# Function Generator for Sequences of Reals

1. Jun 22, 2013

### krausebj0

This paper is still subject to being edited and I am an amatuer mathematician, so there may be some mistakes, typos, and "amatuer" notation. I really cannot believe the result here, so I assume something may obviously be wrong yet, and I just cannot figure it out. I am a CPA, and fairly good at processes, etc.,... maybe just not seeing something right though.

Brief Introduction-

Below are listed steps for purposes of analyzing different Ai. Step 5 is optional and is for testing a cardinality argument. This paper is comprised of 5 steps, a conclusion, and a footnote entitled "Function f and Cantor's Theorem." Steps 1 through 4 are basic definitions and processes. Step 5 is a simple test to check whether function f (defined in step 3) is a surjection or not.

Step 1)

Let D be the set of all dyadic rationals greater than 0 and less than 1 (work in binary). For visually considering results after step 5, also define the sets Fn. Start with a list of countable sets F1, F2, ..., where each Fn is equal to the set of all dyadic rationals containing "n" number of 1's and the union of all Fn would equal D:

F1 = { 0.1, 0.01, 0.001, 0.0001, ...}
F2 = { 0.11, 0.101, 0.1001, ..., 0.011, 0.0101, 0.01001, ..., 0.0011, 0.00101, 0.001001, ...}
F3 = { 0.111, 0.1101, 0.11001, ...,}
.
.
.

Step 2)

Define the set I:

The set I is analagous to the set of all binary strings with infinitely many ones. The set I is equal to the set of all reals greater than 0 and less than or equal to 1 only if all elements are expressed as infinitely large binary strings, where dyadic rationals and the number one are expressed as:

1 = 0.111...
0.1 = 0.0111...
0.101 = 0.100111...

and the unordered set I would look something like:

I = {0.111..., 0.1010101..., 0.111000111..., i in I}.

Define the set RI:

The set RI follows naturally from the set I and is best illustrated by example as opposed to a rigorous definition. For all elements of RI, there is one and only one element of I that naturally corresponds to it. Define RI by example and then let function 'k' be a bijection going from I to RI and let its inverse function 'h' be a bijection going from RI to I:

0.111... <---> { 0.1, 0.11, 0.111, ... }, where k(0.111...) = { 0.1, 0.11, 0.111, ... }
0.10101... <---> { 0.1, 0.101, 0.10101, ... }, where k(0.10101...) = { 0.1, 0.101, 0.10101, ... }
0.1101... <---> { 0.1, 0.11, 0.1101, ... }, where k(0.1101...) = { 0.1, 0.11, 0.1101, ... }
...
i in I <---> ri in RI

Step 3)

Define the sequence Ai:

The sequence Ai can be shown as a set and should have a cardinality equal to or less than aleph null (the smallist infinite cardinal). Each element of Ai should be an element of 'I'.

Ai = i1, i2, i3, ... , i_n, where i_n is in I

Define the function g:

Function g goes from D to RI.

Formally: For any dyadic d in D, g(d) is equal to ri in RI such that d is in ri, h(ri) = i_n in Ai, and there does not exist an element i_(x < n) of Ai such that d is in k(i_(x < n))

I will illustrate through three iterations:

If 0.111... is the first iterated real of Ai:

g(0.1) = {0.1, 0.11, 0.111, ...}
g(0.11) = {0.1, 0.11, 0.111, ...}
g(0.111) = {0.1, 0.11, 0.111, ...}
.
.
.

If 0.10101... is the second iterated real of Ai, and 0.1 has already been related to 0.111... above:

g(0.101) = {0.1, 0.101, 0.10101, ...}
g(0.10101) = {0.1, 0.101, 0.10101, ...}
.
.
.

If 0.101101... is the third iterated real of Ai, and 0.1 and 0.101 have already been related above:

g(0.1011) = {0.1, 0.101, 0.1011, ...}
g(0.101101) = {0.1, 0.101, 0.1011, ...}
.
.
.

... and so on.

