The natural numbers are uncountable
Andrew
the specific identity of the natural number at each position in the
sequence, there is first element in the sequence, a second element in the
sequence, and so on, and so on, hence any simply infinite sequence of unique
natural numbers may be substituted for an infinite sequence of the natural
numbers in their natural order (i.e., 1, 2, 3, . . .). Which must mean that
any such sequence is countable.
On the other hand, if L is the set of all simply infinite sequences formed
from the two symbols '0' and '1'. Let M be the set of all elements of L
which are terminated in nothing but a string of zeros. Reading from left to
right, each unique element of M corresponds with a unique element of the set
of natural numbers. To see this, simply strip off the terminating string of
zeros and each element becomes a natural number in base two - e.g.;-
1000000000. . . = 1 = one
0100000000. . . = 01 = two
1100000000. . . = 11 = three
etc.,
Next, substitute this set M for the set M used by Cantor in his 1891 proof
that the power set of natural numbers are uncountable (which originally used
the set L whereas now we are using a countable subset of L). Cantor's
diagonal method will now create a member of this set M which is guaranteed
to not already be in the sequence.
In order to reveal this unlisted element the diagonal method will invert
symbol one in the first sequence, symbol two in the second sequence, symbol
three in the third sequence, etc., etc. Since each position in the sequence
corresponds to a natural number, and all natural numbers are finite, the
diagonal method can not invert an infinite number of symbols to create the
new element. Hence the new element must be zero terminated - which must
mean that it is a legitimate set of M which was not included in the original
sequence. Hence M is uncountable, but M is also a 1-1 analogue of the set
of natural numbers - hence the set of natural numbers is uncountable.
Therefore, Cantor's own methods and assumptions lead to the conclusion that
there exist sets which are simultaneously both countable and uncountable,
e.g., the set of natural numbers and the set M described earlier. That is
why the continuum hypothesis is unsolvable - because there can be no clean
resolution of these sets into Cantor's first and second orders of infinity -
i.e., Cantor's assumption of the existence of a 'simply infinite sequence'
leads to a clear contradiction and must, therefore, be false.
A more detailed presentation of this argument, entitled, "An Aristotlean
Rejection of the Continuum Hypothesis" can be found at
http://www.cantorsdonutparadox.co.uk
Andrew Stanworth
© 2004 Andrew Stanworth
David C. Ullrich
wrote:
>Given any simply infinite sequence of unique natural numbers, regardless of
>the specific identity of the natural number at each position in the
>sequence, there is first element in the sequence, a second element in the
>sequence, and so on, and so on, hence any simply infinite sequence of unique
>natural numbers may be substituted for an infinite sequence of the natural
>numbers in their natural order (i.e., 1, 2, 3, . . .). Which must mean that
>any such sequence is countable.
>
>On the other hand, if L is the set of all simply infinite sequences formed
>from the two symbols '0' and '1'. Let M be the set of all elements of L
>which are terminated in nothing but a string of zeros. Reading from left to
>right, each unique element of M corresponds with a unique element of the set
>of natural numbers. To see this, simply strip off the terminating string of
>zeros and each element becomes a natural number in base two - e.g.;-
>
>1000000000. . . = 1 = one
>0100000000. . . = 01 = two
>1100000000. . . = 11 = three
>etc.,
>
>Next, substitute this set M for the set M used by Cantor in his 1891 proof
>that the power set of natural numbers are uncountable (which originally used
>the set L whereas now we are using a countable subset of L). Cantor's
>diagonal method will now create a member of this set M which is guaranteed
>to not already be in the sequence.
No, the diagonalization leads to a sequence which is _not_ an element
of M.
>In order to reveal this unlisted element the diagonal method will invert
>symbol one in the first sequence, symbol two in the second sequence, symbol
>three in the third sequence, etc., etc. Since each position in the sequence
>corresponds to a natural number, and all natural numbers are finite, the
>diagonal method can not invert an infinite number of symbols to create the
>new element.
Huh???????????? How does that follow, exactly?
See if you can locate the flaw in the following:
Consider a list of zeroes:
0
0
0
0
.
.
.
Now generate a new list by replacing each 0 by a 1.
Since each element of the original sequence is 0
this cannot add any 1's; thus if we replace each
0 by a 1 we obtain the sequence
0
0
0
0
.
.
.
Does that seem right to you?
> Hence the new element must be zero terminated - which must
>mean that it is a legitimate set of M which was not included in the original
>sequence.
It has to be zero terminated because it's an element of M. It's
an element of M because it's zero terminated. Huh.
>Hence M is uncountable, but M is also a 1-1 analogue of the set
>of natural numbers - hence the set of natural numbers is uncountable.
