where xsi is the i:th binary digit of the real number x in S

Except, of course, for irrational numbers.

yeah it's the same technique used to have bitmaskable enumerations (ie. giving an enumerated item twice the prior value so that you can combine and separate each item).

Then give me a natural for pi.

Then give me a natural for pi.

A cop out claim is that real numbers can have an infinite number of decimals

That's not a cop out. Is it a cop out claim to say irrational numbers have an infinite number of decimals? Or are you not including them in the set of reals?

Regardless this is my favorite post on this sub ever.

That's incorrect, but even assuming it was true, what does this have to do with "the deep state"?

Given that you're trying to make people believe false knowledge, you must be the actual "black op" here.

The rationals already contain all numbers with a finite number of decimals.
If reals couldn't have an infinite decimal expansion what would be the point?

No set can be mapped to its powerset. I wrote a proof for that down in this thread before...

I also proved before that the powerset of the naturals is isomorphic to R.

You are assuming what you want to show here. Why would there be a unique natural number for every subset in P? That's only the case if a bijection exists, so you are already assuming your conclusion.

Your function has to map from natural numbers to sets of natural numbers or the other way around. How else can it be a bijection?

If p is a set then 2^p maps to a set of sets if p is a natural number 2^p is a natural number.

None of this makes any sense.

So first, this isn't a bijection because it isn't surjective, you are missing every natural number that isn't a power of 2.

And more importantly

(I have converted the subsets in P to numbers.)

That is only possible if there is a bijection to the natural numbers or a subset of them. You are assuming your conclusion.

Here's a crazy idea. If you can turn the subsets into numbers, just map every subset to the number. That's exactly what the point of the bijection is.

Yes and if now S has infinitely many elements then there are infinitely many subsets with an infinite number of elements and your ""bijection"" would map all of them to infinity thus not being a function since infinity isn't in N and even less injective.

2N and 3N and 4N and so on are all going to be mapped to infinity.

So what is the value in V corresponding to 2N?

Yes, the set of all even numbers. What number does your bijection map it to?

No, what number in N corresponds to that.

What is that notation supposed to mean? That's not a number in N

That's just plain wrong.

If it represents a natural number it's finite and you can write it down.

That would still only account for the subset of 2^N that are set of the power of a member of N. Where would all the others sets map to in Q?

There is no reason to believe that all of these sums converge to rational numbers.

That is incorrect as many people have pointed out to you before.

By that logic any infinite sum is finite since every finite sum is finite.

No, that makes them cauchy sequences. It doesn't mean they converge in Q because Q isn't complete.

Imagine adding up the numbers 1/n with a small change.

If what you've summed up so far squared is bigger than 2 you subtract the next number otherwise you add normally.

Since you are only adding rational numbers every partial sum is rational. But the limit is obviously sqrt (2), which isn't a rational number. So the limit doesn't exist in Q.

Protip they don't.

Okay, let's take 1/3 and calculate your function: 1/3=0.01010101... in binary. So n(1/3)=2^2+24+26+28, this clearly diverges. So your function n does not map to the natural numbers. And even if you extend the natural numbers to include infinity, you would still get n(1/3)=n(2/3)=infty, so n would not be a bijection because is wouldn't be injective.

So you do not have a bijection.

Your function isn't even well defined, let alone a bijection. What natural number does your function map 1/3 to?

That isn't a valid natural number.

So you admit your function isn't well defined? Great!

Now go and find this mapping to Q.

Your map from P to V is not injective. For example the elements of p {1} and {1/2, 1/4, 1/8, ...} both map to the same element of V, namely 1.

1/2 + 1/4 + 1/8 + ... = 1 exactly. Do you dispute this?

To which fraction does this 0.99999... correspond?

A rational number is ratio of two integers (natural numbers). I mean what two integers correspond to 0.99999...?

If you don't have the integers then how is it a rational number?

No that's not hard at all. I can name any rational numbers and give you the integers associated with it.

No you can't, because that number doesn't exist.

I think it's equal to 1 yes, but you said it was some other rational. So asked which one, but you couldn't name it.

Well since that isn't a rational number your whole bijection falls apart.

If you make an injection between 2^N and Q then you have shown that 2^N is equal or smaller than Q. If you want to show that they are the same size then you need a bijection.

There isn't a such rational number.

1/2 + 1/4 + ... = 1 exactly. For a proof note that 1/2 + ... + 1/2ⁿ = 1 - 2^-n, and 1 - 2^-n converges to as n goes to infinity. Which part of the above do you think is wrong?

You don't need the real numbers to talk about limits, in the rational numbers 2^-n converges to 0 using the standard topology.

Yes it is. Limits of rational numbers are defined as the limits in the usual topology. It's literally the definition of limits.

Then state your definition for an infinite sum of rationals. Be very clear, no hand waving.

I don't know what sum(q_i) means when there are infinite q_i, please define it properly.

You still haven't defined it. At the basic level + only makes sense when dealing with summing two elements. It is easy to show that it makes sense to add up a finite number of elements (using induction), but you need to do a lot more for it to make sense for adding together infinitely many numbers.

So once again you haven't defined a bijection, since your idea for a bijection relied on your own definition of infinite sums which you cannot give.

No matter how you define your sums you will never get injectivity. Have you seen cantors proof that a set is never in bijection with its power set? If not read it. If so where it the error in it?

The same proof holds for any set.

Since the rationals are in bijection with the naturals, any bijection P(N)->Q will give a bijection N->Q, so it's impossible.

Where is the error in cantors proof? Can you point to the specific line he makes a mistake in?

Read it as some number may be paired with ...

Poor wording, but the it still works. It's just that the set of selfish numbers could be empty, which isn't a problem at all.

The empty set is an element of P(N). Indeed, the empty set is a subset of N.

d is the number which is paired to the set of non-selfish numbers, which means that d is paired to the empty set. Since d is not in the empty set it means that it is a non-selfish number. But the set of non-selfish number sis empty, so cannot contain d. Contradiction.

All you have showed there is that if you change the premise then you cannot use the exact same argument. Note that you haven't shown that |M|=|P(M)|, failing to derive a contradiction does not refute the premise.

What you've basically said here is that there is no flaw in the argument, because to find a flaw you have to change the argument itself.

Not puzzling in the slightest that changing the proof makes it fail. In fact it's exactly what I would expect.

P and V do not have the same cardinality because you are mapping all infinite sets in O to infinity.

How many times does that have to be said?

