Different sizes of infinity...?

[pzkpfw]

pkzpfw;
Using the 2 symbols 0 and its complement 1, define an infinite set M0 of infinite sequences beginning with 0, and enter them randomly in a list. leaving an empty row after each.
The complement of any sequence s in M0 is s with all its symbols replaced with their complement. Enter each empty row with the complement of the preceding sequence.
The list now consists of pairs of sequences in a random order.

OK for now.
But note that the complement of the members of M0 are what you've previously called M1.
Together M0+M1 represent M - all the sequences, whatever they start with.

Cantor applies a (u, v) coordinate system to the list. He considers the diagonal elements bn with coordinates (u, u) as a sequence B, and forms its complement E0.

Kind of weird that you now want to use B instead of the D you've used most of the time. But anyway ...

View attachment 7627
fig.1 fig.2

Fig.1 shows B and its complement E0 are in the list.

No.
B could be in the list, but it's not shown in fig 1.
Where u=v=3 the symbol is 0.
Where u=v=4 the symbol is 0.
So B would have xx00... and E0 would have xx11...
Clearly neither of the two rows shown are B or E0.

Fig.2 shows E0 can't be in the list since it would conflict with any element of B,

If Cantors' method was correctly used, we'd have an E0 that conflicts with all elements in the list.
At this point it's not about B, it's about the elements in the list, the row at u=1, u=2, etc.
Yes, E0 would have a different symbol at v=4 than the row u=4 would have at v=4.
(But still, your fig 1 and fig 2 are a bit garbled.)

or B can't be in the list since it would conflict with all sequences.

No.
There's no reason why B cannot be in the list. I've shown examples of that before - so have you.

Cantor concludes E0 can't be in the list and not in the list, a contradiction.

No.
The contradiction isn't the conclusion, it is part of getting to the conclusion.
He shows E0 cannot be in the list.
What would be a contradiction is if E0 were not in M - since M is all the sequences.
The conclusion is that the list, even though it is infinite, is not all of M.

1. M is all the sequences.
2. The List is infinite members of M.
3. E0 is not in the List.
4. E0 is in M.
5. Conclusion: the List, even though it is infinite, is not all of M.

M = [ ... E0 ... List = { ... } ] : the List is infinite, but M still has more.

By definition, fig.1 represents the set M which contains both B and E0, and all sequences and their complements.

Nope.
Cannot be, as shown by correctly generating an E0 that's different to all rows in the list.
Show me any row from fig 1 (telling me its u), and I can show you what symbol a properly constructed E0 would have, at what location, that is different to that row.

The orientation of a sequence is a new property of a sequence introduced by Cantor.
It places a restriction on the position of a sequence in a list, which didn't exist originally.
Cantor is mixing different classes of sequences, with and without orientation.

Orientation has nothing to do with it.
Once again, here are three sequences:
X = 0, 1, 1, 1, 0, 0, 1, 1, 0, ...
Y = 1, 1, 0, 0, 0, 1, 1, 1, 0, ...
Z = 0, 1, 0, 1, 1, 1, 0, 0, 0, ...
Clearly they are all from M0+M1.
But can you tell me which might be B (or D) from some list?
Can you tell me which might be E0 made from some list?
Can you tell me which were generated by flipping coins?
Can you tell me which were taken from the binary representation of some random integer?
You cannot.
They are all just members of M, no different to each other regardless of how they were constructed.

If I removed a sequence from the list, it would still be in the set M.

Yes!
Again: yes!
... and wouldn't that show you: even though the list is still infinite, it's not all of M?

Actions applied to the list do not alter the set M. They are two different and independent things.

Sure.
For example, this is why every time you've said Cantor showed E0 wasn't in M I've told you he never wrote that.

 

[phyti]

"For example, this is why every time you've said Cantor showed E0 wasn't in M I've told you he never wrote that."

Cantor quote from James Meyers:
"from this proposition it follows immediately that the set of all elements of M cannot be put into a sequence such as: E1, E2, …, Ev, …
otherwise we would have a contradiction, that an element E0 would be both an element of M, but also not an element of M."

That sure looks like Cantor is questioning E0 in M or not in M.

 

[pzkpfw]

Cantor quote from James Meyers:
"from this proposition it follows immediately that the set of all elements of M cannot be put into a sequence such as: E1, E2, …, Ev, …
otherwise we would have a contradiction, that an element E0 would be both an element of M, but also not an element of M."

That sure looks like Cantor is questioning E0 in M or not in M.

That sure looks like I was right earlier when I said you don't understand proof by contradiction.