Define Function f: (See Footnote 1 for applicability of Cantor's Theorem)

f = h(g()), where f(d) = i, d is in D, and i is in I

Visually the sets Fn can be "covered." Example, if i1 = 0.111...:

k(0.111...) = {0.1, 0.11, 0.111, ...}, "covering" the sets Fn as follows:

F1 = { [0.1], 0.01, 0.001, 0.0001, ...}, where 0.1 is "covered" by 0.111...,
F2 = { [0.11], 0.101, 0.1001, ..., 0.011, 0.0101, 0.01001, ..., 0.0011, 0.00101, 0.001001, ...}, where 0.11 is "covered" by 0.111...,
F3 = { [0.111], 0.1101, 0.11001, ...,}, where 0.111 is is "covered" by 0.111...,
.
.
.

Step 4)

Define A = { d : f(d) = i1 } U { d : f(d) = i2 } U { d : f(d) = i3 } U ... U { d : f(d) = i_n }

A will always have a cardinality equal to aleph null so long as Ai contains one element. If Ai contains enough elements, A = D.

Define "Complete"

Step 4 is considered complete if and only if A = D.

Step 5)

This step is the optional cardinality test. It is important to realize, for purposes of our test, what we know, what we don't, and what our test can be expected to achieve. We want to test an arbitrary real to see if it has been included in an Ai. The example uses 0.111... for simplicity. The result is a yes or no answer.

First lets run a full scale example of steps 1 through 4 so we can proceed:

Ai = i1, i2, i3, ... (selected "randomly" for purposes of step 5)

i1 = 0.1011... relates to i1D = { 0.1, 0.101, 0.1011, ...}
i2 = 0.0011... relates to i2D = { 0.001, 0.0011, 0.0011..., ...}
i3 = 0.1101... relates to i3D = { 0.11, 0.1101, ...} (where 0.1 was related to i1)
i4 = 0.11101... relates to i4D = { 0.111, 0.11101, ...} (where 0.1 was related to i1 and 0.11 to i3).
.
.
Assume Step 4 is complete as defined in step 4.

Let A = i1D U i2D U i3D U ...

A) We assume yes, 0.111... is an element of Ai, and there is no contradiction. A = D. For now, the possibility that 0.111... was iterated remains open.

B) We assume no, 0.111... is not an element of Ai, and then see if our assumption results in any contradictions. The answer is yes, a contradiction results. How does the contradiction result? This part is probably the trickiest of the proof:

b1) Check what happened to the dyadics relatable to 0.111...:

0.1 was related to i1, so i1 becomes a: i1 = a
0.11 was related to i3, so i3 becomes b: i3 = b
0.111 was related to i4, so i4 becomes c: i4 = c
and so on...

Notice that now b is closer to 0.111... than a, c is closer than b, and so on. The sequence a, b, c, ... is well ordered. Ai must be at most a countably infinite set, so we know that the sequence a, b, c, ... must in turn be a finite sequence with a last term.

b2) We are now free to define sequence (set) Bi and set B:

Bi = Ai U {0.111...} (where Bi has been shown as a set and not a sequence for simplicity)

B = A U {0.1, 0.11, 0.111, ...}

We note that B and Bi are countable, so if Ai is replaced with Bi, there must be dyadic rationals that are related to each element of Bi, and for each element of Bi, there will be at least one (probably an infinitely many) number of dyadic rationals that are related to it and only it, including our test number.

b3) We can now compare Bi to Ai. As an optional step, we can well order Ai and Bi to ensure our tested number is the last element of the set Bi. We know that Bi contains our tested number while Ai may or may not. We necessarily assume Ai does not, so in turn, we must presume that Bi results in at least one dyadic rational having been related to our test number while Ai presumably does not.

b4) Having related at least one dyadic to our test number and to no other real number as a result of iterating Bi, we can now compare A to B. Because we well ordered Ai and Bi for simplicity, we know that 0.111... was the last element of Bi iterated. We know that Ai is a subset of Bi, and that all elements of Ai are related to exactly the same dyadic rationals as Bi, so in turn A is a subset of B. The fact that at least one dyadic is related to our test number when it was the last number iterated is proof that B - A is not empty.

b5) If B - A is not empty, then B contains more elements of D than A does

b6) If B contains more elements of D than A does, we can assume A does not equal D

b7) If A does not equal D, then step 4 was not complete

b8) If step 4 was not complete, then we have a contradiction, as step 4 is necessarily complete before we can assume that our test number was not included in Ai.