Your proof doesn't really make any sense to me. Why would 0.999...=1 imply that there is an infinite decreasing sequence of natural numbers?

I'm assuming that in your understanding 0.999... is supposed to be the closest number to 1.
But obviously that doesn't exist, since (0.999...+1)/2 would be even closer still to 1.
In that case you might say that it's 0.999... again. Then let's say x=0.999... has the property that (x+1)/2=1. So let's see what x is, shall we?
(x+1)/2=1
x+1=2
x=1 Oh woops, something must have gone wrong, or 0.999... was 1 all along.

Because without an infinite decreasing sequence of natural numbers it's never possible to get from 0.999... to 1.

You are asserting that, but as far as I can tell there is no good reason at all to believe so.

True, but I think sum(2-(i+1) ) still is a well-defined number 0.999...

Sure, it's also 1.

Then let's say x=0.999... has the property that (x+1)/2=1.

It doesn't.

Yeah, I corrected it to (x+1)/2=x, my mistake.

It doesn't become equal to 1. It is equal to one. Numbers don't change.

If you write 0.999... down as a series, you can check if the series converges and what value it converges to.

So let's use your sum definition for 0.999... for that.
x = sum(2-(i+1) ) = 1/2+1/4+1/8+...
2x = 2*(1/2) + 2*(1/4) + 2*(1/8) +...
2x = 1 + 1/2 + 1/4 + 1/8 + ... =1+x Because the bit after 1 + is exactly the same as x
2x=x+1
x=1

You have invoked Hilberts Hotel yourself many times.

The stuff after the 1 is exactly the same as before. Nothing is missing ""at the end"" because there is no end.

The elements aren't ""fused"" together.
There are just different ways to write the same number,even though there is only one element in Q that corresponds to that.

0.999... = 1 = 2/2 = 2-1

And no there isn't a problem. There is no element missing. The ""shifted"" series is exactly the same after the "1+" because there is no gap and no element that is missing.

No, it's not R, but it's the Interval [0,2] in R.
Congratulations, you've finally realized what people have been telling you for about 3 days now, the powerset of N is bijective to R.

This again? I have discussed that nonsense with you before.

You are asserting again that there are only as many Dedekind cuts as there are elements in Q. That isn't correct and I've tried to explain to you why at length before, so I won't indulge that again.

You can not define a function into a set that doesn't exist yet.
You can't define R by having a function that maps to R, because that doesn't exist before being defined.
That's why the construction of the reals usually uses Dedekind cuts or equivalence classes of Cauchy sequences in Q, neither of which you seem to have even the tiniest amount of grasp on.

How about you spend some time reading up on this stuff before claiming to have overthrown well established knowledge?

How arrogant do you have to be, to think that for hundreds of years everyone that was presented this and concluded it made sense is wrong and that you, without even understanding basic concepts involved like what a set or a function is, have found the flaw in it that will overthrow mathematics as we know it?

There is no subjectivity in math, things are proven logically. If something is proven to be true, it will never stop being true with new knowledge being discovered or us achieving deeper insights into the universe or a regime change or whatever.

No, it doesn't. How can sqrt(2) be in that set without being defined? It doesn't define anything. A function is defined between two sets. It can't define one of the two sets.

Seriously dude, you don't even understand how functions work, MAYBE take a step back to at least learn what the terms you are using actually mean in mathematics.

Yes, that's not just possible, that's necessary.

But you can't define R with a function, because before being defined R doesn't exist.

You can not define a set from a function.
Is that really so difficult to grasp? You can't define V from Q via a function and neither can you define R from V or R from Q that way...

No, you can't. You can't define a new set by a function before you actually have the set.

But you can't define a function from or to the real numbers if you don't have real numbers.

No, they don't. The sums aren't real numbers and we've been over this repeatedly.

Congratulations!

You've just proposed a worse version of the construction of the real numbers as equivalence classes of Cauchy sequences in Q.

Well, aren't we all lucky that despite that you know enough to prove things about them.

How about you actually learn about the things you try to debunk before trying to debunk them?

So that's just essentially the Cauchy construction except it won't work at all. And it still doesn't prove reals are countable.

No, the rationals are closed under addition. Are you an engineer or something?

Worst definition ever, it won't even work. And it still doesn't prove what you claim; that reals are countable.

This is false. You cannot map the set of all Dedekind cuts to the set of rationals, in a bijective manner.

There is no surjection from Q to the cuts, so good luck with that.

There's no such surjection either.

Because 0.999.. is defined in terms of limits, and this limit is 1.

That sum is still 1 though.

In your mapping from P to V (adding up all the elements), what does the subset of T {1,2,4,8,...} get mapped to? Because the sum of infinite natural numbers is not a natural number. Your map isn't even well defined (yet again), let alone a bijection.

Feel free to keep trying, but cantors proof is not wrong you you will never get a valid proof.

Again, you map more than one element to infinity. So no bijektion.

Then please tell us about the error in the proof.

Those are just the binary representation of the reals in the interval [0,1]

• Those aren't binary sequences.
• Their infinite expansions do not differ, because there is no last digit to differ.

What are you talking about, what do you even mean by fuse? There are only ever two infinite binary sequences that identify the same number.

One with a 0 followed by infinitely many 1s and a 1 followed by infinitely many 0s. Those are the only two sequences that will map to the same value.

And because those two always exist (except for 0) and are just two different respresentations of the same number you can just specify which of the two representations you want to pick.

What do you mean by "fused"? A real number can at most have two different decimal expansions.

It's just the decimal expansion in base two. Actually, you have to forbid decimal expansions ending in only ones because technical reasons. Otherwise you won't have unique decimal expansions.

You can't even define addition for rationals, yet you claim that an infinite addition of rationals will give you a rational?

But i becomes unbounded. How do you define the sum when adding infinitely many rationals?

Actually we can define it in terms of limits.

0.999... is equal to 1.

But 0.999... is defined in terms of a limit, and this limit is indeed 1. So 0.999... = 1.

But i becomes unbounded. How do you define the sum when adding infinitely many rationals?

No one ever claimed that. However, the limit as n approaches infinity will be zero. And you don't need reals for that.

No, you don't need reals. The limit will be 1, even if you only consider rationals.

But 0.999... is exactly 1 and considered as a limit.

That's not a natural number.

No, you it isn't still a bijection. You still map all irrationals to one element.

That's not a natural number.

That's not a natural number.

That's not a natural number.

sum(22i + 1 )

Is not a natural number so it is not even a function.