Cantor quote from James Meyers' translation:	More translation for Phyti:
Then there is a series b1, b2, …, bn, … that can be defined so that bv is also equal to m or w but is different from av.v. That is, if av.v = m, then bv = w.	Applying the scary diagonal technique. Nothing is being put into the list. The list is being looked at to select symbols to make a sequence with ...
Consider the element: E0 = (b1, b2, b3, …) of M	and that sequence ... (A) E0 is a member of M.
One sees straight away, that the equation: E0 = Eu cannot be satisfied by any positive integer u	No member of the list (E4, E99, E576, ...) can be the same as E0. So ... (B) E0 is not a member of the list.
, otherwise for that value of u and for all values of v, we would have: bv = au.v and so we would in particular have: bu = au.u which by the definition of bv is impossible.	(Because the symbols in E0 would be different than all the elements in the list in at least one place. e.g. where u=10 (looking at E10) and v=10 (10th symbol of E10), there would be a symbol that is different than E0 has as its 10th symbol.)
From this proposition it follows immediately that the set of all elements of M cannot be put into a sequence such as: E1, E2, …, Ev, …	So ... (C) The list cannot be all of M.
otherwise we would have a contradiction,	Because that would be a contradiction. Note the "otherwise". He's not saying there is a contradiction, he's saying there would be one if (C) were wrong.
that an element E0 would be both an element of M, but also not an element of M	An element cannot be both in and not in a set. If Eu was all of M, then (B) would be wrong, as E0 would be in there due to (A) (A) would be wrong, as E0 would not be in there due to (B) So Eu is not all of M, (C) is correct

Cantor never says E0 isn't in M, in fact it's the reverse: E0 is in M, and anything that implies it isn't - is wrong.

It's plainly obvious (especially when you don't ignore the line immediately before the bit that you quoted) that the conclusion is not the contradiction. The contradiction is the proof of the conclusion (at least so far in the paper):

The infinite set Eu is not all of M. (Or, M cannot be expressed in such a list.)

 

[phyti]

pkzpfw;

It is impossible for Cantor to put E0 in the list without conflicting with its complement the diagonal elements.
Actions applied to the list do not alter the set M. They are two different and independent things.
The list is incomplete, but that does not imply the set M is incomplete.

 

[pzkpfw]

pkzpfw;

It is impossible for Cantor to put E0 in the list without conflicting with its complement the diagonal elements.

I don't know why you bring it up as we all seem to agree on that. E0 is not in the list.

Actions applied to the list do not alter the set M. They are two different and independent things.

Yes. Have you read post #201?

The list is incomplete, but that does not imply the set M is incomplete.

Yes. Have you read post #201? And #203?

Again (and again and again and again), nobody says M is incomplete.

The times you have said Cantor is saying that ... you are wrong.

(Is English your first language?)

 

[Pinball1970]

This guy will not get Cantor if he cannot get MH

 

[QuarkHead]

You guys have been all round the houses without once mentioning, as far as I can see, the property of the diagonal that makes Cantor's argument solid, namely that, supposing an enumerated list L and that each member of L is an enumerated string of digits, then

the n-th element of the n-th member of L is the n-th element of D the diagonal.
Assume it is in the set in question

Place D on the list and call it the p-th member of L. So now we have that

The p-th element of the p-th member of L is the p-th element of the p-th member of L.
Wow

Call the complement of D as E, assuming E is in the set in question and therefore the q-th member of L. So now the rule above can be modified to

The q- th element of the q-th member of L is not the q-th element of the q-th member of L
That's your contradiction, so the assumption that E can be an enumerated element of the set the question is false

 

[pzkpfw]

You guys have been all round the houses without once mentioning, as far as I can see,

It's been explained multiple times, in detail.

the property of the diagonal that makes Cantor's argument solid, namely that, supposing an enumerated list L and that each member of L is an enumerated string of digits, then

the n-th element of the n-th member of L is the n-th element of D the diagonal.
Assume it is in the set in question

Place D on the list and call it the p-th member of L. So now we have that

It's totally irrelevant to place (or find) D in the list. Cantor doesn't do it.

Occasionally phyti seems to have had an issue that D might be in the list, but it's been explained it doesn't matter either way.

The p-th element of the p-th member of L is the p-th element of the p-th member of L.
Wow

Wow indeed.

Call the complement of D as E, assuming E is in the set in question and therefore the q-th member of L. So now the rule above can be modified to

The q- th element of the q-th member of L is not the q-th element of the q-th member of L
That's your contradiction, so the assumption that E can be an enumerated element of the set the question is false

That's a tortured way to explain that E isn't in the list.