C) Conclusion: If we assume 0.111... was an element of Ai and A = D, there is no contradiction. If we assume 0.111... was not an element of Ai, we end up with the contradiction that step 4 is not complete and A does not equal D. We therefore conclude that 0.111... was in fact an element of Ai. The answer is "Yes".

Overall Conclusion

Function f, as defined in step 3, is a surjection from the set D onto the set I if each element of D is related to an element of I and all elements of I are related to one or more elements of D. Step 5 tests whether an element of the set I could be omitted from a set Ai if A = D and step 4 is complete. Step 5 proves by contradiction that no element of I can be omitted from the set Ai if A = D and step 4 is complete. It is concluded that function f is a surjection from the set D onto the set I.

-Foot Note 1, Function f and Cantors Theorem-

From Step 3: Step 3 defines function f where f(d) = i, d is in D, and i is in I.

Function f has two parts. I will break them into function g and function h so the iteration process can be detailed.

Function g goes from D to RI. It is illustrated again through three iterations:

If 0.111... is the first iterated real of Ai (Ai is defined in part 5):

g(0.1) = {0.1, 0.11, 0.111, ...}
g(0.11) = {0.1, 0.11, 0.111, ...}
g(0.111) = {0.1, 0.11, 0.111, ...}
.
.
.

If 0.10101... is the second iterated real of Ai, and 0.1 has already been related to 0.111...:

g(0.101) = {0.1, 0.101, 0.10101, ...}
g(0.10101) = {0.1, 0.101, 0.10101, ...}
.
.
.

If 0.101101... is the third iterated real of Ai, and 0.1 and 0.101 have already been related above:

g(0.1011) = {0.1, 0.101, 0.1011, ...}
g(0.101101) = {0.1, 0.101, 0.1011, ...}
.
.
.

... and so on.

Function h goes from RI to I:

{0.1, 0.11, 0.111, ...} ---> 0.111...
{0.1, 0.101, 0.10101, ...} ---> 0.10101...
{0.1, 0.101, 0.1011, ...} ---> 0.101101...
.
.
.
ri in RI ---> i in I

Function f can then be defined:

f = h(g()), where f(d) = i, d is in D and i is in I

Let function g go from D to RI.

Define Cantor's A = { d : d is not an element of g(d) }, where d is in D and g(d) is equal to some element ri, ri is in RI.

Cantor's A = {} every time, regardless of the Ai selected. Cantor's Theorem cannot be relied upon to show that function 'g' cannot be a surjection from D onto RI. One must assume that Cantor's diagonal argument then may prevent Ai from being listed so that all elements of the set I are included, but to test that assumption, one should go to step 5.

2. Jun 22, 2013

### mathman

It would help if stated at the very beginning what you are trying to show.

3. Jun 26, 2013

### krausebj0

I see that nobody has noted anything wrong. The result is enormous if there is nothing wrong, so I will point out what I see:

1) I am not trying to prove anything necessarily, as step 5 is optional...

2) Steps 1, 2, and 3, are just basic definitions and are sound, even if some amateur notation is present.

3) Step 4 addresses the notorious “halting problem.” Namely, there will exist a last element of the sequence Ai such that the process halts and A = D. Completeness is defined given step 4 is complete if and only if A = D. This is basically saying, “The sequence (set) Ai is countably infinite, but at the same time has a last element i_x such that the process halts and step 4 is complete upon iteration of i_x.” Alan Turing is famous for work that deals with this sort of problem.

4) Step 5 is a logic argument. It goes line for line, so must be read carefully. Note the sequence a, b, c, … is obviously finite for any element of Ai (or Bi).

5) Ultimately Ai and Bi are sequences, but they are best manipulated as sets... Hopefully my notation above does not confuse anyone. Well ordering Ai and Bi for step 5 is equivalent to invoking the axiom of choice, and in turn the well ordering theorem, if working in ZF set theory. (The work is done in binary and is readily adaptable to ZF or other axiomatic theories)

6) This gets into the work of Goedel (whose work relies on Cantor’s), and many others. Goedel’s first incompleteness theorem shows that a complete and consistent list of axioms, even if countably infinite, cannot be achieved for most theories. Enumerating the reals would obviously involve rewriting some textbooks here...