>
>Therefore, Cantor's own methods and assumptions lead to the conclusion that
>there exist sets which are simultaneously both countable and uncountable,
>e.g., the set of natural numbers and the set M described earlier. That is
>why the continuum hypothesis is unsolvable - because there can be no clean
>resolution of these sets into Cantor's first and second orders of infinity -
>i.e., Cantor's assumption of the existence of a 'simply infinite sequence'
>leads to a clear contradiction and must, therefore, be false.
>
>A more detailed presentation of this argument, entitled, "An Aristotlean
>Rejection of the Continuum Hypothesis" can be found at
>http://www.cantorsdonutparadox.co.uk
>
>Andrew Stanworth
>© 2004 Andrew Stanworth
>
************************
David C. Ullrich
Arturo Magidin
Andrew <sta...@btinternet.com> wrote:
>Given any simply infinite sequence of unique natural numbers, regardless of
>the specific identity of the natural number at each position in the
>sequence, there is first element in the sequence, a second element in the
>sequence, and so on, and so on, hence any simply infinite sequence of unique
>natural numbers may be substituted for an infinite sequence of the natural
>numbers in their natural order (i.e., 1, 2, 3, . . .). Which must mean that
>any such sequence is countable.
>
>On the other hand, if L is the set of all simply infinite sequences formed
>from the two symbols '0' and '1'. Let M be the set of all elements of L
>which are terminated in nothing but a string of zeros. Reading from left to
>right, each unique element of M corresponds with a unique element of the set
>of natural numbers. To see this, simply strip off the terminating string of
>zeros and each element becomes a natural number in base two - e.g.;-
>
>1000000000. . . = 1 = one
>0100000000. . . = 01 = two
>1100000000. . . = 11 = three
>etc.,
>
>Next, substitute this set M for the set M used by Cantor in his 1891 proof
>that the power set of natural numbers are uncountable (which originally used
>the set L whereas now we are using a countable subset of L). Cantor's
>diagonal method will now create a member of this set M which is guaranteed
>to not already be in the sequence.
This is where your argument breaks down. When you started talking
about trailing zeroes I figured this is where you were going, but I
thought I would wait and see.
Note that a sequence of 0's and 1's is in your "list" M IF AND ONLY IF
almost all entries are zeros; that is, it contains at most a finite
number of 1's.
When we use the diagonalization argument on this list, we end up with
a sequence of 0's and 1's, call it S.
Now, certainly, S is not any of the numbers on the list (you have to
be careful about dual representation in binary, but let us set it
aside for a moment).
However, the claim that S SHOULD be on the list is unwarranted: S will
correspond to a natural number if and only if it contains only a
finite number of 1's. That is, you would have to show that in your
original listing, only finitely many of the diagonal entries are 0's.
In fact, this will never occur; the sequence you produce is certainly
not on the original list, but as opposed to the situation with real
numbers (where ANY sequence corresponds to a number), the sequence you
produce DOES NOT correspond to a number that belongs on that list
anyway.
--
======================================================================
"It's not denial. I'm just very selective about
what I accept as reality."
--- Calvin ("Calvin and Hobbes")
======================================================================
Arturo Magidin
mag...@math.berkeley.edu
Babylonian_Astrologer
> 1000000000. . . = 1 = one
> 0100000000. . . = 01 = two
> 1100000000. . . = 11 = three
> etc.,
>
> Next, substitute this set M for the set M
For clarity, let's say, "substitute this set M' for the set M"
> In order to reveal this unlisted element the diagonal method will invert
> symbol one in the first sequence, symbol two in the second sequence, symbol
> three in the third sequence, etc., etc. Since each position in the sequence
> corresponds to a natural number, and all natural numbers are finite, the
> diagonal method can not invert an infinite number of symbols to create the
> new element.
Define the diagonal element d as a binary string such that for all
finite n, the n_th bit of d is the inversion of the n_th bit of the
n_th element of M'.
Then, there is no member of M' which equals d.
> Hence the new element must be zero terminated
there is no finite decimal place of d which is followed by an endless
sequence of 0's because d is not a member of the set of all such
numbers, M'.
J.L. Perez-de-la-Cruz
Acid Pooh
> Rejection of the Continuum Hypothesis" can be found at
> http://www.cantorsdonutparadox.co.uk
>
> Andrew Stanworth
> © 2004 Andrew Stanworth
Oh no! He has a website and a copyright. He's not only cranky but
pretentious. :-(
'cid 'ooh
Bennett Standeven
there exist numbers n such that n > 2^n; a contradiction. Therefore
your reasoning is incorrect.
Andrew
"Bennett Standeven" <be...@pop.networkusa.net> wrote
>
> there exist numbers n such that n > 2^n; a contradiction. Therefore
> your reasoning is incorrect.
Please explain how this impacts my reasoning.
