Different sizes of infinity...?

[pzkpfw]

you
"But the method is applied to an infinite series from M: the series E1, E2, …, Ev, it's not applied to M."

Cantor
"Namely, let m and n be two different characters, and consider a set [Inbegriff] M of elements
E = (x1, x2, … , xv, …)
which depend on infinitely many coordinates x1, x2, … , xv, …, and where each of the coordinates is either m or w."

[He clearly defines the set M composed of sequences En, each composed of coordinates xn, each m or w.
The same exclusion of E0 from M would occur for subsets.

It is plainly obvious that if something isn't in M it cannot be in a subset of M.
But that isn't what Cantors' method is about.

** if you think it is, that's what you need to try to show. Focus! **

Two different orientations cannot coexist in the same list.]

And again with the "different orientations". The diagonal is just a way to [ find / see / look for / choose / select ] characters to make a new element with. That new element is a valid member of M, as M is defined as "all". The question is whether it's a member of the given list (subset).

you
"Cantor then asks if all of M is in E1, E2, …, Ev - by seeing if he can make an E0 (which would be in M as M is all) that clearly isn't in E1, E2, …, Ev."

[He defines the diagonal symbols as a qualified sequence D without knowing if it is a duplicate. He already knows his E0 will differ from D after exchanging symbols which doesn't allow both to coexist in the same list. His action produces the exclusion.

It doesn't matter if D is a duplicate (of what?). It must be a member of M. It might be a member of the subset.
The method isn't about whether D and E0 "coexist" in the same subset (or M).
The point is that E0 cannot be in the subset - because it's different to E1 at character 1, etc.

Given, a pixel display screen with only 8 basic colors,
a green diagonal line from (x=0, y=0) and
a red horizontal line from (x=0, y=10).
What color is the pixel at the intersection of (10, 10) without mixing?

Completely and totally irrelevant. Again, he's not putting two things in one place.
The diagonal isn't there to put a new thing at any coordinate.
The diagonal is just to [ find / see / look for / choose / select ] characters to make a new element with.

The equivalent, using your "pixel display screen with only 8 basic colours" would be more like:
Given "a red horizontal line from (x=0, y=10)."
We create D with red at X=10
So we get E0 with cyan at X=10
So we know E0 cannot be a row on that screen (because of that and its other pixels)

Nobody is putting two things in one place.

He interprets exclusion of E0 as a missing E0.]

Yes, E0 is "missing" from the subset. E0 cannot be any of the elements in the subset, because it's different to all of them by at least one character.
(E0 will be in M.)

you
"3: Finally: this means there are infinite sequences in your binary tree, that are not in the infinite sequence starting at M0. Agree?"

[I have revised the tree for more clarity. M contains all possible sequences.
Cantor can say E0=0101... is in M if he is looking at subset M0 and
not in M if he is looking at subset M1. That is not a contradiction.]

It almost sounds like you've got it here!

you
"Cantor would be happy that a sequence starting with 1 is not in the list of infinite sequences starting with 0."

[He was developing set theory in the form of transfinite sets, thus he would already know the definition of subsets.]

View attachment 7599

Yes!

If the diagonal on some subset of sequences from the tree was the D you highlight in red above, then the resulting E0 would be what you highlight in red.

But - as you show, your tree represents M. Your tree does not represent a subset of M.

We all know D and E0 (and any Ex) are in your tree (M).
The question is whether E0 would be in a subset of your tree (M).

Consider:
1. Pick any infinite subset of your tree, e.g. all the sequences starting 01..., 10..., and 11...
2. Pick any characters from 1. to be D. Don't care how you choose D; flip coins or whatever. All that matters is you take the 1st from one sequence, the 2nd from another, and so on.
3. Generate E by flipping the characters of D.
4. E is obviously in the full tree (M), but is it in the subset you made in step 1?

 

[pzkpfw]

pzkpfw;

The list began as a column of 1 dimensional sequences, randomly formed and randomly added. They were independent of each other.
When Cantor introduced the u,v coordinate grid, the list became a 2 dimensional geometric problem.