7) Is there a line that is wrong above? I would be looking for one to quote the specific line that is mathematically unsound, not just notationally wrong.

Thank you!

4. Jun 27, 2013

### gopher_p

If I'm reading correctly, you claim to have essentially constructed a function which maps the dyadic rationals in (0,1) surjectively onto the reals in (0,1). Cantor's diagonal argument demonstrates that such a function cannot exist. So, yes, your argument/construction/paper is flawed.

The place where your paper first breaks down (in more ways than one) is Step 3. You have not clearly defined what you mean by Ai. It appears that you intend for it to be a "random" enumeration of the elements of I, but you would first need to show that such an enumeration exists (it doesn't). Also, it is not clear that your function g is well-defined (Well-defined is a precise mathematical term. Please do not assume that you intuitively understand what it means). Since the rest of the "paper" relies substantially on the flawed definitions in Step 3, I haven't bothered to check it for correctness. I reckon it falls under the "so bad it's not even wrong" banner.

Protips: If you want mathematicians to read your work and take it seriously, you should probably make an effort to learn the terminology, notation, etc. that is commonly used to express mathematical ideas. Apologizing for your "amateur" notation doesn't make your work any less unreadable (and it is basically unreadable). Also, proof/definition-by-example is a pretty big no-no. While examples are often used to help shed light on precise proofs and definitions that are potentially confusing, they are not to be used in lieu of a proper proof/definition (unless it is extraordinarily clear how the example applies to the general case).

5. Jun 27, 2013

### krausebj0

I am familiar with Cantor's diagonal argument and have addressed it, so I am not following. I do understand that it has, for many years, been used to show that the reals cannot be enumerated.

Ai is nothing more than a sequence of real numbers. It is up to the mathematician to choose them. It is only "random" for step 5, but it does not have to be. A misunderstanding of Ai would lead you to be unable to interpret function g correctly, so I assume this is where we are at.

I can say right away that the function g is defined for each element d of D where there exists an element i_n of Ai such that d is in k(i_n). Function g is "well-defined" for each element of Ai. There is a reason that each element of Ai must be defined prior to function g. If not, Cantor has the advantage.

My notation is not that bad and you can follow. Please do not be naive in the sense that it cannot be cleaned up. Everything from step 1 through 3 has been verified to me as being mathematically sound, including function g.

You must quote the original post if you wish for me to take you seriously.

Last edited: Jun 27, 2013
6. Jun 27, 2013

### krausebj0

The proof goes in order, so the goal would be to find the first line of the proof that is mathematically unsound. I am aware that the notation could be improved, but I also know that it is enough.

7. Jun 27, 2013

### krausebj0

Ai can be random for step 5. Here is why. If one just starts plucking pieces of hay out of a stack, then the person does not have to order the haystack before starting to pluck. The person merely has to start plucking. The process completes when the last piece of hay has been plucked.

In my case, the process completes when A = D. There must exist an element of Ai such that iterating that element results with A = D.

8. Jun 27, 2013

### krausebj0

Ai can be finite. This is like a turing machine, where Ai is the tape.

If step 4 is complete, there exists an element of Ai such that iterating that element results with A = D...

Hope the above helps...

9. Jun 27, 2013

### gopher_p

Cantor's diagonal argument demonstrates that your conclusion is false. There cannot be a surjection from D onto I.

That's all you really had to say. You're not defining anything. You can elaborate to get at your intended meaning; i.e. Let Ai be an arbitrary sequence of elements from I such that no element occurs twice in the sequence. If you want to permit Ai to be finite, you could include "possibly finite" in your description. Anyone capable of reading a purported proof of the countability of the reals knows what a sequence is and knows the limitations on its cardinality. Also, what is the purpose of the i? You're apparently using i's to denote elements of I, but this sequence doesn't seem to have anything to do with a particular i.

So g is really only defined for a subset of D and (possibly) not all of D. FYI, functions are well-defined on their domains or at elements of their domains, not on their ranges/codomains/images or at elements of their ranges/codomains/images.