In any case, it's not the current issue. phyti does seem to accept now that E isn't in the list. The problem is what to do with that knowledge.

 

[phyti]

OuarkHead#207;

You guys have been all round the houses without once mentioning, as far as I can see, the property of the diagonal that makes Cantor's argument solid, namely that, supposing an enumerated list L and that each member of L is an enumerated string of digits, then
the n-th element of the n-th member of L is the n-th element of D the diagonal.
Assume it is in the set in question
Place D on the list and call it the p-th member of L. So now we have that
The p-th element of the p-th member of L is the p-th element of the p-th member of L.
Wow

Cantor shows the start of his list of random sequences of 0's and 1's with a 2-dimensional (u,v) grid. He formed a sequence of 'b' elements each with the coordinates (u, u) but exchanging the 0's and 1's, which he labeled E0.
Since E0 will differ at the diagonal for every row u, E0 can't be in the list.
A sequence has 2 properties, infinite length and direction, similar to a vector with magnitude and direction.
In Cantor's physical list any pair of sequences can't coexist if they have different directions. In the tree representation of the set M, all sequences have the same direction, which avoids the list problem.
If only I could convince pkzpfw the list and the set M are 2 different and independent things.

 

[pzkpfw]

OuarkHead#207;
Cantor shows the start of his list of random sequences of 0's and 1's with a 2-dimensional (u,v) grid. He formed a sequence of 'b' elements each with the coordinates (u, u) but exchanging the 0's and 1's, which he labeled E0.
Since E0 will differ at the diagonal for every row u, E0 can't be in the list.

Yes, as noted, we all now agree that an E0 can be found that isn't in the list. (A)

A sequence has 2 properties, infinite length and direction, similar to a vector with magnitude and direction.
In Cantor's physical list any pair of sequences can't coexist if they have different directions. In the tree representation of the set M, all sequences have the same direction, which avoids the list problem.

No.
All your stuff about vectors and direction continue to be irrelevant. Cantor shows E0 is not in the list. See (A) above.
There is no "problem" that your tree avoids. (Is this "problem" your claim Cantor says E0 is not in M?? No - he doesn't.)

The point (at that stage of the paper) is that the list is an infinite subset of M, yet cannot contain all of M.

Your tree doesn't change anything about that. As I've shown more than once before.

	Cantor	phyti tree more or less equivalent
1	M is all the elements	The full tree (M0 + M1) is all the sequences
2	E1, E2, ..., Ev is an infinite list of elements from M	M0 is an infinite set of sequences from the tree (So is M1)
3	We can find an E0 that isn't in the List	Any member of M1 clearly isn't in M0 (And vice versa)
4	Since E0 is in M, but isn't in the list: M has more elements than the list - even though the list is infinite.	Members of M1 are not in M0, so the full tree has more sequences than M0 (or M1) - even though M0 [and M1] is infinite. (And vice versa)

Especially note 1. Your tree represents M - all of the elements/sequences. It does not represent the list.
(It contains all of the list, so does M.)

If only I could convince pkzpfw the list and the set M are 2 different and independent things.

This isn't about me. You are alone here (and in other forums over years) arguing against Cantor.

You are forgetting:

If E1, E2, …, Ev is any infinite series of elements of the set M, then there always exists an element E0 of M, which cannot be the same element as any element Ev.

(Bolded bits: E0 is in M. Cantor never says it isn't. What he shows is it isn't in the list - it cannot be Ev which would be a member of E1, E2, ...)

The list is made of members of M. We are not talking apples and oranges here.
The list is infinite. That sounds like a lot.
But the list isn't all of M - as we can find an E0 which isn't in the list.

This was all to show something about infinity.

 

[James R]

This isn't about me. You are alone here (and in other forums over years) arguing against Cantor.

It's the same story with the Monty Hall problem and phyti. Years arguing for an incorrect analysis, making the same error over and over again.

It appears that phyti is quite inflexible in his thinking.

 

[phyti]

pzkpfw;




The diagonal D is red. The cda forms E0. The v coordinates of E0 differ from the v coordinates of each diagonal element for all u. There is no row for E0.
Inspection reveals u6 is the same as D. Thus D was present before the application of the cda. The list also shows both u9 and its complement u10 in the list.


Referring to the binary tree, D is an element of Mo.
By symmetry, if D is present, so is its complement E0.

Cantor's conclusion:
"From this proposition it follows immediately that the totality of all elements of M cannot be put into the sequence [Reihenform]: E1, E2, …, Ev, … otherwise we would have the contradiction, that a thing [Ding] E0 would be both an element of M, but also not an element of M."