No it did not become a "geometric problem". It's all just a way to be very precise when talking about elements.
Each element is a member of the list, numbered, 1, 2, 3, ...
Each element is a string of characters, numbered, 1, 2, 3, ...
Two sets of index numbers. So they get called coordinates.

Now he can define a 2 dimensional sequence

No. D is diagonal, and you've made the claim before that that makes it 45 degrees or 2D etc. But that's irrelevant.
D is found by the diagonal, but it simply is a string of characters (m/w or 0/1, whatever you feel like today).
e.g. D = { m, w, w, w, m, m, w, m, w, ... }
... looks pretty "1 dimensional" to me.
D is no different in that respect than the members of the subset. After all - D must be a member of M, so how could it be different?

that affects the location of another sequence which was not possible originally. He has manipulated the list to support his idea of the set M having a greater cardinality than the set N.

Yes, pretty nice result.

A counter example.
Each sequence of the set M can be represented by a symbol Sn. The new list would remain 1 dimensional, and begin

1. S1
2. S2
3. S3
⋮

Since N is infinite there is always another row.
The cardinality of M would equal that of N.

It's not about M, it's about an infinite subset of M.
So ignoring that you imply your list is M (which must include all elements), we can still describe a process:
1. Start a new proposed sequence Z.
2. Add to Z the inverse of the first character of S1
3. Add to Z the inverse of the second character of S2
4. Add to Z the inverse of the third character of S3
5. Continue ...

That would show that Z - which must be a member of M - cannot be a member of the subset.

 

[phyti]

pzkpfw;
“The parallel in this thread is your need to show Cantor was wrong,”

The binary tree representing M shows there is no missing sequence.

Set M is an abstract mathematical construct, represented by a tree style graph.
A list is a 2-dimensional surface containing symbols in some form of text.
The sequences in the tree graph are 1-dimensional and parallel to each other.
The sequences in the list are 1-dimensional and parallel to each other.
Cantor defines a u,v coordinate grid for the list.
He uses the diagonal elements u,u which we label for convenience as D, to form a sequence E0. D is the only sequence that intersects all other sequences in the list, thus having a symbol in common with each. Cantor knows in swapping 0’s and 1’s, E0 will differ with each one. He can declare E0 is not in the list.
The list is not the set M. Missing in the list does not translate to missing in M.
Do you comprehend the difference?

The N tree contains all possible integer sequences.
Just as M contains itself at every branch point, N does the same.
There are 10 subsets.
Put a decimal point at the start of each sequence and you have all the 'real'
numbers in the interval 0 to <1.
The example is the random sequence starting at 4520...

 

[pzkpfw]

pzkpfw;
“The parallel in this thread is your need to show Cantor was wrong,”

You don't really address this.

The binary tree representing M shows there is no missing sequence.

There is no missing sequence in M. Nobody said there was.
I've even mentioned multiple times that D and E0/E from any enumeration/list we're looking at is in M.

Set M is an abstract mathematical construct, represented by a tree style graph.
A list is a 2-dimensional surface containing symbols in some form of text.
The sequences in the tree graph are 1-dimensional and parallel to each other.
The sequences in the list are 1-dimensional and parallel to each other.
Cantor defines a u,v coordinate grid for the list.

Dimensionality has absolutely nothing to do with any of this.
Different representations of the sequences do not change what the sequences are.
Your peril sensitive sunglasses do not remove the peril.

He uses the diagonal elements u,u which we label for convenience as D, to form a sequence E0. D is the only sequence that intersects all other sequences in the list, thus having a symbol in common with each. Cantor knows in swapping 0’s and 1’s, E0 will differ with each one. He can declare E0 is not in the list.

Yes.

The list is not the set M.

Yes. The list turns out to be a subset of M. Even though it (the list) is infinite.

Missing in the list does not translate to missing in M.

Yes. This is correct!
I've written this several times too.

Do you comprehend the difference?

Of course, as I've mentioned this multiple times myself.
(Are you actually reading, or just skimming my posts?)