Andrew
Andrew
"David C. Ullrich" <ull...@math.okstate.edu> wrote
The argument works by assuming that the diagonal method is the sole
mechanism whereby novelty from any other element may be engendered. Since
all positions in the list correspond to natural numbers, and all natural
numbers are finite, there is no way to modify an infinity of symbols - the
properties of the natural numbers themselves forbid it. Hence, the infinity
of zeros (i.e., the similarity with every other element in the sequence) can
never be totally eradicated and so the novel sequence is guaranteed to be
zero terminated regardless of at what specific position.
>
> See if you can locate the flaw in the following:
> Consider a list of zeroes:
>
> 0
> 0
> 0
> 0
> .
> .
> .
>
> Now generate a new list by replacing each 0 by a 1.
> Since each element of the original sequence is 0
> this cannot add any 1's; thus if we replace each
> 0 by a 1 we obtain the sequence
>
> 0
> 0
> 0
> 0
> .
> .
> .
>
> Does that seem right to you?
>
It looks like the same sequence to me.
> > Hence the new element must be zero terminated - which must
> >mean that it is a legitimate set of M which was not included in the
original
> >sequence.
>
> It has to be zero terminated because it's an element of M. It's
> an element of M because it's zero terminated. Huh.
No, I do not say it has to be zero terminated because it's an element of M.
That is something you have made up for yourself. I say that because it can
never be anything other than zero terminated, based upon the properties of
the place positions which correspond to natural numbers, it is an element of
M.
Andrew
Andrew
"Arturo Magidin" <mag...@math.berkeley.edu> wrote
> This is where your argument breaks down. When you started talking
> about trailing zeroes I figured this is where you were going, but I
> thought I would wait and see.
>
> Note that a sequence of 0's and 1's is in your "list" M IF AND ONLY IF
> almost all entries are zeros; that is, it contains at most a finite
> number of 1's.
The list is not M. M is the set from which the simply infinite sequence is
drawn.
I agree that an element is in M only if it contains a finite number of 1's -
though I do not see the point of saying "at most a finite number", what is
less than finite?
As for you saying "IF AND ONLY IF almost all entries are zeros" that does
follow since every element must have an infinity of zero's but only a finite
number of 1's.
>
> When we use the diagonalization argument on this list, we end up with
> a sequence of 0's and 1's, call it S.
My core point is that we can not escape ending up with a sequnce of zeros
and ones terminated with an infinite sequence of zeros by using the diagonal
method.
>
> Now, certainly, S is not any of the numbers on the list (you have to
> be careful about dual representation in binary, but let us set it
> aside for a moment).
I agree, S is definitely not on the list.
Dual representation in binary? How?
>
> However, the claim that S SHOULD be on the list is unwarranted: S will
> correspond to a natural number if and only if it contains only a
> finite number of 1's. That is, you would have to show that in your
> original listing, only finitely many of the diagonal entries are 0's.
I agree that "S will correspond to a natural number if and only if it
conatains only a finite number of 1's".
However, it must be made clear that S is not a natural number - I only agree
that it corresponds to a unique natural number. In order for it be a
natural number it is necessary to strip off the infinite string of zeros
which remain unaltered from any other entry in the sequence. After which
the remaining zeros since they are interlaced between a finite number of 1's
(if indeed zeros exist in the number at all) must be finite. I find the
last line of your statement to be unintelligible.
>
> In fact, this will never occur; the sequence you produce is certainly
> not on the original list, but as opposed to the situation with real
> numbers (where ANY sequence corresponds to a number), the sequence you
> produce DOES NOT correspond to a number that belongs on that list
> anyway.
I agree that not every infinite series of 0's and 1's belongs in the
sequence. Only those which are zero terminated.
The argument works by assuming that the diagonal method is the sole
mechanism whereby novelty from any other element may be engendered. Since
all positions in the list correspond to natural numbers, and all natural
numbers are finite, there is no way to modify an infinity of symbols - the
properties of the natural numbers themselves forbid it. Hence, the infinity
of zeros (i.e., the similarity with every other element in the sequence) can
never be totally eradicated and so the novel sequence is guaranteed to be
zero terminated regardless of at what specific position.
Hence, the novel element produced is a legitimate member of M, and after the
infinity of trailing zeros are stripped off it is also a legitimate natural
number.
Andrew
Andrew
"Babylonian_Astrologer" <starof...@yahoo.com> wrote
> > 1000000000. . . = 1 = one
> > 0100000000. . . = 01 = two
> > 1100000000. . . = 11 = three
> > etc.,
> >
> > Next, substitute this set M for the set M
>
> For clarity, let's say, "substitute this set M' for the set M"
Ok.
>
> > In order to reveal this unlisted element the diagonal method will invert
> > symbol one in the first sequence, symbol two in the second sequence,
symbol
> > three in the third sequence, etc., etc. Since each position in the
sequence
> > corresponds to a natural number, and all natural numbers are finite, the
> > diagonal method can not invert an infinite number of symbols to create
the
> > new element.
>
> Define the diagonal element d as a binary string such that for all
> finite n, the n_th bit of d is the inversion of the n_th bit of the
> n_th element of M'.
