12th Archivist
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This concept just occurred to me, and I wanted to share it. It has probably already been thought of before, and might even overlap completely with aleph-n. Still, known or unknown, I might as well mention it and my thoughts on it.

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How many distinct integers are there on the real number line bounded by zero and infinity? Well, if we consider a "distinct integer" to be one or two or sixty or somesuch, there will be an infinite number of these distinct integers. How about I call this "infinity with a degree of one", or "degree one" for short.

How many infinitesimal (but still distinct) values are there bounded by zero and one? 0.1, 0.01, and 0.001 are all infinitesimal, distinct values between zero and one, and since I can keep adding zeros to the front of that 1, not to mention any other numbers that could appear (and still be distinct), there are an infinite amount as well. Call this "degree zero".

Degree zero is contained within a single, essentially infinitesimal range within degree one. To elaborate, all of degree zero is bounded by zero and one, which is but one of the infinite number of values in degree one. What does this mean?

This can be taken a step further. If degree zero is an infinitesimal within degree one, there must be an infinitesimal bounded by zero and the very first value of degree zero, right? How many of those are in there? Again, an infinite amount, so I shall call this "degree negative-one".

I can even take it a step in the opposite direction, too. Zero to infinity in degree one could be a single value of a larger set, this particular set having an infinite number of these single values. I can call this "degree two", since it contains an infinite number of degree ones, just like degree one contains an infinite number of degree zeros, and so forth.

My logic is probably clear now, so where will I go with it? Why, to the logical extreme, of course! What are the intricacies of degree infinity, negative or positive? Are such mighty numbers even capable of being understood by the human mind? I have a possible answer, so I think so, at least from an outside perspective.

So, in that case, what is the lowest possible number? That is, the very first number of degree negative-infinity. Since I do not want to count down an infinite number of degrees, I will simply use induction. The first number of degree one is zero. The first number of degree zero is zero. This is pretty certain, so it should hold true that zero is the first number of degree negative-infinity. The same concept applies to degree positive-infinity, as well. Since the final number of degree zero is one, the final number of degree infinity is probably also one.

This does not state that 1 = infinity, where 1 is the second value of degree one. To more-or-less prove it, say 1 = infinity in degree zero (they all work, so this is just an example) when measured in degree negative-one. When measured from degree zero, 1 = 1, and when measured from a higher degree, 1 = infinitesimally small value. A person asserting that 1 = infinity will first measure 1 as infinity in degree zero, then proceed to perform math in degree one as if the assertions made in degree zero still applied. For example, such a person might state that "if 1 = infinity in degree zero, 2 = 2 * infinity. However, since 2 * infinity is still infinity, 1 = 2." This is wrong, since "1 = 2" is being measured in degree one (where the assumptions no longer hold), where 1 != infinity, and in degree zero, infinity = infinity, so it all works.

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There. I am finished, many hours later. To be honest, I feel kind of childish to have shared this, since this has probably been covered for most of you reading it, or maybe I made a big error somewhere, or was really, really unclear. So, if you have criticisms, I would prefer it if you kept them as non-personal as possible. (The same should also apply to praise.) EDIT: Oh, and no facepalms, please. That kind of stuff is really, really annoying to me.

Discuss this or point out flaws in my reasoning if you wish. I am tired of typing right now, so I just want to get this post out.

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[Last edited by 12th Archivist at 06-24-2011 06:09 AM]

Rabscuttle
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You are not using the word "infinitesimal" correctly. http://en.wikipedia.org/wiki/Infinitesimal

eg 0.1, 0.01, 0.001 are not infinitesimal.

But you can think of the set of numbers between 0 and 1 (not inclusive). However, it does not make sense to take the first (smallest) element of this set - there is no such number, because you can always pick a smaller number.
So extending this to a smaller set (what you're calling degree -1) is not possible.

I'm not sure I quite follow what you mean going in the other direction, but you might want to looking into infinities of different cardinalities.