It's pretty bad. And if can be cleaned up, clean it up. It's your job to make it readable, not the readers job to figure out what the hell you're talking about.

While your intentions may be mathematically sound, what you have written is not. You have very explicitly written that g is a function on D when, in general, it is not.

I am taking my valuable time to point out to you why your argument is wrong and why your mathematical writing is bad. You came here asking for feedback on a convoluted proof of a claim that is patently false. No one else even bothered to offer actual feedback. I don't need to do this. I certainly won't be wasting any more of my time with you if you persist with this attitude. I have taken the time to make some sense of your original post (i.e. "fix" the first four parts so that the fifth actually makes sense). I know what is wrong, and I am willing to share that with you. But this is not going to be an argument. You are wrong. If you're not willing to accept that, there is not point in me trying to convince you.

10. Jun 27, 2013

### krausebj0

I agree that my conclusion must be false "according to Cantor's Theorem". That is the point of this thread. We agree (you just think we disagree).

Ai is the tape. Step 3 defines the turing machine. You are very confused here. The purpose of this is to analyze different Ai. The mathematician must select different Ai, probably starting with some finite sequences, before they can tell what is going on here. You have not come close to doing this yet. You think Ai must be rigorously defined in step 3 as x, y, z, ...

If A does not equal D, then function g would only be defined for a subset of D, correct.

You do not have to read this. It is your job to figure out my notation. This is not a published, edited paper, and you are not an editor.

I wish you would not take your valuable time to clog my thread until you have thought this through more, actually. I appreciate what you are trying to do, but not what you have done.

11. Jun 27, 2013

### krausebj0

The weight of these words broke your jaw gopher_p. Mine too though at first, so please do not think that I am degrading you. Until you have completely made sense of the following, I respectfully ask you to not post again and also thank you for your time thus far:

"There is a reason that each element of Ai must be defined prior to function g. If not, Cantor has the advantage."

Let's really think this through. To do so, consider what the difference is between a codomain and a range when we are talking about a function. Cantor's proof says an element of the codomain is not in the range. By defining the range (as being equal to the codomain) before the function, we can conform the function so that it does in fact output to the desired range. Cantor's diagonal cannot obstruct an output to a pre-defined range if Cantor's diagonal is in fact included in that range. I know that was a mouthful...

All elements of Ai will be included in the range of function f (even Cantor's diagonal, should it be included in the Ai).

Ai can have one element. Ai can be infinitely large. I suggest starting with a few finite Ai and switching the order up to see what happens.

Always remember, Ai is the tape, and step 3 the turing machine.

12. Jun 27, 2013

### krausebj0

Here is a real brain teaser:

Let Ai = 0.10111..., 0.110111..., 0.1110111..., ...

Is Ai an infinite sequence?

Elementary mathematics of course teaches us yes. The answer is actually no. Why not? A does not equal D. Ai is only an infinitely large sequence if A = D.

Proof:

Consider 0.111...

Show Ai as a set:

Ai = { 0.10111..., 0.110111..., 0.1110111..., ... }

Ai is well ordered.

Define Bi = Ai U {0.111...}

Well order Bi.

If we now again show Bi as a sequence. It will look like Ai, but it is finite, because it has a last term which is equal to 0.111...

If Bi is finite, then Ai must also have been finite.

Last edited: Jun 27, 2013
13. Jun 27, 2013

### kielejocain

Note: To say a set $S$ is 'countable' is to say you can set up a bijection from $S$ to the natural numbers $\textbf{N}$; in layman's terms, you can list them all. Listing them all isn't as easy as it may seem.

If two sets are in bijection, you can think of them logically as the same, in the sense that renaming all the elements of one gives you the other. The "renaming" function is precisely your bijection. Thus, all countably infinite sets are, as sets, no different than $N$. This seems strange to non-mathematicians when we say things like, "the integers $\textbf{Z}$ and the integers $2\textbf{Z}$ are the same." Most people would say that one is a subset of the other. Yes, strictly speaking, leaving them as named, they are not equal; one is strictly contained in the other. But up to a renaming (up to bijection), they are the same. On the other hand, the set $\{1, 2, 3, 4\}$ and the set $\{2, 4\}$ are not in any way the same; there is no relabeling of one that gives the other. This is not surprising. What is surprising is that there is no relabeling of the reals ℝ that gives $N$; you cannot count the reals.