Ok.
> Then, there is no member of M' which equals d.
Good argument. I like it!
This is similar to the paradox Cantor recognised when thinking about 'the
set of all sets' - Cantor's answer was simply to not think of this as being
a legitimate set definition. I suppose that is why Cantor does not set his
diagonal method loose on sets directly, he only ever lets them loose on "any
simply ininifte sequence" drawn from the set (in this case the set being M).
>
> > Hence the new element must be zero terminated
>
> there is no finite decimal place of d which is followed by an endless
> sequence of 0's because d is not a member of the set of all such
> numbers, M'.
I agree with you, principly because this is so confused that what you
describe can not be a member of M. For a start off, the members of M are
not numbers, they are infinite sequences of 0's and 1's. Also, decimal
places play no part in the set M, or any of its members, whatsoever.
Andrew.
David C. Ullrich
wrote:
Uh, right. This is nonsense, by the way.
>> See if you can locate the flaw in the following:
>> Consider a list of zeroes:
>>
>> 0
>> 0
>> 0
>> 0
>> .
>> .
>> .
>>
>> Now generate a new list by replacing each 0 by a 1.
>> Since each element of the original sequence is 0
>> this cannot add any 1's; thus if we replace each
>> 0 by a 1 we obtain the sequence
>>
>> 0
>> 0
>> 0
>> 0
>> .
>> .
>> .
>>
>> Does that seem right to you?
>>
>It looks like the same sequence to me.
The question is does it seem to you that that is the
sequence we get by starting with the original and changing
all the 0's to 1's.
>> > Hence the new element must be zero terminated - which must
>> >mean that it is a legitimate set of M which was not included in the
>original
>> >sequence.
>>
>> It has to be zero terminated because it's an element of M. It's
>> an element of M because it's zero terminated. Huh.
>
>No, I do not say it has to be zero terminated because it's an element of M.
>That is something you have made up for yourself. I say that because it can
>never be anything other than zero terminated, based upon the properties of
>the place positions which correspond to natural numbers, it is an element of
>M.
>
>Andrew
>
>
************************
David C. Ullrich
Will Twentyman
> Given any simply infinite sequence of unique natural numbers, regardless of
> the specific identity of the natural number at each position in the
> sequence, there is first element in the sequence, a second element in the
> sequence, and so on, and so on, hence any simply infinite sequence of unique
> natural numbers may be substituted for an infinite sequence of the natural
> numbers in their natural order (i.e., 1, 2, 3, . . .). Which must mean that
> any such sequence is countable.
>
> On the other hand, if L is the set of all simply infinite sequences formed
> from the two symbols '0' and '1'. Let M be the set of all elements of L
> which are terminated in nothing but a string of zeros. Reading from left to
> right, each unique element of M corresponds with a unique element of the set
> of natural numbers. To see this, simply strip off the terminating string of
> zeros and each element becomes a natural number in base two - e.g.;-
>
> 1000000000. . . = 1 = one
> 0100000000. . . = 01 = two
> 1100000000. . . = 11 = three
> etc.,
Notice that each of these will have only a finite number of 1s.
Moreover, the sequence has every representation with a finite number of 1s.
> Next, substitute this set M for the set M used by Cantor in his 1891 proof
> that the power set of natural numbers are uncountable (which originally used
> the set L whereas now we are using a countable subset of L). Cantor's
> diagonal method will now create a member of this set M which is guaranteed
> to not already be in the sequence.
Ah, but that "number" will have an infinite number of 1s in the
representation, so it is not actually a natural number.
> In order to reveal this unlisted element the diagonal method will invert
> symbol one in the first sequence, symbol two in the second sequence, symbol
> three in the third sequence, etc., etc. Since each position in the sequence
> corresponds to a natural number, and all natural numbers are finite, the
> diagonal method can not invert an infinite number of symbols to create the
> new element. Hence the new element must be zero terminated - which must
> mean that it is a legitimate set of M which was not included in the original
> sequence. Hence M is uncountable, but M is also a 1-1 analogue of the set
> of natural numbers - hence the set of natural numbers is uncountable.
But it is *not* zero terminated. In fact, it will be 001111111111....,
which is not a number.
You have a more fundamental problem. You missed the fact that your
change of representation is simply the following: f:N->N by f(n)=n,
which is a bijection. This proves, by definition, that N is countable.
--
Will Twentyman
email: wtwentyman at copper dot net
Peter Webb
> Hence, the novel element produced is a legitimate member of M, and after
the
> infinity of trailing zeros are stripped off it is also a legitimate
natural
> number.
>
> Andrew
>
So what is it? What is Natural number that your diagonal constrauction
produces?
Arturo Magidin
Andrew <sta...@btinternet.com> wrote:
>
>"Arturo Magidin" <mag...@math.berkeley.edu> wrote
>
>> This is where your argument breaks down. When you started talking
>> about trailing zeroes I figured this is where you were going, but I
>> thought I would wait and see.