For example, the set of integers is countably infinite
But the set of real numbers between zero and 1 is uncountably infinite.
(there are "more" of them - this basically means that there is no way to define a map from (integers -> (0, 1)) such that every number in (0, 1) will be pointed at.))

for more info, maybe start here
http://en.wikipedia.org/wiki/Cardinality

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[Last edited by Rabscuttle at 06-24-2011 06:34 AM]

Banjooie
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Isn't this just the difference between countable and uncountable infinities?*facepalm*

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[Last edited by Banjooie at 06-24-2011 06:46 AM]

skell
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I think you are overcomplicating things. Or I can't understand you correctly, it has quite a stream-of-consciousness feel to it (but Lucky is still better at it). 1 cannot be infinite number, because, as the name implies, it has no finish, no end, never, nada. No two infinities are the same infinities, we had quite a lot of fun reducing expression with infinities on polytechnics to make them from: inf / inf ^2 to 0.5 (just a non-working example).

That being said, 1 cannot be an infinity in degree zero, because degree zero is a set of infinite numbers with finite boundaries. Taking the logic further, you can pick ANY two Real numbers and declare that they are infinite+ and infinite-. <0, 1> won't work, but <0,1) will do very well. You still won't get an infinite number, rather, it will be infinitely and infinitesimally close to 1.

I think I miss my math classes on polytechnics. Your fault!

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Someone Else
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Rabscuttle wrote:
For example, the set of integers is countably infinite
But the set of real numbers between zero and 1 is uncountably infinite.

That doesn't make sense to me. I can make a one to one correspondence between numbers between 0 and 1 to positive integers. Simply reverse the digits, adding or subtracting the zero point at the start. For example,

12357 <-> 0.75321
234900 <-> 0.009432

Therefore, because there is a one to one correspondence between integers and real numbers between 0 and 1, the latter set should also be countably infinite.

TripleM
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Someone Else wrote:

Rabscuttle wrote:
For example, the set of integers is countably infinite
But the set of real numbers between zero and 1 is uncountably infinite.

That doesn't make sense to me. I can make a one to one correspondence between numbers between 0 and 1 to positive integers. Simply reverse the digits, adding or subtracting the zero point at the start. For example,

12357 <-> 0.75321
234900 <-> 0.009432

One to one means the same has to apply in reverse. What positive integer do you associate with pi - 3?

da rogu3
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There are flaws in your argument.

Firstly, if degree 0 is between 0 and 1 (non-inclusive), then degree -1 wouldn't exist as it should be part of degree 1. (as mentioned above)

I don't really understand the point you're trying to make in that last paragraph. Infinity isn't a fixed real number. There are an uncountably infinite amount of real numbers.

As for lowest possible number - for whole numbers, it's zero; for integers and real numbers, it's -infinity.

west.logan
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Judging from the comments, perhaps I'm misunderstanding what 12th Archivist is saying but... I don't think he's equating 1 to infinity at all, just saying there are an infinite number of steps between 0 and 1, for example.

Yes, I've thought of things like this. It's an interesting mental exercise but remember that infinity is a concept we use to make calculations easier, it's not something you actually multiply or add (adding 1 to 1 is two, not two infinite numbers, though there is an infinite number of steps between them). As you suggest, you could go the other way and think of "infinity" as being something like an integer, then you would have a "second infinity" which would be like two, and so on.

I suggest you read about Knuth's up-arrow notation. It makes large numbers simpler and it's kind of mind-boggling to think about

http://en.wikipedia.org/wiki/Knuth%27s_up-arrow_notation


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12th Archivist
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west.logan is saying the right things here. When typing my post, I was more concerned with getting the idea out than I was about making it as clear as possible. Right now it sounds like a bad example of a teacher presenting to a class, so I will get to work reorganizing everything so it sounds better.

As for the second-to-last paragraph, zero to one are inclusive, but I really did not want to bring it up, since having them non-inclusive brings up a whole mess of problems.

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Tahnan
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Someone Else wrote:

Rabscuttle wrote:
For example, the set of integers is countably infinite
But the set of real numbers between zero and 1 is uncountably infinite.

That doesn't make sense to me. I can make a one to one correspondence between numbers between 0 and 1 to positive integers. Simply reverse the digits, adding or subtracting the zero point at the start. For example,

12357 <-> 0.75321
234900 <-> 0.009432

Therefore, because there is a one to one correspondence between integers and real numbers between 0 and 1, the latter set should also be countably infinite.