The cda creates a false conclusion of a missing sequence in the list, which does not alter the set M. The list of horizontal sequences cannot contain a diagonal sequence. He assumes the failure of the list implies a contradiction of E0 being in M and not in M.
The list and the set M are two different and independent things.

 

[pzkpfw]

pzkpfw;

View attachment 7652


The diagonal D is red. The cda forms E0. The v coordinates of E0 differ from the v coordinates of each diagonal element for all u. There is no row for E0.

Yes, we all know this. (But it's good you have a better diagram now than the faulty one you presented in post #200.)

Inspection reveals u6 is the same as D. Thus D was present before the application of the cda.

Yes.
Previously you've tried to claim this is some kind of problem, but it has nothing to do with Cantors' method - which is about showing E0 isn't in the list (which is an infinite set of elements from M).
D may or may not be in the list, it makes no difference.

The list also shows both u9 and its complement u10 in the list.

Yes, no reason some row and its complement cannot be in the list. It's only E0 that the method shows isn't in the list.

View attachment 7653
Referring to the binary tree, D is an element of Mo.
By symmetry, if D is present, so is its complement E0.

Your full tree represents M, not the list. So, as noted several times before, it's not at all a surprise that some D and E0 are in the tree.

All sequences, including any D and any E0 from any list, will be in your full tree. They are also in M. This has been explained to you many times now.

Cantor's conclusion:
"From this proposition it follows immediately that the totality of all elements of M cannot be put into the sequence [Banana]: E1, E2, …, Ev, … otherwise we would have the contradiction, that a thing [Ding dong] E0 would be both an element of M, but also not an element of M."

Yes. He showed E0 cannot be in the list. Since E0 is in M, the list, even though it is infinite, cannot represent all of M.

The cda creates a false conclusion of a missing sequence in the list,

No, it's not false.

As shown by the method, E0 cannot be in the list. You've agreed with this several times now (even here in post #202 where you wrote "The v coordinates of E0 differ from the v coordinates of each diagonal element for all u. There is no row for E0").

(You often seem to get confused about whether you are thinking about the list or M. Maybe that's the whole base of your inability to understand this?)

which does not alter the set M.

Correct. M is all the sequences. E0 not being in the list does not mean it isn't in M.

M = [ ..., E0, ..., List = [ E1, E2, ... ] ]

The list (any list), and all D's and all E0's (from any list) will be in M. But it's proven that not all of M can be in the list.

The list of horizontal sequences cannot contain a diagonal sequence.

Rubbish.
Horizontal and diagonal have nothing to do with it. Even here in post #202 you show the diagonal is also row u6 in your diagram. Nothing prevented that.

As I wrote in #201 which you ignored as usual:

Once again, here are three sequences:
X = 0, 1, 1, 1, 0, 0, 1, 1, 0, ...
Y = 1, 1, 0, 0, 0, 1, 1, 1, 0, ...
Z = 0, 1, 0, 1, 1, 1, 0, 0, 0, ...
Clearly they are all from M0+M1 (i.e. M).
But can you tell me which might be D from some list?
Can you tell me which might be E0 made from some list?
Can you tell me which were generated by flipping coins?
Can you tell me which were taken from the binary representation of some random integer?
You cannot.
They are all just members of your tree (i.e. M), no different to each other regardless of how they were constructed.

He assumes the failure of the list implies a contradiction of E0 being in M and not in M.

No, he doesn't.
It is very very simple logic.

	Cantor	Explained to phyti
1	From this proposition it follows immediately that the totality of all elements of M	All of M ...
2	cannot be put into the sequence [Banana]: E1, E2, …, Ev, …	... cannot be in the list. (i.e. M is not a countable infinity.)
3	otherwise	"otherwise" means: If 1 and 2 are wrong then 5 would be true. Since a contradiction cannot be true then 1 and 2 must be true.
4	we would have	"would". He's not saying the contradiction exists. He's saying it would exist if 1 and 2 were wrong.
5	the contradiction, that a thing [Ding dong] E0 would be both an element of M, but also not an element of M.	He's not saying E0 is both in and not in M. He says right here that would be a contradiction. Since this cannot be true, 1 and 2 are.

The list and the set M are two different and independent things.

You've said this before.
They are not totally independant, as the list is an infinite set of elements from M.

 

[phyti]

The set M is a logical abstraction or mental construct. It is infinite or without a boundary.
The list is a real world physical container (of symbols).
The list has boundaries, so is finite.
It is impossible to contain an infinite set in a finite container.