No, it is impossible because 1/3 and 2/3 both map to infinity. Thus n is not injective, by the definition of injectivitiy. Any bijection is surjective and injective, so n is not bijective.

Can you provide some links of this you caught my interest.

It hasn't, what cantor showed there is a quick and intuitively easy way that reals have higher cardinallity. The proper proof is more set theoretical.

So an 1800s mathematician conspired to destroy mathematics. Meanwhile, in one line, you completely disproved him.

If you were confident in this, then you'd write a paper. Not post here.

I see that there is debate over it, but it doesn't appear to be a meaningful one.

Just post on math.stackexchange.com and see what people have to say.

I'd love to discuss it with you, but I don't really understand deep maths like this.

Next time you struggle on your homework you should ask for clarification not assume there's a big conspiracy to keep math hard.

HAHAHAHAHAHAHAHAHAHAHAHAHAHAHAHAHAHA!

Refutation is here.

Ooorrr again, it can be that you are ignorant and don't see the blantant error in your shit. It is quite trivial to show that R is uncountable, cantors is one of them with the diagonal arguement which shows that no matter what supposed list you have, he can find elements you missed.

Other is cantors theorem which is more general that shows that there is no surjection from S to P(S), and hten showing |P(N)|=|R|, both not difficult.

Knock yourself out: neither does OP.

One idea I came to think of is to look at all the Dedekind cuts, and at most there can be only one cut between each rational number pair with minimal difference between the numbers.

There is no such thing as "a rational number pair with minimal difference between the numbers". If you give me a pair of distinct rational numbers a and b, then I can find a rational number c between them, so that the difference between a and c and the difference between c and b is smaller than the difference between a and b (For example c=(a+b)/2 has this property). So what do you mean when you talk about a "rational number pair with minimal distance between the numbers".

If you have two distinct rational numbers, then there are infinitely many real numbers between them. And similarly if you have two distinct real numbers then there are infinitely many rational numbers between them. So I don´t see how "mapping the rationals to the even numbers and mapping the irrationals to the odd numbers" is supposed to work.

That doesn't work. There are bijection between the rational numbers and the natural numbers, but they are not monotonous.

As an example let´s just take a list of all rational numbers from a random website.

https://proofwiki.org/wiki/Rational_Numbers_are_Countably_Infinite#Proof_1

In this list the first entry is 0/1. The second entry is 1/1. As far as I understand you, you suggest that in this situation 0/1 and 1/1 have minimum distance. But 1/2 is inbetween them, and has smaller distance to both of them, so the distance between is not minimal. If you want your suggestion to work, then you need to find a list of all rational numbers where the entries are always getting bigger and never smaller, and that doesn´t exist.

And why is this an issue?

Wi th logic like that I'm surprised you don't post on holofractal.

Is that supposed to be an insult?

Yes that entire cult is a joke.

"I'm smarter than most people. Some people are as smart as me, but I'm braver than them." - you

Mathematician here.

HAHAHAHAHAHAHAHAHAHAHAHAHAHAHAHAHAHAHAHAH!

Er, sorry, but this is SO dumb.

Refutation at the top level.

But for a century they have had to defend and whitewash it in order to protect their credibility.

I'm sorry, but any mathematician that was capable of completely rewriting portions of mathematics as we know it would never have to worry about their credibility. The problem is in the capability to do so. You're more than welcome to try, but literally everything you've mentioned here is so ill reasoned, I'm not sure you'll get very far.

OOorrr a more possible explination is that you are too stupid to understand basic anything.

I'm very confused by this, would you mind clarifying some things for me? You say n(x) = sum(2^xsi ) where xsi is the ith bit of a real number in [0,1]. Since xsi is always 0 or 1, 2^xsi is always 1 or 2 and n(x) diverges for all x. Did you mean something else?

Oh, I see what you mean now. So to be more precise, n(x) = sum(2^xsi*i ). However I must still take issue with this. Since n(x) diverges for all x in [0,1], as I see it, n(x) does not form a bijection between N and R as you claim, it maps all of R to infinity. N is infinite, but infinity is not in R.

I do not have a problem with that. I have a problem with n(x) being many-to-one by virtue of mapping all irrationals to infinity. Bijections are by definition one-to-one, so n(x) is not a bijection.

Yes, any irrational number will have infinitely many non-zero digits in any base, since irrationals have terminating expansions by definition. But as your example shows, we don't even need to consider the irrationals to show that n(x) is not a bijection between N and R. Any rational with a denominator that is not a power of 2 will have infinitely many 1 bits in its expansion, and thus map to infinity, which again shows that n(x) is many to one. e.g. n(1/3) and n(1/5) both map to infinity.

I'm afraid my expertise does not extend to Dedekind cuts. But I would imagine that someone who is versed enough to be familiar with Dedekind cuts would be capable of addressing my simple argument on the lower terms I have employed. This strikes me as similar to invoking calculus in an effort to prove that 1+1=2.

True that 1/3 has infinitely many 1 bits in its binary expansion, as do all irrationals. But the crux of my argument is that all reals with infinitely many 1 bits in their binary expansions, whether rational or irrational, will map to infinity given your n(x). If more than one x maps to infinity, then n(x) is many to one, and therefore cannot be a bijection.

I guess I'm confused again. Saying that all reals terminate is equivalent to saying that all reals are rational. Yet you mentioned Dedekind cuts, so you must believe irrationals exist. Of course R is countable if all reals are rational.

But irrational numbers do have an infinite number of non-zero digits in their expansions:
>It can be shown that irrational numbers, when expressed in a positional numeral system (e.g. as decimal numbers, or with any other natural basis), do not terminate, nor do they repeat, i.e., do not contain a subsequence of digits, the repetition of which makes up the tail of the representation. For example, the decimal representation of the number π starts with 3.14159265358979, but no finite number of digits can represent π exactly, nor does it repeat.

"do not terminate" is not the same as infinite length.

In this context, yes it is.

The size of the set of natural numbers N is the infinite ordinal number ω.

Well, no, ordinals don't denote size, they denote order type. The size of the set N is aleph_one. This happens to be the same set as ω so you aren't too far off here.

But ω doesn't appear in the set N itself.

True, but so what? Nobody is claiming anything like this happens with irrational numbers. Each real number (rational or irrational) has infinitely many digits to the right of the decimal point (one digit for each natural number). Nobody is saying there's an "ω-th digit".

You agree at least I take it that infinite ordinal numbers are not natural numbers

Yes

yet the claim is that real numbers can have an infinite number of decimals. That's inconsistent.

