Let’s consider the binary tree. You say the infinite bitstring 111… is in it somewhere. I’d just like to know where.
Here’s level 1, consisting of all the length-1 bitstrings. 0, 1
And level 2, all the length-2 bitstrings: 00, 01, 10, 11
Level 3: 000, 001, 010, 011, 100, 101, 110, 111
In general, level n contains the 2^n bitstrings of length n.
Just tell me, where is 111…, the infinite bitstring? It’s not in level 1, it’s not in level 2, it’s not in level 3, etc. So where is it?
That’s the only number that doesn’t get enumerated in the binary tree sequence…
Since it is rational, it gets added with those.
I actually explained this in the emails I sent out, that it’s only an illusion that the bottom number gets enumerated, but since it is a rational, it still gets counted.
Ok, then you agree that 111… was not on the list after all.
Ok, so you build your tree, which you now agree only contains all the finite-length strings.
Now you add in all the infinite-length rationals that you forgot. That’s ok actually, because there are only countably many rationals, and adding a countable set (of numbers in the tree) to all the rationals you missed, still leaves you with a countable set.
But notice that you didn’t just leave off 111… You also left off the binary representation of 1/3 = .01010101010101… Where was that on the tree? It’s not. You had to add that in too. You left off ALL the infinite-length rationals, not just one.
But now where are all the infinite-length irrationals like the binary representation of pi, sqrt(2), etc.? You left those off too.
To put this more concretely, let’s think about the decimal tree. You have 3.1, 3.14, 3.141, 3.1415, … in the tree. But where is pi? Did you need to add that at the end too?
So now you have two lists. You have the list of numbers that are already on the tree – those are the finite-length strings. And now you have another list of numbers that must be added in at the end. But how many of those are there? Well, there are uncountably many. That’s the flaw in your argument. You didn’t just leave off the infinite-length rationals. You also left off all the irrationals.
I did not received one of those emails. You left me off your list. Just as you left all the irrationals off your list. Thus proving that I’m irrational 
How can it be an illusion? You just agreed that 111… is not on the list. And I just pointed out that you left ALL the irrationals off your list. That’s not an illusion. That is the central fact of the infinite binary tree. It has only countably many nodes. It leaves off all the infinite-length rationals; and it leaves off all of the irrationals.
And what of the binary representation of all the irrationals? They all have infinite-length strings, so NONE of them are on your list. You have uncountably many irrationals left over that aren’t on the tree.
Your mind is still stuck on the list being finite, the algorithm is infinite, however, the last number on the strong never gets counted, which just happens to be 111… In my sequence. Numbers like 010101… Are evidently being counted very early in The sequence in fact.
I’ll post it all later, I’m with someone right now …
Let me ask you this though, you picked up on the last part of the string never being enumerated, what about the first part or the second part??
Do you think 000… Isn’t being counted?
As I said earlier, I agree that the algorithm goes on forever.
At stage 1 it produces 0 and 1.
At stage 2 it produces 00, 01, 10, and 11.
At stage n it produces all the bitstrings of length n.
So if you claim a bitstring like 10101010101010… is produced by the algorithm, please tell me at what stage it’s produced.
Ok, 111… never gets counted. What about 0101010101…?
At what stage? How? Every string produced by the algorithm is a finite string. Their lengths are unbounded in the sense that there are arbitrarily large finite length strings. But there are no infinite length strings.
I’m in no hurry. After all we are discussing the infinite.
Last part, first part, second part? Not sure what you mean.
However I do realize I’ve been a little unclear about something. I don’t care about the digits to the left of the decimal point. That is, everything I can say about pi = 3.14159… is much easier to say about pi - 3 = .14159…
So when I think of a bitstring like 01010101010101… I am REALLY thinking about the number between 0 and 1 whose binary expansion is .0101010101…
So: All my bitstrings have an implied binary point in front of them. All the real numbers I care about are between 0 and 1. If we are using decimals then I’ll just talk about pi - 3 = .14159…
If you had a procedure to enumerate just the reals between 0 and 1, you could easily extend that to all the reals. Everything to the left of the decimal point is irrelevant.
Was that what you meant by first part and second part?
I’m willing to agree that any bitstring ending in an infinite sequence of 0’s can be truncated. So if 1 is on the list then 100000000… is on the list. It doesn’t affect my argument.
Also, any sequence ending in 1111… doesn’t affect the argument either, since any such number is rational. I’m more interested in irrationals that never have any pattern and that do not have trailing sequences of only 1’s or only 0’s.
Wow!!!
I can get it with a
0101…
00110011…
000111000111…
0000111100001111…
Column pattern, but that’s besides the point… I assumed it was on the list because I never manually did step 5.
I’m going to look closer at sequences (if there’s a columnar sequence or whether each one needs to be done!
I’ll reply tomorrow, thank you wtf!!