Cantor's proof that this is impossible is still one of the prettiest mathematical proofs I know. (Of course, as TripleM pointed out, your technique here can't work anyway; rather than pi-3, I think I'd go with "1/3", which is "0.33333333..." and thus there's no positive integer that it could correspond to.)

Mind you, the number of rational numbers between zero and one, i.e. those that have a fractional representation, is countably infinite.

Cantor's proof, by the way, is more or less the following: suppose you think there is a one-to-one correspondence. Then you could list them out in a numbered list, e.g.

 1 0.500000000000000000...
 2 0.123456789123456789...
 3 0.333333333333333333...
 4 0.100000000000000000...
 5 0.141592653589793238...

and so forth. In that case, I can name a number you haven't listed: the nth position after the decimal point in my number is one more than the nth position after the decimal point in the nth number on your list (turning 9 into 0).

 1 0.500000000000000000...
 2 0.123456789123456789...
 3 0.333333333333333333...
 4 0.100000000000000000...
 5 0.141592653589793238...

So my number will start "0.63410...". And it can't possibly be on your list already, because it's different from every number on your list in at least one of its digits. And if you add my number to your list, I can just do it again; if you try rearranging your list to number them in some other way, I can still use the diagonal trick. (In fact, there are lots of numbers you left off--I think, though I'm not sure, that instead of "your digit plus one", Cantor actually used "1 if your digit is 0, otherwise 0" or something like that. Basically, anything that's composed of an alteration of the digits along the red diagonal.)

Therefore, no matter how much you think you can correspond the numbers between 0 and 1 to the counting numbers, it's always possible to demonstrate that there are numbers you left out. And that's what makes them uncountable.

Seriously, I find this stunningly beautiful. I'd probably have a picture of it on my desk at work next to my wife's picture, if only I had a job....

The spitemaster
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What I find interesting is infinite sets of numbers. For example, prime numbers. If positive integers is countably infinite what does that mean for prime numbers which are a degree of magnitude less that those numbers?

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da rogu3
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The spitemaster wrote:
What I find interesting is infinite sets of numbers. For example, prime numbers. If positive integers is countably infinite what does that mean for prime numbers which are a degree of magnitude less that those numbers?

Also countable, by linking each number n with the nth prime number.
1 --> 2
2 --> 3
3 --> 5
4 --> 7
etc...

TripleM
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Tahnan wrote:
..rather than pi-3, I think I'd go with "1/3", which is "0.33333333..." and thus there's no positive integer that it could correspond to.

But pi is the official number of the Eighth!

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But of course there is! It's infinite, but that doesn't stop it from being a number.

Trickster
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Tahnan wrote:
Therefore, no matter how much you think you can correspond the numbers between 0 and 1 to the counting numbers, it's always possible to demonstrate that there are numbers you left out. And that's what makes them uncountable.

Seriously, I find this stunningly beautiful. I'd probably have a picture of it on my desk at work next to my wife's picture, if only I had a job....

Diagonalization is the most common argument used because it's intuitively obvious, and it can be generalized to other situations (like when we diagonalize on Turing machines to prove a particular language is unrecognizable; that is to say, to prove no Turing machine or computer program can solve a certain problem).

But I rather prefer Cantor's original argument, which is even more elegant (to some). And brief! The actual proof can be given formally in six or seven lines of text, but I'll expand it here to make it more accessible.

Assume you have an enumeration of R: sequence A = <a0, a1...>. We will use A to deterministically construct two disjoint infinite sequences from A: B and C. As we do this we will prove (no matter what order A appears in) that there must exist a real number that is not in A: a contradiction of our assumption.

Let b0 = a0, and let c0 = the first a in A such that b0 < c0. Alternate choosing the next b and c (b1, c1, b2, c2, ...). To choose each b or c, always take the first value in A (the ak with least k) where its value is larger than all the b's so far, and smaller than all the c's so far.

When you finish, you have a subset of A, in a total linear order, like so: <b0, b1, b2, ... ... c2, c1, c0>

Now consider the supremum of B: the smallest value greater than or equal to all b's. Since the reals are complete, every supremum of a set of reals is also a real number. Where is this number in the sequence A, we ask?

It cannot be in B, because B has no upper bound. (Let it be bk. Then b(k+1) is larger than bk, so bk cannot be supremum of B.)