>>
>> Note that a sequence of 0's and 1's is in your "list" M IF AND ONLY IF
>> almost all entries are zeros; that is, it contains at most a finite
>> number of 1's.
>
>The list is not M. M is the set from which the simply infinite sequence is
>drawn.
Sigh. The set M contains an element for each natural number. You have
simply used a different representation. Now you are simply engaging in
sophistry and bait-n-switch.
Again:
A sequence of 0's and 1's will represent a natural number under your
scheme if and only if it contains only a finite number of 1's.
Every element in M is of this form.
The sequence you construct via the diagonal process will NOT be of
this form. Thus, even though it will not be in your original list, it
is also NOT the representation of a natural number. Your conclusion
that the list is incomplete or that natural numbers are "uncountable"
does not follow.
>I agree that an element is in M only if it contains a finite number of 1's -
>though I do not see the point of saying "at most a finite number", what is
>less than finite?
Your grasp of the English language is rivaled only by your grasp of
logic, apparently. "At most a finite number" as opposed to a possibly
infinite number. E.g., the sequence
0,1,0,1,0,1,0,1,0,1,0,1,....
contains an infinite number of 1's, and does not represent (under your
scheme) a natural number.
>> When we use the diagonalization argument on this list, we end up with
>> a sequence of 0's and 1's, call it S.
>
>My core point is that we can not escape ending up with a sequnce of zeros
>and ones terminated with an infinite sequence of zeros by using the diagonal
>method.
Your assertion may be that, but it is completely
unjustified. Performing the diagonal method on your list will produce
a sequence which does NOT "terminate with an infinite sequence of
zeros". It will produce a sequence which has an infinite number of
1's.
>> However, the claim that S SHOULD be on the list is unwarranted: S will
>> correspond to a natural number if and only if it contains only a
>> finite number of 1's. That is, you would have to show that in your
>> original listing, only finitely many of the diagonal entries are 0's.
>
>I agree that "S will correspond to a natural number if and only if it
>conatains only a finite number of 1's".
>However, it must be made clear that S is not a natural number - I only agree
>that it corresponds to a unique natural number.
This is double talk nonsense. "S is not a natural number", yet you
agree that "it corresponds to a unique natural number"?
Your original list pretended to be a list of natural numbers. You
produce a new sequence. You claim this sequence somehow forces the
conclusion that natural numbers are uncountable (the subject of this
thread); which means that you are ASSERTING IMPLICITLY that the
sequence you obtain, which is not in your original list, "ought to be
there". That is, after all, the thrust of Cantor's Diagonal Argument:
that if you start with a list of real numbers, you can produce a REAL
number not on the list.
Your S does NOT correspond to any natural number. It will contain an
infinite number of 1s.
>The argument works by assuming that the diagonal method is the sole
>mechanism whereby novelty from any other element may be engendered.
Repeating the same nonsense over and over does not make it right. It
only makes you more deluded.
>Since
>all positions in the list correspond to natural numbers, and all natural
>numbers are finite, there is no way to modify an infinity of symbols - the
>properties of the natural numbers themselves forbid it.
This is doubletalk nonsense again.
Ross A. Finlayson
Will Twentyman wrote:
> But it is *not* zero terminated. In fact, it will be 001111111111....,
> which is not a number.
>
> You have a more fundamental problem. You missed the fact that your
> change of representation is simply the following: f:N->N by f(n)=n,
> which is a bijection. This proves, by definition, that N is countable.
>
> --
> Will Twentyman
> email: wtwentyman at copper dot net
How about if that number is negative three? That is to say, the values
"terminating" with repeating ones represent the negative of the binary complement
that ends with all zeros.
Then, zero is dually represented. There are a variety of places where the issue
of dual representation leads to sophisticated or not rejection of the Cantor
hypothesis.
It was interesting to note Magidin say that why yes, dual representation does
lead to the complications, where he did not explicitly say that it allows the
alternative hypothesis to get that one element that is otherwise unmapped in the
powerset, and he may or may not believe so. I speak for myself, he speaks for
himself.
In a model of ubiquitous ordinals, the powerset of a set is the order type is the
succcessor is a set.
In a model of ubiquitous ordinals, infinity is zero, the "empty set". In
another, it is not.
Regards,
Ross F.
Arturo Magidin
Ross A. Finlayson <r...@tiki-lounge.com> wrote:
>It was interesting to note Magidin say that why yes, dual representation does
>lead to the complications,
Actually, I was being dense and confused. I was thinking of binary
expansions of real numbers, which may have dual representation. But
the issue at hand was a representation of natural numbers in binary by
writing them in opposite order and with trailing zeros. In that
situation, there is no problem of dual representation.
Bennett Standeven
Your reasoning entails a contradiction, so it must be wrong.