Why is that inconsistent? What does the number of digits in a real number have anything to do with whether infinite ordinals are natural numbers?

Yes they are the same element.

Because by claiming that real numbers can have an infinite number of decimals, it's impossible to form a set of real numbers.

Wait, what? You are going to really need to elaborate on what you say here. Why can't I have a set of real numbers? Is [0,1] not a set? What does this have to do with the number of digits?

For example are 0.999... and 1.0 the same element in R?

Yes, so?

If you think Zeno's paradoxes show anything, do you also think movement is impossible?

"I don't know enough to actually refute that, and I refuse to admit that I'm wrong, so I'll just toss out a non-sequitur"

Have a nice day dumbass.

Zeno's paradoxes say the same things about Q and R. If you consider Zeno's paradox to prove any statement about one of these things, it proves the same thing about the other.

Real numbers don't have an infinite number of decimals.

Yes, they do.

What makes you think that there are only countably many dedekind cuts?

The set of rational numbers is dense in R, but does not have the same cardinality.

I get that that is counter intuitive. Nevertheless it is true.

If they had the same cardinality you'd be able to list the real numbers in a specific order, thus creating a bijection N <=> R, which Cantor has proved impossible quite conclusively.

As for some of the other things I read in this chat. Are you saying that no irrational numbers exist, or are you saying that they have terminating representations?

It'd be great if you did, alas you didn't.
The reason your function isn't a bijection has been pointed out to you repeatedly I think.

And Cantors theorem is something else than the Cantor diagonalization argument. It works just as well though.

We can pretty easily show that the powerset of the natural numbers is bijective to the real numbers in the interval [0,1].
The powerset is the set of all subsets of N.
Now every subset S of N corresponds to the number that has a 1 in it's n-th binary place exactly if S contains the n-th number in N.
So N maps to 1, the empty set maps to 0 and {1,2,3} maps to 0.111.
This is pretty clearly a bijection, so we know that P(n) has the same cardinality as [0,1].

Now Cantors theorem states that the cardinality of any set is strictly smaller than that of its powerset.
So since R and P(n) have the same cardinality N has to have smaller cardinality than R.

The proof for that is one I actually prefer to the diagonalization argument:
Assume a bijection f:N->P(N) exists, then it is also surjective.
Consider the set B that contains all numbers n that are mapped to a set that doesn't contain them. So B = {n in N | n not in f(n)}.
Since f is surjective it has to map to every subset of N, which contains B, so there is some x with f(x) = B. Now if f(x)=B, then B can not contain x, but if B doesn't contain x, then it should contain x.

This is clearly a contradiction, so such a bijection can not exist.

Therefore the cardinality of N is strictly smaller than that of P(n) and also R.

I literally wrote the proof down for you.
If you claim that it's false disprove it, don't just declare it and invoke a completely unrelated thing.

The three steps in Hilberts Hotel prove that:
1. N is isomorphic to N+1
2. N is isomorphic to 2N
3. N is isomorphic to NxN where x is the cartesian product.

None of that has anything to do with the powerset.

No N isn't isomorphic to 2^N because that would be the powerset and I proved literally 4 comments up this chain that they can not be isomorphic.

Hilberts Hotels last stage is having infinitely many busses with infinitely many guests. So for every number in N there is a bus containing one guest for every number in N.
That makes it NxN not 2^N.

Yes there is a bijection between those two sets because they have the same number of elements.

I'll try to explain it more simply, but first answer this question:
Do you think that there can ever be a bijection between S and 2^S ?

Your mapping is not a bijection and people have pointed out the reason for that repeatedly, so I'm not even going to engage that.

So let's take a look at 2^N . It's the set of all subsets of N. Any subset of N either contains or doesn't contain the n-th natural number. So for example the set that contains all of them is N and the set that contains none of them is the empty set.

Now it's easy to see that this is bijective to [0,1].
Let's say that for a set A, A_n is 0 if A doesn't contain n and is 1 if it does.
Then f(A) = Sum A_n*2^-n is obviously a bijection between P(A) and [0,1].

Are you with me so far?

As a variety of other people have already pointed out to you n(x) isn't an injection not even from the rationals, since it doesn't even map 1/3 or 2/3 and if it did it would map them to infinity, which isn't part of the natural numbers, hence it doesn't even map them at all.

You say I insist, but there are obviously even rational numbers with infinite decimal expansions. Like 1/3.

The decimal representation of a number isn't the number, it's a representation. 0.25 = 1/4 =2/8. You can have lots of representations of the same number.

And yes, a real number can have infinitely many digits, because it is still bounded. Almost all real numbers between 0 and 1 have an infinite decimal representation. But none of them are bigger than 1. If natural numbers had infinitely many digits however then they would become infinitely large and not even be numbers anymore.

The problem is that your supposed bijection is not a bijection. For starters, you would map all irrationals to infinity. The same is true for rationals with an infinite decimal representation. What you have shown is that a subset of the rationals, rationals p/q where q contain all the prime factors of 10, is countable.

But almost all reals have an non-terminating non-repeating decimal expansion. Hence, almost all reals would be mapped to infinity.

The set of all Dedekind cuts is not countable. Between any two rationals, you can always find uncountably many irrational numbers. You assumption is false.

The difference is not well-defined, because the cut contains rationals "infinitely close", so to speak.

We know perfectly fine what real numbers are.

Each cut only defines one unique number, and a pair of rationals with minimal difference are the end of A and the beginning of B:

You keep talking about this idea: "a pair of rationals with minimal difference".

There is no such thing. Pick an irrational number like the 1/sqrt(2) - or, honestly, any irrational you like - now, what exactly are these "pair of rationals with minimal difference"?

This is impossible to do. Given any such pair or rationals a and b, it is trivial to find a rational number c between those two

Prove me wrong - let's see you actually do this calculation for any irrational number of your choice.

Now you are assuming what you are trying to prove. That's a real no-no!

Here's your statement:

Take two rational numbers a and b, a!=b, and construct a pair (a, b) such that a < b, and for all rational numbers c, c =!a and c != b, c is either smaller than a or larger than b.

Well, my claim is that this is impossible because I can always find many examples of c that breaks your rules. An easy example is the average c = (a + b) / 2, which is a rational number that's always between a and b.

So I know that for any pair of numbers a and b you give me, there are some numbers c so such that a < c < b. So you cannot construct such a pair a, b.