If 'diagonal' has nothing to do with it, why is it referred to as 'the diagonal argument'?

 

[pzkpfw]

The set M is a logical abstraction or mental construct. It is infinite or without a boundary.

Sure.

The list is a real world physical container (of symbols).
The list has boundaries, so is finite.

No.
Where did this desperate rot come from?
The list is infinite - it has infinite members of infinite length.
... and the interesting thing is that being infinite doesn't mean "all".

If the list was finite, much of Cantors' paper would be pointless. That might have been noticed by mathematicians in the hundred years since.

(The more or less equivalent in your tree is noted in post #210 (and also before). The set M0 from your tree is infinite. Yet it doesn't contain all of your tree. In fact none of M1 - which is also infinite - is in M0. And vice versa. Even your own tree shows that an infinite subset doesn't mean "all".)

If E1, E2, …, Ev is any infinite series of elements of the set M, then there always exists an element E0 of M, which cannot be the same element as any element Ev.

If you really thought the list was finite, why have you not said this before?

It is impossible to contain an infinite set in a finite container.

Luckily that's not what this is all about.

If 'diagonal' has nothing to do with it, why is it referred to as 'the diagonal argument'?

Don't be so dishonest.

You know that comment is from a reply to your weird assertion that D being made from a diagonal somehow makes it different to the elements in the list. It's not. But you ignore the points explaining that.

The "diagonal" is merely a way to select (complements of) symbols to put in a new element. It comes from the u=v "formula" - which as shown before, isn't the only formula that could be used.

For that matter, referring to Cantor's paper, the word "diagonal" doesn't even appear in the English translation, other than in the title and notes. As James Meyer writes (my emphasis): "Note: The above is the part of Cantor’s paper which is relevant to what we now call Diagonal proofs.". Heck, Cantor's paper doesn't even make a "D". He goes straight to making E0 after defining where the symbols would come from.

You, phyti, are wrong when you write stuff like: "The list of horizontal sequences cannot contain a diagonal sequence.". (You even showed this as wrong by having a list in post #212 where "D" was also u6.

 

[pzkpfw]

)

 

[phyti]

pzkpfw;
"You, phyti, are wrong when you write stuff like: "The list of horizontal sequences cannot contain a diagonal sequence.". (You even showed this as wrong by having a list in post #212 where "D" was also u6."

It was to show The diagonal elements he used to form E0 were already in the list, which implies E0 must also be in the list. They occur in pairs.
He has no method of comparing sequences for duplication, since they have no last element!
--------------------------------------------------------------------------------
All the previous analysis of the Cantor diagonal argument is for nothing.
Cantor wanted to promote the idea of transfinite numbers. He thought he could 'imagine' a complete and finished infinite set.
He did successful work in other areas of mathematics. Measurement was based on a world of objects with boundaries, thus could not be applied to infinite sets.

 

[pzkpfw]

pzkpfw;
"You, phyti, are wrong when you write stuff like: "The list of horizontal sequences cannot contain a diagonal sequence.". (You even showed this as wrong by having a list in post #212 where "D" was also u6."

It was to show The diagonal elements he used to form E0 were already in the list,

That's never been a surprise as noted many times before.

(And the reason I noted it there was to show again that "D" being constructed by looking at the "diagonal" doesn't make it or E0 different to the other elements of M in any "incompatible" way. Counter to all your 1D/2D, 45 degrees, horizontal/diagonal claims. I note you ignore that and just fling another wild claim at the thread.)

which implies E0 must also be in the list.

No.

This is very odd as you've agreed many times that the method shows E0 is not in the list. To quote you (different thread where you are equally wrong): "where is the consistency?". You flip flop more than a politician.

Your list in post #212 shows u6 ("D") in the list.
Try showing E0 in that list!
"Implies" does not beat the clear logic in Cantors' paper.

They occur in pairs.

As M (and your full tree) are all elements, then the complement of every sequence will be in them.

But as shown by the method, E0 for some list cannot be in that list. Many many pairs may be, like the u9/u10 in #212 ... but not D/E0.

If you think E0 can be in the list in post #212, show it.

He has no method of comparing sequences for duplication, since they have no last element!

It is perfectly normal maths to show something works then continue to infinity. e.g. See "Induction".

E0 cannot be u1 because the 1st symbol will be different.
E0 cannot be u2 because the 2nd symbol will be different.
E0 cannot be u3 because the 3rd symbol will be different.
...
You cannot weasel out of the proof by simply saying "but when we get to u999999999999 it'll suddenly be the same symbol"!