Will Twentyman
>
> Will Twentyman wrote:
>
>
>>But it is *not* zero terminated. In fact, it will be 001111111111....,
>>which is not a number.
>>
>>You have a more fundamental problem. You missed the fact that your
>>change of representation is simply the following: f:N->N by f(n)=n,
>>which is a bijection. This proves, by definition, that N is countable.
>>
>>--
>>Will Twentyman
>>email: wtwentyman at copper dot net
>
>
> How about if that number is negative three? That is to say, the values
> "terminating" with repeating ones represent the negative of the binary complement
> that ends with all zeros.
He made a few claims. First, that the diagonalization would produce a
number that ends in an infinite number of 1's. Second, that his mapping
corresponded to the natural numbers. He would be correct if he said
there are an uncountably infinite number of infinite binary strings.
Those are a different scenario entirely, however.
Also, what would propose that 10101010101... represents?
Ross A. Finlayson
Will Twentyman wrote:
Hi,
I wasn't suggesting that the author was correct, except in spirit. If you get me
talking about it, I go off about pruning and augmenting sets and hierarchical tier
levels with the reflected pyramid, necessarily addressing non-cofinite sets in
mapping.
Hausdorff did say that an uncountable set was a countable union of countable sets.
It's convenient to have a model of the numerical system where there are functions
between each element of the naturals and each element of the unit interval of reals.
Do you think the impulse function is rectangular, triangular, sawtooth, or sine wave?
The natural numbers are by definition countable. It's convenient to have models where
half of the integers are even integers, a third multiples of three, etcetera, and log
|N|/|N|, where |N| is the amount of elements of the set N of naturals integers, prime.
Are the normal and abnormal numbers each uncountable?
What possible, even fantastic, results might arise from the use of a model where there
rationals and irrationals alternate in the reals? Is anything in the universe
discrete or continuous?
Please explain what is the impulse function, and why it evaluates to have area when
it has no width, and semi-infinite height.
Regards,
Ross F.
Chairman of the Ozzy Osbourne Appreciation Society
Ok, I'll bite.
What is semi-infinite?
> Regards,
>
> Ross F.
>
--
Roman to Arabic to reply by email
Poker Joker
> Given any simply infinite sequence of unique natural numbers, regardless
of
> the specific identity of the natural number at each position in the
> sequence, there is first element in the sequence, a second element in the
> sequence, and so on, and so on, hence any simply infinite sequence of
unique
> natural numbers may be substituted for an infinite sequence of the natural
> numbers in their natural order (i.e., 1, 2, 3, . . .). Which must mean
that
> any such sequence is countable.
Then why is the subject: "The natural numbers are uncountable?" I assume
that you want to argue that the naturals are not countable, so why is this
your first argument? People who are convincing usually come out with their
strongest argument first. Is this the best you can do? Argue the wrong
side
of your own argument?
Will Twentyman
Ok.
> Do you think the impulse function is rectangular, triangular, sawtooth, or sine wave?
I'm undecided between rectangular and a verticle line.
> Please explain what is the impulse function, and why it evaluates to have area when
> it has no width, and semi-infinite height.
Since it's the limit of rectangles with area 1, it has area 1. When
dealing with infinity, things get... odd. I've always been partial to
the integral of e^-x from 0 to infinity myself.
Will Twentyman
infinite in one direction. Think the non-negative half of the real
number line.
Babylonian_Astrologer
> "Babylonian_Astrologer" <starof...@yahoo.com> wrote
>
> > > 1000000000. . . = 1 = one
> > > 0100000000. . . = 01 = two
> > > 1100000000. . . = 11 = three
> > > etc.,
> > >
> > > Next, substitute this set M for the set M
> >
> > For clarity, let's say, "substitute this set M' for the set M"
>
> Ok.
>
> >
> > > In order to reveal this unlisted element the diagonal method will invert
> > > symbol one in the first sequence, symbol two in the second sequence,
> symbol
> > > three in the third sequence, etc., etc. Since each position in the
> sequence
> > > corresponds to a natural number, and all natural numbers are finite, the
> > > diagonal method can not invert an infinite number of symbols to create
> the
> > > new element.
> >
> > Define the diagonal element d as a binary string such that for all
> > finite n, the n_th bit of d is the inversion of the n_th bit of the
> > n_th element of M'.
>
> Ok.
>
> > Then, there is no member of M' which equals d.
>
> Good argument. I like it!
Then, do you admit that the "diagonal method" CAN invert an infinite
number of symbols?
>
> This is similar to the paradox Cantor recognised when thinking about 'the
> set of all sets' - Cantor's answer was simply to not think of this as being
> a legitimate set definition. I suppose that is why Cantor does not set his
> diagonal method loose on sets directly, he only ever lets them loose on "any
> simply ininifte sequence" drawn from the set (in this case the set being M).