QED. :-D

For any two rational numbers a and b, a != b, a<b, it is always possible to construct a rational number c = (a+b)/2 that is greater than a and less than b. Thus, there cannot exist a "pair of rationals with minimal distance"

What do you mean by "each rational number"? You can't mean "each successive rational number", because without the concept of minimal distance pairs, there's no such thing as successive rational numbers. And you can't mean "at most one cut between any two rational numbers" either, because then for any two cuts, you could find a rational number greater than the upper cut and a rational number less than the lower cut.

If you mean to say "For any two cuts there exists a rational number between them", that that's just a restatement of the theorem that there exists at least one rational number between any two irrational numbers, and I don't see how that helps your proof.

Can you explain dedekinds cut to me?

No, the answer is at most |P(Q)| since both A and B are subsets of Q and not elements, as has been pointed out to you repeatedly.

You can't describe the split in two with a finite number of rational numbers, that's the whole point.

As an exercise for you, which rational number do you think splits the rationals in two around the square root of 2? I can easily define the two sets, but there is no closest rational to the square root of two from either side. So which rational number would identify that?

Yes, for every real number there is exactly one split, that's why the dedekind cuts are used to define the reals. Not for every rational one. Otherwise some rational number could describe sqrt(2).

So you can dedekind cut the real numbers between A < 1 and B >= 1. And this is valid because A is less than all elements of B, and A doesn't have a largest element.

Because if I said .99 is the largest element, you could just tack on another .009 and get .999 which is larger.

And then you can do this for all rational numbers, |Q|?

And then you can do this for sqrt(2) so the sqrt(2) is rational.

I guess I'm confused on how to get from the construction step, to at most |Q|.

The square root of 2 isn't rational.

You don't get to at most |Q|. /u/MrNeoson doesn't know what he's talking about.

What do you think a subset is? What do you think A and B are?

A and B are subsets. The fact that you need subsets to define the cuts is the point.

Your mistake is that you think there is a next rational number for any given number. There isn't.

Assume for some rational number x there was a next rational number y. Then (x+y)/2 would be in between them, so y can't be the next one.

You can't "walk" the rational numbers. There are more ways to split the rational numbers into two halves than there are rational numbers.

The real numbers are those splits. You are assuming the result you want by claiming that they can't be split in more ways than there are elements in Q. That doesn't prove anything, you are just asserting this over and over again.

I wish I could give you a more intuitive example to wrap your head around it, but there isn't one.

Many people in this thread have spent time to write down proofs to show you that you are wrong, but you just go: "Oh that's too hard for me, don't get it"

Meanwhile you claim to have disproven an accepted mathematical fact, without any proof and with a ""bijection"" that isn't even a function and even less bijective.

I agree that it seems false. It's certainly unintuitive.

But there are many ways to prove it. I wrote down a proof for that in our conversation here.

Just, you know, turn down your arrogance a tad. Those are basic concepts that anyone with a higher math education knows and you think you are overthrowing established knowledge without fully grasping it.

Sure.
But A has to contain infinitely many elements to do that.

Of course there is only one possible split, because there is only one square root of 2. If there was more than one, Dedekind cuts would do a poor job of defining the reals.

But if picking an element from Q isn't enough to define where that split is, then why do you think that there can only be as many splits as elements in Q?

Think about this. For any two distinct elements from Q that you select, there are infinitely many elements of Q between them.

Again, there is no largest number in A and no smallest number in B, so even if it goes against your intuition you need more than just a finite amount of rational numbers to define where the cut is.

Your intuition is right for the finite case, but the finite case doesn't always carry over to the infinite case.

If you think it does then surely a powerset of a set S having 2^|S| elements ought to convince you that no set can have the same cardinality as its power set, which is actually true even in the infinite case.

If a finite number of elements in A and B were sufficient there'd be a bijection to Q, that's why it's relevant that you need an infinite number of elements to define them.

What does splitting the naturals into odd and even have to do with anything? We aren't talking about the natural numbers.

With the process you are describing it will never become uncountable and it will never become the set of real numbers.

If you think in any kind of process it won't get you anywhere. If you can add things step by step then you can number the steps and the result will necessarily be listable.

The real numbers are the set of ALL dedekind cuts in the rationals.

It's not clear what you mean. You can find one unique Dedekind cut for each rational, yes.

This is simply not true, since there are more cuts than rational numbers.

I'm not sure what you mean. A cut always contain rational numbers.

He wants to make a counting argument because of the total order of Q and all a in A being smaller than all b in B.

But that doesn't work for irrationals.

I know that, but he thinks every cut is uniquely determined by the rational ""in the middle"".

I've been down this road, sadly.

But there may not be a rational number in "the middle". So his argument fails.

Yeah, I know.

Since a cut is defined by creating subsets of Q, the number of possible cuts must be less than equal to the number of subsets of Q, which is |P(Q)|. It turns out for this case it's equal, and since |P(Q)| is larger than |Q|, the number of possible cuts is larger than |Q|.

Why? You can "cut" the rational line for each irrational. For example, sqrt(2) will "cut" the real line into two pieces and a such cut won't be considered by your argument.

Yes, it does. Consider a cut for sqrt(2). This cut doesn't have rational "in the middle" as you claim.

You can't do that. Given any two different rational numbers a and b, it's easy to show that (a + b)/2 will lie between a and b.

a pair of rationals with minimal difference are the end of A and the beginning of B:

Look up the definition of cut. Depending on whose definition, either A has no "end" or B has no "beginning" so either way what you wrote here isn't true, by definition.

My idea is that there can at most be only one cut between each rational number.

And your idea would be incorrect.

"Between each rational" doesn't even make sense, think about the words you are using before you hit submit.

You should probably learn what a cut really is before saying more.

A and B are specially defined sets of rationals, not rationals themselves.

Your first sentence is critically wrong. Suppose a, b are rationals. Then (a+b)/2 is a rational between them. Then there exists an irrational between a and b, between a and (a+b)/2, and between (a+b)/2 and b. In fact, there are infinitely many rationals between any two rationals, and infinitely many irrationals between any two rationals.

That's why dedekind cuts are identified by sets and not by rational numbers.

The first sentence is incoherent. Please rephrase your argument.

Dedekind cuts are not defined in terms of two rational numbers, but two special sets of rationals. However the contradiction is obvious even with your incorrect premise. Suppose an ordering gives a,b adjacency. Then let x be irrational s.t. a < x < b. Now suppose another ordering, since your statement should hold for all orderings since your choice was arbitrary, gives the ordering a,c,b. Then there exists an irrational y so a<y<c<b, i.e., a second irrational between x and y.