This is more complex than I thought…
I may be able to do the columns in a binary procession to make it work, rather than using the counting procession I just used. I’m still looking at it.
One thing I know for sure is that algorithms cannot generate a random sequence, and godel used an algorithm to generate a random sequence, so I know incompleteness theorems and their cousins, uncountability theorems are false.
It’s trivial to say that you don’t observe something until you observe it, but if you know what it does when you do and don’t observe it, you have a binary completeness set.
I’m not really babbling here, but I am not demonstrating the power of that statement here.
I’m still looking at all of this …
A binary procession of columns means…
0101…
00110011…
0101…
00110011…
000111000111…
0101…
00110011…
000111000111…
0000111100001111…
0101…
I’m still looking at it…
Wtf, solved it
I figured out what the problem was, took me until the afternoon to solve it though.
I can’t believe I missed what you were talking about, I felt like such an idiot!
Now that I know how to solve your problem, I’m quite confident, that I have solved real number enumeration… I’ll post it on math forums first and here second. I’ll be busy a lot today so it may be a while - this evening.
I’m a little confused about your column idea and the lists of infinite bitstrings you’'ve posted. I assume they’re infinite since they all end with dots. That has nothing to do with what we’ve been talking about, which is attempting to enumerate the reals using the infinite binary tree. In the binary tree, each node represents a finite bitstring.
If you feel like explaining your column idea, I’d be interested to hear about it.
If anything I’ve said has brought you clarity, I’m happy to have been of service.
To the extent that anything I said has inspired you to believe the real numbers are enumerable, I have failed miserably.
Let me know how it goes on the other boards.
I must tell you, you cannot enumerate the reals. But even more importantly, your columns of infinite bitstrings don’t seem to have anything to do with the infinite binary tree as far as I can tell, so your recent posts have indeed left me confused.
Anyway if you have more thoughts, I’ll be interested, and if you’re busy, it can wait. The discussion of infinity is as old as Aristotle.
oh…so that’s what enumeration is. I think I can solve it now…But not until i’ve hada good soda, root beer float. I think I get it now, but I will still give Ecmandu partial credit because he gave me the motivation and the “cool factor” to do it.
This para of yours is very confused and I wanted to unpack it a bit for you. If you think I’m wrong that’s cool, but let me just articulate some counterpoints.
Yes, this the heart of the matter. The set of infinite bitstrings is uncountable, and the set of computable bitstrings is countable.
That’s not possible, since we just agreed that you can’t use an algorithm to generate a random sequence. You are misunderstaning/misapplying Gödel’s theorems.
It seems that you’ve heard of them but not understood them. That’s perfectly ok, a lot of people fall into that category, myself included. But your statements do not support the thesis that you have understanding of the work of Gödel. It’s extremely technical, it’s not at all about whatever people heard.
This in no way follows from Gödel even if you happen to disbelieve in uncountable sets. Uncountable sets do follow logically from the standard axioms of set theory. What that means philosophically is of course a more personal set of beliefs.
Now you are confusing math with physics. Math has nothing to do with physics. Math has been independent of physics since the discovery of non-Euclidean geometry in the 1840’s.
Please define “binary completeness set.”
These two paragraphs of yours jumped out at me as being exactly that.
YES!! I can agree. Me too. Infinity is interesting to study. I’m still looking at all of this myself. I wholeheartedly support your endeavor.
Would it be helpful if you thought of standard mathematical infinity as simply a logical consequence? Rather than trying to fight it because it conflicts with your intuitions about the world?
NOBODY claims there are real numbers in the physical universe. This is abstract math only. We can just take it on its own terms.
Random numbers algorithms generate a set of numbers. Time is the missing factor. Every number in the sequence follows the pattern of the algorithm, which is set in stone, until you multiply the seed based on time, which is ever changing.
You are saying that if we could introduce a random element, such as time, we could then generate randomness. But we can only do that because you allowed us to INPUT randomness in the first place.
So the question is, where did your computer obtain true randomness from? The cpu time in a computer is completely deterministic. It’s good enough to introduce fake randomness for programs that need it, but it’s not really random in the sense that it could ever be used to generate a truly random sequence of numbers.
This is a hard philosophical and physical problem. Is there anything at all in the universe that’s truly random? Nobody knows. Maybe we’re all a computer program. Some people believe that. I don’t. At least I hope we’re not all a simulation.
Tsk, tsk. 
Well if the algorithm gets so big that it is impossible to compute the next number within the given time slot, then it is random, no?
Not sure what you mean. A set of numbers is computable or not depending on whether the numbers can be cranked out by an algorithm. Time is not relevant.
What i mean the universe is a gian computer, it computes the state of things, but it is such a big algorithm that things appear random and unpredictable.
I think we’re in agreement then that there are two different things: randomness, and the appearance of randomness.
Whether there is any true randomness in the universe is an open question.