It cannot be in C, because C has no lower bound. (Let it be ck. Then c(k+1) would be smaller than ck, yet still larger than all b's, so ck cannot be supremum of B.)

Finally, it cannot be in the elements of A which were not taken into B and C. (Let it be ak. Then after a *maximum* of k steps, ak, being between B and C, would have been the next element taken into either B or C, depending on whether the step was odd or even.)

By contradiction, A is not an enumeration of R, and no such enumeration can exist.

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Someone Else
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That is quite a bit more elegant than diagonalization, IMO. Clearly, I am wrong. I think that if I saw it formally, I'd be able to understand it easier...

That's what wikipedia is for. EDIT: Wikipedia doesn't have a formal version of it. I'll keep looking.

EDIT2:
However, both these proofs fail.
Firstly, assume the list:
0.0000...
0.1000...
0.1100...
With the underlined digits being replaced, and limit 1/9. Then the replacing them all with 1 yields the next number in the sequence.

Secondly, assume the list:
y(v)=((-1)^(v))/v -> 0
Then, the proof says that it's uncountable because 0 is never in the list. However, it's clearly countable because 0 <-> 1 and y(v) <-> (v+1).

So therefore, I must conclude that I'm still right. I don't see why you can't have an infinite number correspond to an irrational one.

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[Last edited by Someone Else at 06-25-2011 07:38 AM]

TripleM
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Someone Else wrote:
However, both these proofs fail.
Firstly, assume the list:
0.0000...
0.1000...
0.1100...
With the underlined digits being replaced, and limit 1/9. Then the replacing them all with 1 yields the next number in the sequence.

If you replace them all by 1, you get precisely the value 1/9. Since you're saying the real numbers are countable, 1/9 must correspond to a particular term in the sequence - say term n. But that number has n-1 ones followed by a 0, so can't equal 1/9.

Secondly, assume the list:
y(v)=((-1)^(v))/v -> 0
Then, the proof says that it's uncountable because 0 is never in the list. However, it's clearly countable because 0 <-> 1 and y(v) <-> (v+1).

The proof assumes you have defined a list that contains every real number, and then proceeds to show that a particular real number isn't in the list. It doesn't say that if you start with some other sequence that doesn't contain every real number, that sequence is uncountable.

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[Last edited by TripleM at 06-25-2011 08:20 AM]

skell
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Math and English don't mix well for foreigners, despite all my love for the math
Maybe it is because it is early in the morning but I can barely understand anything from what you all are writing now.

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Someone Else
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For the first one, the diagonalization of my list for n terms equals term n+1. Therefore, because term n+1 is in my list, the proof fails.

For the second, I showed that given a list containing a set of countable numbers, the proof claimed that it was uncountable. Therefore, the proof is unreliable and fails.

TripleM
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Someone Else wrote:
For the first one, the diagonalization of my list for n terms equals term n+1. Therefore, because term n+1 is in my list, the proof fails.

The proof doesn't use the number formed by the first n terms. It uses the number formed by all terms. That number is well-defined and not on the list.

For the second, I showed that given a list containing a set of countable numbers, the proof claimed that it was uncountable. Therefore, the proof is unreliable and fails.

No, the proof didn't claim it was uncountable. The proof says if you provide a list that is countable and contains all real numbers, it doesn't contain all real numbers. Doesn't say anything about your list at all.

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[Last edited by TripleM at 06-25-2011 10:09 AM]

Trickster
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Someone Else wrote:
Firstly, assume the list:
0.0000...
0.1000...
0.1100...
With the underlined digits being replaced, and limit 1/9. Then the replacing them all with 1 yields the next number in the sequence.

You're confused about what is being shown.

If you have two infinite sets, it is always possible to pair them so that you have items left over. You can even do this with a set and itself. Consider two copies of the natural numbers. Pair each number in the first copy with its square in the second; you have items left over. However: we can also pair each number in the first set with each number in the second without leaving any unpaired. This is what we mean when we say two infinite numbers are the same "size": if it is possible to produce such a pairing.

The two proofs above each show that no matter what order you try to pair the naturals with the reals, you can prove a contradiction from the result. This means there is no possible function that maps the naturals to the reals in a bijection (one-to-one and onto; meaning no numbers on either side left unpaired). To prove something like this you cannot choose the order of the real numbers--you have to prove that there exists no possible order which is enumerable.