You can do that but then your dedekind cuts don't create the real numbers. Instead you defined Q as a subset of R.

If you can't see that this is nonsensical there's no point in discussion. As far as I can tell you're literally assuming without justification that there exists exactly one irrational for every rational.

Again, this is entirely without justification. You should review the definition of the reals using Dedekind cuts. Alternately, study Cantor's proof that there cannot exist a bijection from N to R.

But a Dedekind cut is a set of rational numbers and cannot be identified by two specific rationals.

Of course. But the claim that there exists at most 1 irrational between any two rationals is contradicted by the fact that there are infinitely many rationals between any two rationals as I demonstrated.

No, it isn't. Between any real numbers a and b, assuming we don't have a = b, you can find infinitely many rationals as well as irrationals.

I did just say that was true for rationals. It's true for all reals too, as I implied in my previous comment, but it's enough to say it about rationals to disprove the claim about irrationals.

Between any two irrational numbers, you can fine infinitely many rationals and irrationals.

I never disputed it

I read your post wrong. Yeah, I agree with you. There are more than one irrational between two given rationals.

But a Dedekind cut is a set of rational numbers and cannot be identified by two specific rationals.

But a Dedekind cut is a set of rational numbers and cannot be identified by two specific rationals.

But a Dedekind cut is a set of rational numbers and cannot be identified by two specific rationals.

But a Dedekind cut is a set of rational numbers and cannot be identified by two specific rationals.

But a Dedekind cut is a set of rational numbers and cannot be identified by two specific rationals.

But a Dedekind cut is a set of rational numbers and cannot be identified by two specific rationals.

But a Dedekind cut is a set of rational numbers and cannot be identified by two specific rationals.

But a Dedekind cut is a set of rational numbers and cannot be identified by two specific rationals.

Your supposed bijection is not a bijection, since you effectively map all irrationals to infinity.

It doesn't matter how you defined the real numbers. 0.1010101... and 0.11011011... are still real numbers

Not trying to make this an insult, but I encourage you to seek some therapy. You seem very paranoid, and this might be a symptom of some larger psychological issue that you'll want to get checked out.

All the best

Logical consistency how about that?

Yes, that is perfectly logically consistent. Feel free to produce a contradiction and collect your fields medal.

And what is inconsistent with that? Nothing about an element in a set says it cannot have infinite decimals.

That is because 0.4999.. and 0.5 are the same element. That is ONE element in the real numbers, it is ONE element with MULTIPLE REPRESENTATIONS, not two elements.

Just as 2/3 and 4/6 are the same element, just different representation.

This is because all elements in reals (and in almost any mathematical strucutre) is not an actual individual element, but a whole set of other elements that are collectively called an equivalence class. These equivalence classes are the the elements in the reals.

Cantors diagonal argument works still because you can just pick the representations that are most favourable to work with and discard the representations that are not, as you are only discarding representations, not elements.

An alternative method is to use cantors theorem where |S|<|P(S)| and then show that |R|=|P(N)|, which can be done in steps by showing that |P(N)|=|[0,1]|

A counter argument can be made by representing the real numbers in the interval [0, 1) as a binary tree, where each path from the root and down represents a binary sequence. And then the natural numbers are assigned from the root of the tree left to right level after level. Hence |N| = |R|.

No, cause you only get FINITE digits, and none of the infinite digits, ergo it doesn't work as it is not surjective.

Yes, you are correct. But what if I add a cut between each node in the (infinite complete) binary tree? And the cuts represent infinite number of digits.

Then it is no longer to the natural numbers. It doesn't matter how you try, Cantors thing showed it cannot be done.

The beauty of cantors theorem is that no matter what list you think you can come up with in any fashion, NO MATTER WHAT LIST!, he can produce an element you missed. It doesn't matter how good you think the list is, you have missed some elements.

To even have a chance to defeat it you would have to show cantors theorem wrong. Something you cannot as it is possible to prove in multiple different methods.

Yeah no, that is a nutorious crank and you shouldn't listen to him.

Your P can be put into a surjection with the real numbers, but not a bijection, as it contains duplicates.

There is where youre definition fails, because in your P, two sets represent the same real number.

The guy you linked to is a finitist and a crank. Please do not use him, you will only do yourself a disfavour.

it contains duplicates because it has the subset {1} and the subset {0,0.5,0.25,0.125,....} and {0.5,0.25,0.125,....} which all represent the same number

No, because zero is not a member of T. And I found this:

Even then you got 2 sets being the same real number so the issue persists.

So then it's possible to construct a new set D which is the limits of the sums of the rational numbers in each element in P. And then for example 0.4999... = 0.5. And then D is exactly the real numbers in the interval [0, 1] (If I start the natural numbers from 1 instead of 0). That's a much cleaner definition than using Dedekind cuts!

Actually it isn't as that assumes a craptonnes of things, for example it assumes a priori 1: The limit existing, which, if you only have rational numbers, most don't 2: It assumes you can do a sum of inifnitely many terms, which in rational numbers you cannot so there is an issue there.

To name a few, it is an increadibly ugly definition as it assumes a craptonne of things.

When you start going to the root of things for limits and all, defining what a limit is, you still suffer rom the limit not existing as an element. So to create this we need to use other things. This is where dedekinds cuts come in, we have the ability, by ZFC, to create sets like that and can show it will create the neccisery limits we desire.

We can also use cauchy sequences and do the same.

It is because the limit, when you only have rationsl, most often does not exist.

So when you ask for the limit, there is no limit to pick.

Hmm... So you mean for example that the irrational number e = sum(1/n!) is a case where a limit is possible to find, but that in general there is no way of determining such limits for all irrationals?

No, I say that without constructing it PRIOR to doing the limit, the number e doesn't exist!

You asking for an object that you have not yet constructed mathematicly which is non-sense. It is like asking for the sum of all natural numbers squared, it doesn't make sense to proclaim something as "the limit" without first constructing the object before and then show it satisfies the criteria of being the limit.

You are assuming the limit exists already.

Again, you assume that the limit exists which it may not do and in the rational number cases, msot don't.

it is trivial to determine that 0.999... is not in [0,1) and that "axiom" is problematic in on itself, you are trying to define real numbers there but your definition will fail several key properties we have decided that the real numbers must satiesfy.

For starters, basic operations like multiplication, addition and division don't work properly.

That doesn't work unfortunately as ordering of sets is only a partial ordering, it is not a total ordering.

(trichotomy law): For any a,b in S, either a<=b or b<=a.