>
> >
> > > Hence the new element must be zero terminated
> >
> > there is no finite decimal place of d which is followed by an endless
> > sequence of 0's because d is not a member of the set of all such
> > numbers, M'.
>
> I agree with you, principly because this is so confused that what you
> describe can not be a member of M. For a start off, the members of M are
> not numbers, they are infinite sequences of 0's and 1's. Also, decimal
> places play no part in the set M, or any of its members, whatsoever.
>
> Andrew.
There is no finite position in d which is followed by an endless
sequence of 0's because d is not a member of the set of all such
sequences, M'.
Poker Joker
"Babylonian_Astrologer" <starof...@yahoo.com> wrote in message
news:e8c4027f.04070...@posting.google.com...
What does the term "infinite number" mean? Why would anyone admit
that some process involves an "infinite number" of anything?
Babylonian_Astrologer
> What does the term "infinite number" mean?
It means, "unlimited quantity".
> Why would anyone admit that some process involves an "infinite number" of
> anything?
Because the definition of d inverts an unlimited quantity of symbols.
David Bandel
> >>
>>
WM
> wrote:
>
> >Given any simply infinite sequence of unique natural numbers, regardless of
> >the specific identity of the natural number at each position in the
> >sequence, there is first element in the sequence, a second element in the
> >sequence, and so on, and so on, hence any simply infinite sequence of unique
> >natural numbers may be substituted for an infinite sequence of the natural
> >numbers in their natural order (i.e., 1, 2, 3, . . .). Which must mean that
> >any such sequence is countable.
> >
> >from the two symbols '0' and '1'. Let M be the set of all elements of L
> >which are terminated in nothing but a string of zeros. Reading from left to
> >right, each unique element of M corresponds with a unique element of the set
> >of natural numbers. To see this, simply strip off the terminating string of
> >zeros and each element becomes a natural number in base two - e.g.;-
> >
> >0100000000. . . = 01 = two
> >1100000000. . . = 11 = three
> >etc.,
> >
> >that the power set of natural numbers are uncountable (which originally used
> >the set L whereas now we are using a countable subset of L). Cantor's
> >diagonal method will now create a member of this set M which is guaranteed
> >to not already be in the sequence.
>
> of M.
>
> >symbol one in the first sequence, symbol two in the second sequence, symbol
> >three in the third sequence, etc., etc. Since each position in the sequence
> >corresponds to a natural number, and all natural numbers are finite, the
> >diagonal method can not invert an infinite number of symbols to create the
> >new element.
>
Unfortunately set theorists are regularly unable to see such simple facts.
> See if you can locate the flaw in the following:
> Consider a list of zeroes:
>
> 0
> 0
> 0
> 0
> .
> .
> .
>
> Now generate a new list by replacing each 0 by a 1.
> Since each element of the original sequence is 0
> this cannot add any 1's; thus if we replace each
> 0 by a 1 we obtain the sequence
>
> 0
> 0
> 0
> 0
> .
> .
> .
>
> Does that seem right to you?
Regards, WM
WM
> In fact, this will never occur; the sequence you produce is certainly
> not on the original list,
Regards, WM
George Greene
> On the other hand, if L is the set of all simply infinite sequences formed
> from the two symbols '0' and '1'.
> Let M be the set of all elements of L
> which are terminated in nothing but a string of zeros.
It is in 1-1 correspondence with the class of all finite strings over the
same alphabet, which is (also, obviously) countable. You can just map
every string in M to the finite string that you get from REMOVING the infinite
tail of 0's.
> Reading from left to
> right, each unique element of M corresponds with a unique element of the set
> of natural numbers. To see this, simply strip off the terminating string of
> zeros and each element becomes a natural number in base two - e.g.;-
The class of all finite strings over any (given, fixed) finite alphabet is
countable. That's true regardless of the size of the alphabet as long as it's
finite; you don't have to use base 2 to see it.
> Next, substitute this set M for the set M used by Cantor in his 1891 proof
> that the power set of natural numbers are uncountable (which originally used
> the set L whereas now we are using a countable subset of L). Cantor's
> diagonal method will now create a member of this set M which is guaranteed
> to not already be in the sequence.
Every member of M ENDS WITH AN IFNINITE NUMBER OF 0s!
Cantor's method here will create strings WITH 1's, ARBITRARILY far out!
WM
> Cantor's method here will create strings WITH 1's, ARBITRARILY far out!
Regards, WM
Arturo Magidin
I have no interest in anything you have to say; I do not reply to your posts. Do not reply to mine; especially not one made over 10 years ago. Especially not by LYING about it via removals that are not marked, thus attempting to falsify the record and removing context. Just how deep is your dishonesty, anyway?
>
> Regards, WM
There are no regards. What there is your pathetic attempts at starting an argument. Stick them where the sun don't shine.