In other words, you've set up bijections from the set $D$ and subsets $F_n$ to $\textbf{N}$. We're good so far; $D$ is countable, and subsets of countable sets are countable.

Now you've redefined the interval $(0,1]$ in a very slight way (usually we take the finite choice) in hopes that it will be useful in the new form; no problems yet.

Notation: Your last line here is a bit strange; it's unclear what you mean by "i in I.'' I'm assuming at the moment you mean that $i$ is to represent a generic element in $I$. When you put a list between curly brackets, you're implying everything listed is an explicit element of the set. If you want to give a label to represent an arbitrary element, just say, "let $i\in I$" (or, "let $i$ be in $I$").

Not to a mathematician; if you can't define it properly, you can't use it. Surely it can be useful to give example to clarify a tedious definition, but the tedious definition is what's most important to the logic of your argument.

Ok, let's define this properly. Let $i\in I$. You want to create the set $R_i$ which is the subset of the dyadic rationals $D$ whose finite presentation as a binary decimal matches that of $i$.

Let $i\in I$, with $i=d_1 d_2 d_3\ldots$, $d_j\in\{0,1\}$ the binary representation of $i$. Define the set $R_i\subset D$ to be the set of dyadic rationals

$R_i=\{d_1d_2\ldots d_n\ |\ n\in N\}.$

Let $RI=\{R_i\}_{i\in I}$; then there exists a bijection $k:I\to RI$ with inverse $h$ (note: all bijections between sets have inverses).

Notation: Notice how my sets look here. My definition of $R_i$ gave a specific, generic element, and simply noted after a | where the ambiguity arose and how ambiguous it was. Then in defining $RI$, I used $I$ as an indexing set; alternatively, I could've used the same notation as for $R_i$:

$RI=\{R_i\ |\ i\in I\}$.

All you've really done right now is give a really complicated 'name' to each number in $(0,1]$; my guess is that this is what will lead to your confusion.

So you've defined a (countable) subsequence $A_i$ of elements of $I$, and a function $g:D\to RI$. The sequence is fine as presented for now, but your function $g$ is your first major problem. Given an arbitrary sequence $A_i$ of elements in $I$, there is absolutely no reason for there to be an element that matches a dyadic rational $d$ for its entire length. For example, here's my sequence:

$A_1=.01111111\ldots$
$A_2=.00111111\ldots$
$A_3=.00011111\ldots$
etc.

What is $g(.1)$? It's going to have to be an $R_i$ where the first decimal digit of $i$ is 1 (so that $.1\in R_i$), but then $h(R_i)=i$ is not in the sequence $A_i$ by construction.

It is possible to CONSTRUCT a sequence $A_i$ that satisfies these conditions; in particular, just pick an ordering $D=\{d_1, d_2,\ldots\}$ for $D$, then define $A_i$ to be all the digits of $d_i$with infinitely many 1's following. Then for any $d\in D$ there will be at least one $A_i$ with $d\in R_{A_i}$; choose the first such one to be $g(d)$ (there will definitely be more than one).

Based on a previous comment of yours, I think YOU think the sequence $A_i$ hits all of $I$; it does not and can not. To write all elements of a set in a sequence is precisely the same as to define a bijection between that set and $\textbf{N}$. Since you know $I$ is uncountable, you know your process of "picking" hay pieces (that is, of assigning natural numbers to elements of $I$) will never "end" (not even in countably infinitely many steps). You are necessarily going to run out of your countably infinite numbers before you run out of uncountably infinite hay pieces.

Whenever you describe a process by which you do something one step at a time, you are basically counting.

Sure, it works for those three elements. Without describing the full sequence $A_i$, there's no way to guarantee it continues to do so.

Yep: mathematically, you're simply saying $f=h\circ g$, the composition of $h$ and $g$. Certainly $f:D\to I$.

"Covered" in what way? Are you saying that for every $d\in D$ there is an $A_i$ with $d\in k(A_i)$? If so that's true, but has nothing to do with the $F_n$. You built the sequence $A_i$ precisely that way. Each $F_n$ is just a subset of $D$, and so obviously each element belongs to some $k(A_i)\in RI$ by construction.