Trichotomy is a property in an ordered set that says for any 2 elements a, b in S, that a<b, a=b or a>b, and only one at a time.

But yes, it is different than the set of real numbers, because in my definition for example 0.4999... is a different number than 0.5 whereas in the set of real numbers they are the same (due to the use of limits I assume).

Which also kills the supremum property, your system doesn't have it and CANNOT have it.

Instead of the set of real numbers R it can be called the set of standard numbers D, by forming numbers z.f where z is an integer and f is a member of P (the power set of T I defined earlier) D = {z.f | z∈Z, f∈P}

Again, why would you do this? This satisfies none of the properties we want and have wonky things such as as not all elements are comparable and even your definition z.f is non-sense. You haven't said what kind of structure it really is. You can go with ordered pair (z,f) but how do you compare then? (z,f)>(x,g) when what?

z>x and g is subset of f?

You have illdefined things and you cannot throw any number of axioms at something and hope it works, your thing is incoherent selfdefeating.

I ask you, why are you doing this?

Ok that's a reasonable question. If you suppose a map from the natural numbers to R and follow cantors argument to obtain a new real number which wasn't mapped to, the new one has only 0s and 1s. The trick is that whenever there was a '9' in the binary expansion of the nth real number mapped to, we have to be sure to make the nth digit of our new number 1 rather than 0. if we do this then there is no way that the new number was already in the sequence. (because we would have to have repeating 9s which is the same as a 0, not a 1).

That's a good question though, and a subtlety of the argument which is important to notice.

I am a mathematician. Your proof is wrong from the get-go.

You define S to be [0, 1] - the set of all the real numbers from 0 to 1 - and then claim that the function n is a bijection from N onto S.

However, the very first example you give, n(x) = 2^0 + 2^3 + 2^4 + 2^6 + ... does not in fact give you a number in S, i.e. a real number between 0 and 1 - indeed, the sum seems to be unbounded ("goes to infinity").

So you're obviously badly confused already.

In order to go further, I'd have to try to guess what "bijection" you really meant to talk about, which I'm not going to do. My idea is that if you cleaned up what you meant, you would eventually get a function n into [0, 1] but that it would be an injection - a function that's one-to-one but not onto.

As you can imagine, I'm very very very very doubtful you have a proof that the real numbers are countable. Were this true, all of measure theory and so basically all of calculus would be grossly inconsistent. But calculus does actually work - bridges stay up, airplanes don't fall out of the sky and so on.

However, if you're interested in going on, if you clean up your "bijection" I'll take a look at it for you.

This new function is a mapping from the natural numbers to all decimals of a finite length. It is an injection - indeed, it misses almost all numbers in [0, 1].

For example, which natural number x gives you 1/sqrt(2) as a result? (Heck, which natural number gives you 1/3 as a result?)

You seem to assume that a real number can have an infinite number of decimals. I claim that that's false.

Wait, what? What's the decimal expansion of 1/3 then? Or π? If reals can only have finitely many decimals, then they'd all be rational...

The sum of two natural numbers is always finite. So if I sum together infinitely many natural numbers, their sum must also be finite.

See some similiarities and an obviously wrong conclusion?
Induction does not apply to the limit case.

Exactly!

Induction allows you to prove something for every natural number, which notably doesn't include infinity.

So what makes you think that your argument for e being rational should be true? Where is the difference between your line of arguing and my argument that an infinite sum of natural numbers is finite?

What on earth does "finite yet unending" even mean?

That's an oxymoron if I've ever seen one.

Yes and since it's finite it's ending. Explicitly not unending.

"Finite" and "unending" are contradictory.

The same is therefore true for this sum tending to infinity.

Why?

You have that for any n, 1/1! + 1/2! + 1/3! + ... + 1/n! is a rational number. My question is: why does that imply that the result of the infinite sum is a rational number?

For example you also have that for all n, the set {1,2,3,...,n} is finite. But that does not imply that the set {1,2,3,...} is finite.

So your claim is that 1/1! + 1/2! + 1/3! + ... + 1/n! is rational for any n.

Not that 1/1! + 1/2! + 1/3! + ... is rational.

You what? An "uncomplete" infinite series is called a partial sum. An infinite series reaches infinity, hence infinite series.

But e is equal to the infinite sum 1/1! + 1/2! + 1/3! + ...

It is not equal to any 1/1! + 1/2! + 1/3! + ... + 1/n!, whichever n you choose.

An series is always a limit. Otherwise, it's just a sum.

An series is always a limit. Otherwise, it's just a sum.

You're assuming that something true for finite sums is also true for infinite sums, which is not true.

Other than that, you haven't answered what the decimal expansions of 1/3 or π are (and we can add e to that list) since they're finite according to you.

Infinite sums of natural numbers always results in a natural numbers, and all natural numbers have a finite number of digits.

Can you name one example of when this is the case? It's not possible unless you mean rational numbers instead of natural numbers, and even then it's not guaranteed. Infinite sums of rational numbers are often not rational.

You seem to think that inductance can tell you anything beyond finite cases, which is not true. As soon as you 'go to infinity' inductance goes out the window.

You misunderstand me. Can you give me an infinite list of natural numbers and then tell me what they sum up to?

For example, what is 1+1+1+1+...?

Interpret it however you like. You said it equals a natural number. which natural number? If you can't tell me then I have no reason to take you seriously.

ω-1 does not exist. If it did, it would imply that there is a largest natural number, which would imply that there are only finitely many natural numbers. So what is the sum equal to, if it is a natural number? If the sum is a natural number, then it must have a finite number of digits, so must be expressible in decimal (or binary) representation.

From a philosophical viewpoint,

This is math, not philosophy.

At each moment in time there is a defined natural number ω-1. However, in the next moment ω-1 has increased. So from a mathematical perspective we can take a snapshot in time and deal with ω-1 as a uniquely defined natural number.

There is no definition of the natural numbers that has them depend on the time. None of this makes any sense. I asked you to produce a natural number that is equal to 1+1+1+1+..., and you keep dancing around the issue. The number either exists, in which case it can be written down in decimal notation, or it cannot. The fact that you keep resorting to new symbols or pseudo-philosophy to describe this sum tells me that you are unable to evaluate it as a natural number.

there are no ended infinite processes.

Why not? Math processes don't take time, they're just done. Can you give me a good reason why 1+1/2+1/4+1/8+... doesn't equal 2?

If everything follows sound logic and works out to be consistent, then why shouldn't we be able to consider things like infinite sums?

