

Here is **my main claim**: there are *countably many nodes*, but *uncountably many sequences of nodes*. (I'll be very interested in arguments against it. In fact, I might even accept one, if i…

8:20 PM

@BelsaZarkin: I have another suggestion: to make the discussion as meaningful as possible, we should first define what we mean by countable/uncountable.

1 hour later…

9:25 PM

@Dejan: By countable set I mean a set that can be put in bijection with N or a set which is a subset of a countable set.

Welcome back. I can agree with that definition.

So you claim that there is a bijection between N and the set of all sequences of nodes?

Consider a tree which has not all nodes. Node (n,m), the mth node on the nth level counted from the left side and the root node as level 0, respectively, may be lacking. Then you can recognize that all paths which pass through this node are not in the tree.

I agree. I shall pose a very concrete question.

Suppose we have the list of FISs of 0.333...:
3
33
333
3333
etc.

I claim that there is a bijection between N and all finite paths that have an infinite tail like 000... appended. There is even a bijection between N and all infinite tails that can be defined (by a finite definition - others are not definitions that can be used in mathematical discourse). Hence all finite paths with all possible infinite tails belong to a countable set.

Now we create a string 44444... Does this new string differ from all elements of the list? (Note that 4 is not 3.)

9:34 PM

Of course it differs from all elements of the list.

Ok. Now note that this can be done for any such list:
123534543...
342345623...
562876453...
346256244...
etc.

I can take the diagonal elements and change them.

I shall change 1 to 6, 2 to 7, 3 to 8, 4 to 9, 5 to 0 and vice versa.

The diagonal of the list of strings above is 1422...

This means the string I get will be 6977...

Of course. My list with diagonal 111... only was an example to show that 111... does not differ from all its FISs. This holds for every string. And the complete tree contains all FISs of every string by definition.

Ok ... But the main point of Cantor's diagonal argument is precisely that this does not happen. He takes the diagonal and changes every single element. This way he obtains a string that differs from all strings in the list. Do you agree?

Remember, the complete tree, whethet decimal or binary or terbary or otherwise based, cannot be diagonalized. It contains all FISs by definition. And it can be constructed like every countable set without AC and other peculiarities.

No, the diagonal differs from every string in the list, not from all strings, if the list is as complete as the complete tree. And that is possible as Cantor himself has shown by counting the rationals. Here we need only a subset of the rationals, namely the terminating decimals.

Ok, let's try this. The FISs can be naturally ordered (let's assume binary): 0000..., 1000..., 0100..., 1100..., 0010..., 1010..., etc. Do you agree?

9:41 PM

Yes, of course.

Ok, then taking the diagonal and changing every element gives us 1111... This string differs from every element in the list.

(Ok, I guess we agree so far.)

Which FIS of 111... should be missing in the list?

Therefore this string is not a FIS. Therefore this string does not appear in the list of all FISs.

No FIS is missing. The string itself, on the other hand, is.

(Do we still agree?)

We had agreed already that every infinite string is nothing but its FISs. If you want to deviate, please state the difference.

Do you believe that the infinite strings somewhere pass through the tree and can be distinguished beyond every finite level? How that?

At least this cannot be done by nodes as my 1000$ bet shows.

The string 1111... is represented by the set of FISs {1000..., 1100..., 1110..., 1111...}. But this set is not an element of the list. A set of FISs cannot be in the list. Only single FISs appear there. Or is this also wrong?

Exactly: I believe the infinite strings pass through the tree.

9:51 PM

Only single FISs appear there. Nevertheless there is no node of 111... that is missing in the list. You cannot distinguish 111... from all its FISs by nodes because all nodes at finite places are in the FISs.

I absolutely agree. This is because 111... itself is not a node.

The infinite strings pass through the tree such that beyond any FIS there is something that distinguished them? What is that? It cannot be a node or set of nodes. All nodes reside in the tree.

111... is not a node. But all its nodes are in the tree, i.e., in its FISs.

Correct. But this means that there might be more strings than there are nodes.

Might be. But if you insist on uncountably many strings that can be distinguished by nodes, then tell me at least one of them that is missing in the tree constructed with countably many paths. And remember, Cantor's list works by digits, i.e., nodes only.

Cantor's argument shows the following statement: for every list L of strings there exists a string A that differs from every element of L. Do we agree on this? (Note that I haven't used "all".)

9:57 PM

Yes, Cantor proves "every".

(Maybe I should say "term" instead of "element".)

Ok, I'm very glad we agree on this.

Cantor proves this for the domain that is covered by the infinite tree. If the tree is complete, we get a contradiction when "every" is identified with "all".

Then Cantor goes on to say that this means that there is no complete list of strings. Because for every list L we can build a new string A that is not on the list. (Therefore there is always at least one string missing.) Do we still agree?

Cantor even says that there is at least one string missing that differs from every listed string at a finite place d_n =/= a_nn for an n that is in a FIS. Differing beyond all finite places is not possible and does not help. Therefore the tree contradicts Cantor if every and all is confused.

In the infinite this is not allowed as we see from the simple example: Beyond every FISON there follow infinitely many naturals. Beyond all FISONs there follows nothing natural.

I'm not sure Cantor says that. But what (I believe) matters now is the following: do you agree with Cantor's argument that every list L of strings is incomplete?

10:11 PM

No. The complete tree is complete by definition. And all its FISs can be listed because they are countable. If we define numbers by digits (nodes) alone, then that list is not incomplete although the diagonal differs from every entry. In fact all real numbers are not defined by digits or nodes but by finite definitions. But that is not at all considered by the diagonal argument.

So the diagonal differs from every element in the list, but is still in the list?

That is a paradox caused by the infinite, in particular by the finished or actual infinity. As Weyl already observed: We cannot transfer finite logic to infinite sets. I have to get up early. Therefore it's time for me to leave. Have a good night! Hope to continue tomorrow with fresh mind.

(The reason I ask - the definition of a complete list is the following: a list L is complete if every string appears in it (at a finite place).)

Ok, sleep well. Bye bye.

If completed infinity and "countability" are meaningful notions, then this paradox appears unavoidably. I would beg you to think over again and again the tree constructed with countably many paths. (I did it very often, for years.) You cannot find anything missing by means of considering nodes. Bye bye.

10:27 PM

(Nevertheless, I'll post my thoughts here, before I forget them.)

So, if I understood correctly, you claim the following are all true:

1. The binary tree is complete, i.e. it contains all possible strings.
2. There is a string that differs from every element of the binary tree.
3. Despite this, this string is still in the binary tree.

I am beginning to wonder: why don't we reject real numbers as paradoxical then?

Or should we perhaps reject the binary tree?

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Jan '1320

$Mathematics$ Room for Dejan Govc and Wolfgang Muec…

Continuation of a discussion about cardinality from math.SE

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