I'm not sure what you mean by, "If Ai contains enough elements, A = D." I've shown you that there are sequences for which you can show $A\neq D$, and I've shown how to construct a sequence $A_i$ such that $A = D$ for sure. Perhaps you intended $A$ to be the smallest FINITE union such that $A = D$; if that's the case, you will never succeed. Choose the first power $k$ of 2 such that $2^k > n$; there are more than $n$ dyadic rationals that have their final 1 in the ($k+1$)-st position (as there are $2^k$ of them).

Obviously there must be SOME flaw with this argument, as it would conclude that arbitrary reals (by which you mean in the interval $(0,1]$, but those are of the same cardinality) have been listed in a sequence.

I think this may be a major flaw in your logic; you seem to be preparing to show that every element in $I$ is an element in $A$, but you defined $A$ to be a subset of $D$. What you're really testing is whether every $i\in I$ is in the sequence $A_i$; you don't need $A$ for that.

We definitely don't know that; there is no a priori reason why your sequence of dyadic rationals couldn't have been related to elements of $A_i$ farther and farther along the chain. What you define as $a, b, c,\ldots$ are simply the first element in the sequence $A_i$ that starts with 1, 2, 3, etc. 1's. This sequence will be finite precisely if your $.1111\ldots$ belongs to it. You've sneakily begged the question.

A sequence and a set are not the same; a set is simply a 'relatively small' collection of elements; a sequence is a countably infinite ordered set. Of particular issue here; where does $.111\ldots$ fall in the order of $B_i$, and what order do you intend the sequence $B$ to run in?

You've not demonstrated that every element of $A_i$ is related to some dyadic rational, nor have you demonstrated that $.111\ldots$ is related to some dyadic rational. Surely there are dyadic rationals that COULD be related to it, but you've not demonstrated that any ARE. Any dyadic rational COULD be related to an uncountably infinite number of elements of $I$, and even to a potentially infinite number of $A_i$. You only get to choose one of them; $f(d)$ is that choice. What if $.111\ldots$ is never that choice? What if $.111\ldots$ isn't even in $A_i$, and couldn't possibly be that choice?

I'm stopping here, as I hope I've demonstrated sufficiently that your argument has unfixable holes. Also, the lights came back on. Hi, gopher_p.

Surely you must know your argument has such holes, as its conclusion contradicts an established fact and therefore must be unsound.

You usually learn more getting things wrong than you do getting them right.

Last edited: Jun 27, 2013
14. Jun 27, 2013

### krausebj0

I am very familiar with cardinalities, etc. I trust you are just trying to ensure we are both on the same page?

I have addressed Cantor's Theorem in the proof and above, so I will disregard your putting words into my mouth by saying I cannot count the reals.

Thanks for the notational suggestions, especially for I and RI.

You, like the other first poster, have really messed up Ai. Think of this as a turing machine where Ai is the tape. That is the best way I can explain it.

g(0.1) will equal something if A = D. You are way off here... not sure what to say.

You assume that I assume Ai hits all elements of I. This I do not do.

N has nothing to do with this, other than you could note that D is countable.

The "covering" is just a visual aid because the point is to analyze different Ai. So are the sets Fn. Once you understand the above, you'll get the covering part too. It would be good if I could show how i1 covers the sets Fn, then how i2 does, then how i3 does, and so on. I was trying not to be messy. Ai can contain only one element too though.

The rest of what you say is just so far off because you did not get the above, that I really did not read it. Sorry. I know you are just trying to help, but there is no way anything you have suggested so far, save for notational suggestions which I do appreciate, has been shown as being mathematically wrong with my proof.

Again, I am looking for someone to show the first line that is wrong with the proof. Just one line will do. No notational lesson is needed here.

15. Jun 27, 2013

### krausebj0

Kielejocain,

Have you ever heard of the phrase, "My mind is open." You cling to the old textbooks above...

Here is a real brain teaser:

Let Ai = 0.10111..., 0.110111..., 0.1110111..., ...

Is Ai an infinite sequence?

Elementary mathematics of course teaches us yes. The answer is actually no. Why not? A does not equal D. Ai is only an infinitely large sequence if A = D.