As an example that may shed more light: it's easy to see that, under the standard ordering, the rational numbers don't pair up in a bijection with the naturals. This is obvious, because there is no "next" rational number...the rational numbers are "dense", in that between any two rationals lie an infinite number of rational numbers. But it's easy to pair them up; we just need to describe some way of doing this so that for every natural number I name, you can name the corresponding rational number, and vice versa. We only need to find one possible way of doing it to prove it.

An easy way to imagine this is to think of a 2d grid of numbers with numerator horizontally and denominator vertically:

... [-3] [-2] [-1] [ 0] [ 1] [ 2] [ 3] ...
[1] 15 14 1 0 5 6 27
[2] 16 13 2 3 4 7 26
[3] 17 12 11 10 9 8 25
[4] 18 19 20 21 22 23 24

So the progression goes 0 = 0/1, 1 = -1/1, 2 = -2/1, 3 = 0/2, 4 = 1/2, 5 = 1/1, 6 = 2/1, 7 = 2/2, 8 = 2/3, 9 = 1/3, 10 = 0/3, 11= -1/3, etc. (Now there's a sticky problem of the fact that some rational numbers are duplicated, but this actually makes the pairing harder, and it turns out if we can create an surjection in both directions, a bijection must also be possible. Or, we could modify the progression to skip duplicates, since those are calculable, if slowly.)

That's the basic idea, anyway. Your intuition is correct but what we mean by "quantity" is very different in the infinite realm. It's not about what always happens, it's about what can happen, and most infinite quantities we deal with can be placed in a bijection with the naturals. It just so happens that the reals can't, and in a sense, they are "larger" because no function can exist which maps them together. (This may seem silly but it actually provides us with a lot of useful intuitions about math, language, and computer science.)

I'll post a message tomorrow with some basic ideas that should make this more clear and intuitive.

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west.logan
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Man, what you jokers talking 'bout? Twenty-four is the highest numba.

http://www.youtube.com/watch?v=f3ek85X2uOE


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Someone Else
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TripleM wrote:

Someone Else wrote:
For the first one, the diagonalization of my list for n terms equals term n+1. Therefore, because term n+1 is in my list, the proof fails.

The proof doesn't use the number formed by the first n terms. It uses the number formed by all terms. That number is well-defined and not on the list.

No. That's incorrect. All I need to do is to show that it fails in one specific instance, because a general contradiction is needed to show there is no possible bijection between the real and natural sets.
Therefore, using my counter-example, consider changing n>6 digits. The diagonal number is:
R(n)=0.111...111=r(n)+1
Where r(n) is the nth number in my set. Because this situation does not change when approaching infinity, the diagonal number is always in my set.

For the second, I showed that given a list containing a set of countable numbers, the proof claimed that it was uncountable. Therefore, the proof is unreliable and fails.

No, the proof didn't claim it was uncountable. The proof says if you provide a list that is countable and contains all real numbers, it doesn't contain all real numbers. Doesn't say anything about your list at all.

Based on this proof alone, we cannot conclude that R is uncountable. We already know that it fails in an alternating converging sequence. Only from some other information do we know that my set is countable. How do we know that some other information won't prove the reals countable? My counter-proof does not show the countability of the reals, but it does show that this proof is not enough to go on.


Trickster, I understand what you're saying. That's why I know that you need to prove it for every conceivable possibility, whereas I only need to find one contradiction. I don't think I can prove the reals countable (unless you allow me to say that ...a3a2a1 <-> 0.a1a2a3...); I do think that you can't prove them uncountable.

In addition, TripleM, at your previous statement, you don't get precisely 1/9. Because to do that, you'd need to have n elements in the list, where n+1 is 1/9. As you said, having precisely 1/9 in the list is impossible. And the proof does use the first n terms, where n is the length of the list. Therefore, my proof still stands.

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[Last edited by Someone Else at 06-25-2011 07:07 PM]

da rogu3
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Someone Else wrote:
I do think that you can't prove them uncountable.

Why not? Isn't this be proved by contradiction using Cantor's Diagonal theorem?

Someone Else
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da rogu3 wrote:

Someone Else wrote:
I do think that you can't prove them uncountable.