It's a similar problem with the claim that real numbers can have an infinite number of decimals while at the same time claiming that infinity is not a number. Another false conspiracy construct.

That's not a problem at all. You can have infinite numbers, but these are usually ordinal numbers like you mentioned earlier, or cardinal numbers, such as in this case. Real numbers can have up to an Aleph_0 amount of digits, and there's no reason why they shouldn't.

While we are not required for things in mathematics to conform to reality, let's consider a real world example of an infinite number of steps that can be completed in finite time.

Imagine you are on one side of a room. Your first task is to make it to the middle of the room. Your second task is to make it half way between that point and the far wall. Your third is to make it half way again, and so on. Clearly you have infinite tasks, but these can be completed in finite time by just crossing the room.

So even in the real world (which math is not limited to) there is no problem with infinite sums/processes.

If you can describe how space is quantized go get your nobel prize in physics right now!

Is there any evidence that space is quantized?

Do you still think this junk?

Who cares about reality?

Please, actually read up on this instead of spreading misinformation.

You can claim that ω-1 exists all you want if you can't prove or define it in a consistent way it doesn't matter.

The natural numbers aren't in a state of flux. They are a set that is static and unchanging. They are described by an infinite process, but the actual set is the result of the finished process.

This isn't philosophy, this is math. And even if it were philosophy then that obviously contains a notion of completed infinite processes as well. You yourself invoked Zenos paradoxes earlier. The fact that you can move shows you that you can complete an infinite process.

It is used everywhere in mathematics, that's what limits are, that's what the natural numbers are. There are extensive proofs for when talking about a completed infinite process makes sense and when it doesn't.

Your lack of basic mathematical knowledge doesn't mean there is a grand conspiracy, it means that you have a lot to learn and would rather believe that there is a massive conspiracy than that your intuition might be wrong.

This is trivially false unless you consider 0 a natural number. Either case, limits don't have to be rational. So you're double wrong.

Well, consider for example e = sum(1/n!).

No, just answer his very simple question: what is the decimal expansion of 1/3? If you can't answer that, then there's no point discussing anything with you. Don't avoid a very simple question by diverting attention to another one.

Who said mathematics was an empirical science? Show me the number 1 please.

Congratulation, welcome to mathematics. We do not care about physical reality.

This is false. A rational Cauchy sequence does not have to converge in the field of rational numbers.

Well, consider for example e = sum(1/n!). Adding two rational numbers results in a rational number. The same is therefore true for this sum tending to infinity. So e is then, actually a rational number.

What is true for any amount of finite cases does notm ean it is true for the infinite case.

For any given finite set, if I attatch another element to this set, i get another finite set. This happens for any number of finite amount of itmes I do it, but if I do it infinitely, I have an infinite set!

MrNeson you moron. Real numbers CAN have infinite decimals, that is one of their properties. Those with finite decimal expansions are 10-adic rationals which is a subset of rational numbers which is known to be countable.

So you have not proven anything new, but your own ignorance.

For your proposition to be true, you must find a rational number that gives squareroot of 2, but that doesn't exist so squareroot of 2 CANNOT have finite decimal expansions.

WHat natural number do you get from \sqrt(2)/2?

HAHAHAHAHAA

Read this. http://lesswrong.com/lw/j8/the_crackpot_offer/

That's incorrect, but even assuming it was true, what does this have to do with "the deep state"?

Okay, let's take 1/3 and calculate your function: 1/3=0.01010101... in binary. So n(1/3)=2^2+24+26+28, this clearly diverges. So your function n does not map to the natural numbers. And even if you extend the natural numbers to include infinity, you would still get n(1/3)=n(2/3)=infty, so n would not be a bijection because is wouldn't be injective.

Knock yourself out: neither does OP.

My idea is that there can at most be only one cut between each rational number.

And your idea would be incorrect.

"Between each rational" doesn't even make sense, think about the words you are using before you hit submit.

You should probably learn what a cut really is before saying more.

That's why dedekind cuts are identified by sets and not by rational numbers.

The first sentence is incoherent. Please rephrase your argument.

Dedekind cuts are not defined in terms of two rational numbers, but two special sets of rationals. However the contradiction is obvious even with your incorrect premise. Suppose an ordering gives a,b adjacency. Then let x be irrational s.t. a < x < b. Now suppose another ordering, since your statement should hold for all orderings since your choice was arbitrary, gives the ordering a,c,b. Then there exists an irrational y so a<y<c<b, i.e., a second irrational between x and y.

You have that for any n, 1/1! + 1/2! + 1/3! + ... + 1/n! is a rational number. My question is: why does that imply that the result of the infinite sum is a rational number?

For example you also have that for all n, the set {1,2,3,...,n} is finite. But that does not imply that the set {1,2,3,...} is finite.

Can you explain dedekinds cut to me?

The square root of 2 isn't rational.

You don't get to at most |Q|. /u/MrNeoson doesn't know what he's talking about.

I agree that it seems false. It's certainly unintuitive.

But there are many ways to prove it. I wrote down a proof for that in our conversation here.

Just, you know, turn down your arrogance a tad. Those are basic concepts that anyone with a higher math education knows and you think you are overthrowing established knowledge without fully grasping it.

That's just plain wrong.

Well since that isn't a rational number your whole bijection falls apart.

This is simply not true, since there are more cuts than rational numbers.

But that doesn't work for irrationals.

(trichotomy law): For any a,b in S, either a<=b or b<=a.

Trichotomy is a property in an ordered set that says for any 2 elements a, b in S, that a<b, a=b or a>b, and only one at a time.

But yes, it is different than the set of real numbers, because in my definition for example 0.4999... is a different number than 0.5 whereas in the set of real numbers they are the same (due to the use of limits I assume).

Which also kills the supremum property, your system doesn't have it and CANNOT have it.

Instead of the set of real numbers R it can be called the set of standard numbers D, by forming numbers z.f where z is an integer and f is a member of P (the power set of T I defined earlier) D = {z.f | z∈Z, f∈P}

Again, why would you do this? This satisfies none of the properties we want and have wonky things such as as not all elements are comparable and even your definition z.f is non-sense. You haven't said what kind of structure it really is. You can go with ordered pair (z,f) but how do you compare then? (z,f)>(x,g) when what?

z>x and g is subset of f?

You have illdefined things and you cannot throw any number of axioms at something and hope it works, your thing is incoherent selfdefeating.

I ask you, why are you doing this?

And also 1.999... which makes it not unique