Proof:

Consider 0.111...

Show Ai as a set:

Ai = { 0.10111..., 0.110111..., 0.1110111..., ... }

Ai is well ordered.

Define Bi = Ai U {0.111...}

Well order Bi.

If we now again show Bi as a sequence. It will look like Ai, but it is finite, because it has a last term which is equal to 0.111...

If Bi is finite, then Ai must also have been finite.

16. Jun 27, 2013

### krausebj0

RI is a subset of the infinitely large elements of the powerset of dyadic rationals. I am not so sure your above notation checks out Kielejocain. I take that back. Let's just keep things simple, as I DO TRULY feel that the visual representation of RI is best for this. It is clearly obvious what RI is, is it not?

17. Jun 27, 2013

### krausebj0

This thread is going to turn into a nightmare if people do not start doing the one thing I have asked. Go in order. Pick the first statement you think is wrong. Ignore everything else. Quote the line of the op you think is wrong and nothing else. Discuss only it.

This is really getting ridiculous. All of you people are trying to help, yes, but none of you are. Follow the simple instructions please.

18. Jun 27, 2013

### kielejocain

Why is Bi finite? Because it has a 'last term?' That isn't the mathematical definition of finite. If you redefine finite to mean this, then you aren't doing what most people call math anymore; you're doing your own version of math. There's nothing inherently wrong with that, but you can't then ask me to show you where you're wrong; you're creating your own system of mathematics. Of course, no one will take it seriously until it has the predictive/descriptive power of what most of us call 'math' (what you call 'old textbooks'), a model of the universe in which, among other things, a 'finite set' is one with cardinality in the natural numbers. Bi doesn't satisfy that property. If you want to write new textbooks, then godspeed to you; no one on this forum will be of any use to you.

I have an open mind, I assure you. I also have quite a long view of the consequences of your definition of finite; I'll stick with the one in the textbooks for now.

19. Jun 27, 2013

### krausebj0

Again, you are clinging to the old textbooks, but I will acknowledge your argument by accepting that you think the cardinality of Bi is greater than that of aleph null.

A finite set is a recognized term. It will have a cardinality less than aleph null. If I say that both Ai and Bi have finite cardinalities, you could question my logic, sure. I would too. Same if I say they are finite sequences (by this I mean they have a last term). Ok. Fair enough.

No two elements of Ri are the same. They are all disjoint. The further any two elements of I are from each other, the further their relative sets ri in RI are from each other. You have not understood what RI is yet, so all of your above comments are null.

I have no intention of rewriting textbooks and ignoring others. I only ignore those who do not listen to me. By posting the above, you clearly indicated that you have not understood RI, function g, etc.

I suggest you run a few Ai that are finite. Change the order of the seqences. Afterwords, start building until you can ensure A = D. I can and do assert that no Ai is infinitely large until A = D. Once you get that, you can go back to giving me the basics on cardinality.

Again, sorry, but first you must listen before you can teach.

Last edited: Jun 27, 2013
20. Jun 27, 2013

### krausebj0

Sorry, ri is in RI. There is only one set RI.

I wish to define RI better for you. I consider it your quoted line of the proof (your one per post please).

0.111... <---> { 0.1, 0.11, 0.111, ... }, where k(0.111...) = { 0.1, 0.11, 0.111, ... }
0.10101... <---> { 0.1, 0.101, 0.10101, ... }, where k(0.10101...) = { 0.1, 0.101, 0.10101, ... }
0.1101... <---> { 0.1, 0.11, 0.1101, ... }, where k(0.1101...) = { 0.1, 0.11, 0.1101, ... }
...
i in I <---> ri in RI

The cardinality of RI is equal to the cardinality of I. All elements of RI have a cardinality equal to aleph null. All elements of RI will have one and only one element from each set Fn. There is one and only one element of I that naturally corresponds to each element of RI. This natural correspondence, I truly feel, is best illustrated by example. It can be rigorously reproduced in ZF set theory using infinitely large elements of P(N) in place of I and the related sets ri in RI would follow naturally. The set RI is a subset of the infinitely large elements of the powerset of dyadic rationals.

I hope we are clear on what RI is.

Last edited: Jun 27, 2013