Why not? Isn't this be proved by contradiction using Cantor's Diagonal theorem?

Well, it would be, if the diagonal theorem worked. Which, as I have shown, it does not.

da rogu3
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Oh, I haven't studied set theory in too much detail, I just assumed that Cantor's argument was reliable enough. I didn't quite understand your counter-argument much.

I suppose it's one of those things where it's easy to see the results, but difficult to prove them.

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[Last edited by da rogu3 at 06-25-2011 06:54 PM]

TFMurphy
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Someone Else wrote:
No. That's incorrect. All I need to do is to show that it fails in one specific instance, because a general contradiction is needed to show there is no possible bijection between the real and natural sets.
Therefore, using my counter-example, consider changing n>6 digits. The diagonal number is:
R(n)=0.111...111=r(n)+1
Where r(n) is the nth number in my set. Because this situation does not change when approaching infinity, the diagonal number is always in my set.

Let's look at this set of yours.

Your set is defined completely by r(n). By definition, this is countably infinite. So you're trying to use the diagonal argument to prove it uncountably infinite, and you're running into problems. Well, of course you are.

When you create a diagonal number, the diagonal number is, again by definition, not a part of the set. This means one of two things:
(1) The diagonal number cannot be part of the set due to the rules of creating the set. For example, choosing 0.222~ as the diagonal number is obviously not going to be a part of your r(n) set. So we've picked a number that has no place in the set, and thus have proven nothing.
(2) The diagonal number *should* be part of the set, but you left it out. This is trivial to see for the set of Reals because any decimal expansion we pick is going to be a Real number. Adding it to the set somewhere just forces us to try again with a new diagonal number, which is again not in the set, and leads to our contradiction.

You've picked 0.1~ as the diagonal number. Either this is supposed to be part of the set and you left it out, or it's not supposed to be part of the set and it proves nothing about the set's cardinality. In this case, it's the latter: 0.1~ is 1/9 (your set is an *infinite* set, not a finite one), and the limit is not part of the set.

(And in any case, the diagonal method was used for the bijection between Integers and Reals, not Integers and 'any particular subset of the Reals'. It can and has been generalised to show that bijections between a set 'X' and Power Set of 'X' (a set containing all possible subsets of 'X') are impossible, but using it outside of those conditions isn't going to prove or disprove much of anything.)

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[Last edited by TFMurphy at 06-25-2011 07:01 PM : No such term as 'countably finite': corrected terms.]

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@da roug3
Exactly. Except for the fact that I think that there are ways to count the reals. For example:
Take the real number n.a(1)a(2)a(3)...a(m)... (n is any number of digits, a(m) is one digit for each a(m)). It corresponds to the counting number P(n+1)^(...a(m)...a(3)a(2)a(1)), where P(n) is the nth prime.

Note that this may not be a good, or valid way of counting the reals. If that's the case, I don't know how to do it, but I still believe it possible.

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[Last edited by Someone Else at 06-25-2011 07:19 PM]

Someone Else
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The diagonal number in this case is part of the set. However, let's not debate the finer points. I understand now where I went wrong. Specifically,

the diagonal method was used for the bijection between Integers and Reals, not Integers and 'any particular subset of the Reals'

Your other points, however, are flawed with the version of the proof you used. If my numbers ended in ...9999~, then it could possibly be equal to the diagonalized number. A stronger version of the proof would be that the number is replaced by a digit x such that 1<=x<=8, to avoid that identity.

But I admit defeat in this debate. You have argued well and persuasively, my good sir. By the way, the largest reference I used was http://arxiv.org/ftp/math/papers/0306/0306200.pdf

TripleM
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Someone Else wrote:

The proof says if you provide a list that is countable and contains all real numbers, it doesn't contain all real numbers. Doesn't say anything about your list at all.

Based on this proof alone, we cannot conclude that R is uncountable.

If you have a set which contains all real numbers and is countable, then the set does not contain all real numbers. Therefore the real numbers are not countable. I'm not quite sure where you see the flaw in that.

The proof is showing whatever countable set you start with, it is missing a real number. It doesn't fail on your sequence; your sequence is missing the number 0. Therefore the set you started with cannot be all the real numbers.

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[Last edited by TripleM at 06-25-2011 09:37 PM]