--
Arturo Magidin
WM
> On Saturday, October 7, 2017 at 6:24:41 AM UTC-5, WM wrote:
> > Am Dienstag, 6. Juli 2004 01:37:24 UTC+2 schrieb Arturo Magidin:
>
> > If there are aleph_0 digits, then the sequence has a tail of aleph_0 zeros because every natural number indexing a digit belongs to a finite initial segment
> and leaves aleph_0 digits zero.
>
>
> If a post was made 13.25 years ago, then how clueless is the guy replying to it, or how desperate is he to get into an argument with his betters?
> I have no interest in anything you have to say;
"He claims that q′ will always exist 'because there is always a rational between any two real numbers'. But this assertion is empty: the existence of such a rational does not imply the existence of a rational that has not yet been chosen. The author never actually establishes the existence of such a q′." [Arturo Magidin, loc cit]
This shows that you have not understood that all rationals that have been chosen already have been removed from the set and therefore cannot be chosen for a second time.
Why don't you correct your mistake and apologize for your wrong accusation? Here you can learn how such a text would read: https://www.hs-augsburg.de/~mueckenh/Transfinity/Transfinity/pdf, p. 271.
Regards, WM
Arturo Magidin
> Am Montag, 9. Oktober 2017 02:19:45 UTC+2 schrieb Arturo Magidin:
> > On Saturday, October 7, 2017 at 6:24:41 AM UTC-5, WM wrote:
> > > Am Dienstag, 6. Juli 2004 01:37:24 UTC+2 schrieb Arturo Magidin:
> >
> > > If there are aleph_0 digits, then the sequence has a tail of aleph_0 zeros because every natural number indexing a digit belongs to a finite initial segment
> > and leaves aleph_0 digits zero.
> >
> >
> > If a post was made 13.25 years ago, then how clueless is the guy replying to it, or how desperate is he to get into an argument with his betters?
>
> I have only corrected a mistake of yours.
> Or don't you believe in proofs?
> >
> > I have no interest in anything you have to say;
>
> Why then do you say so much nonsense about what you don't understand in my work?
Because you are an idiot and I have no interst in anything you have to say.
If you want to talk about correcting errors and apologizing, then you can start whenever you are ready, because you are an incorrigible liar, and an idiot to boot. That is all, Mucker. So, go ahead and whine some more so that everyone can see just how far you are willing to go to try to provoke your betters into arguing with you. Lies. Deception. Falsification of the record. And dredging posts that are over a luster and over a decade old in a pathetic attempt at pretending you are relevant.
You aren't. You are an idiot, and a liar. And I have zero interest in your lies. Or your statements. Or what you tiny non-functioning brain seems to think is a "proof".
--
Arturo Magidin
WM
> On Monday, October 9, 2017 at 4:11:25 AM UTC-5, WM wrote:
> > Am Montag, 9. Oktober 2017 02:19:45 UTC+2 schrieb Arturo Magidin:
> > > On Saturday, October 7, 2017 at 6:24:41 AM UTC-5, WM wrote:
> > > > Am Dienstag, 6. Juli 2004 01:37:24 UTC+2 schrieb Arturo Magidin:
> > >
> > > > If there are aleph_0 digits, then the sequence has a tail of aleph_0 zeros because every natural number indexing a digit belongs to a finite initial segment
> > > and leaves aleph_0 digits zero.
> > >
> > >
> > > If a post was made 13.25 years ago, then how clueless is the guy replying to it, or how desperate is he to get into an argument with his betters?
> >
> > I have only corrected a mistake of yours.
>
> You only re-stated your silliness, in a pathetic attempt at drawing my into an argument about your nonsense. It ain't happening, because you aren't worth the time.
>
> > Or don't you believe in proofs?
>
> I know for a fact that you wouldn't recongize a valid proof
> if it bit you in the ass, as they have often done. You notion of "proof" is as nonsensical
> as the rest of your stupidity. I have no interest in it, and you are doing nothing but lying your ass off.
> > >
> > > I have no interest in anything you have to say;
> >
> > Why then do you say so much nonsense about what you don't understand in my work?
>
> You mean, why did I make a comment six years ago when you were posting lies and pretending to be someone else?
> Because you are an idiot and I have no interst in anything you have to say.
>
> If you want to talk about correcting errors and apologizing, then you can start whenever you are ready, because you are an incorrigible liar, and an idiot to boot. That is all, Mucker. So, go ahead and whine some more so that everyone can see just how far you are willing to go to try to provoke your betters into arguing with you. Lies. Deception. Falsification of the record. And dredging posts that are over a luster and over a decade old in a pathetic attempt at pretending you are relevant.
>
> You aren't. You are an idiot, and a liar. And I have zero interest in your lies. Or your statements. Or what you tiny non-functioning brain seems to think is a "proof".
You seem to be a prime specimen of matheology with really striking arguments.
Thanks for exhibiting and underlining this feature.
Regards, WM