The thing is, that's not what Cantor was trying to show.
At least twice now I've suggested this might be the bit you need to show, if it's what you think. ("Focus".)

 

[phyti]

Why does Cantor conclude this?

"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."

He is referring to M and not the list.
A thing cannot be in a container and outside the container simultaneously.
Which is it?

 

[pzkpfw]

Why does Cantor conclude this?

"From this proposition it follows immediately that the totality of all elements of M cannot be put into the sequence [Reihenform]: E1, E2, …, Ev, …

Yes:
A: the totality of all elements of M
B: cannot be put into
C: the sequence E1, E2, …, Ev, …

"Put" is awkward language here (one reason I suggest using modern treatments and not getting hung up on translations of the original paper), but this is:
A: not all of M
B: is in
C: the list

This was shown by finding E, which must be in M by definition, that cannot be in the list.
The list, even though infinite, is a subset of M.

You wrote "Missing in the list does not translate to missing in M.". But nobody says this.

If M is "[]" and the list is "()":

[...E...(...x...)]

E is in M {by definition}, it is not in the list {shown by the method}.

Any x in the list, is of course in M. {D will be in M, it might be in the list}.

 

[phyti]

Cantor
"Namely, let m and n be two different characters, and consider a set [Inbegriff] M of elements
E = (x1, x2, … , xv, …)"

[As you said, "focus".
You added 'from' to his quote.]

Cantor
"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."

[He concludes (above) the set M > the set N, since there was no place in the list for E0. He assumes the list is complete (contains all of M) and matches the cardinality of N. Then uses the cda to fabricate the E0 sequence that can't be in the list. He concludes, if the list is incomplete then the set is also incomplete and must have a greater cardinality than N. That's his goal. He wouldn't waste his time with random subsets, knowing they all would fail, as noted in the next quote.

"4. E is obviously in the full tree (M), but is it in the subset you made in step 1?"

[No, because it can't be in the same list containing the diagonal elements used in its formation since it will differ at each diagonal element. This is a limitation of the list and has no effect on the set M. You get the same results for any size subset so it’s independent of the number of rows.
The list is a 2-dimensional (u, v) flat array of cells, thus the sequences can be represented by horizontal or diagonal lines. The sequences are 1-dimensional but their orientation relative to the origin is 2-dimensional. The rule for plane geometry is 2 parallel lines do not meet but if not parallel they will meet.]


[The circled elements are the diagonal elements used to form E0.
They represent D, an existing sequence, 3rd from the top.
By symmetry E0 is 3rd from the bottom.
The cda sequences are not sequences of the tree, but isolated elements from various subsets. They would appear contiguous in the list which is another of its limitations.
The sequence would move among various subsets.
The ordering of the tree requires a sequence remains in the subset where it originates.

The impossibility of D and E0 coexisting in the physical list has no effect on the abstract set M. Cantor can't prove any difference in cardinality of M and N using lists. He is mixing apples and oranges or practicing voodoo. He could burn a paper list and the set M remains.
The image of John is not John. The list for M is not the set M.
Focus!

 

[pzkpfw]

Cantor
"Namely, let m and n be two different characters, and consider a set [Inbegriff] M of elements
E = (x1, x2, … , xv, …)"

[As you said, "focus".
You added 'from' to his quote.]

Yes, focus. Above you quote part of the setup of M. But "from" is in (implied if you must) the setup of the list.

Using: https://jamesrmeyer.com/infinite/cantors-original-1891-proof

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.

1. "E1, E2, …, Ev" is the list. It is infinite.
2. "any infinite series of the set M" the list is made of elements of M. Note also the "any". If the list were defined as all of M, what would "any" mean? Different orderings? Why would that be relevant? The list is made of elements from M, it's not defined to be all of M.
3. "then there always exists ... " If the list were all of M, then the rest of this very proposition would be nonsense. It would self-contradict, as it proposes there's an element in M that's not in the list.

If you claim that 2. above is wrong, there's really no point discussing anything further! (Focus)

Please show how the list ("N" now?) is defined by Cantor to be all of M.

( See Note 3 in the wikipedia article: https://en.wikipedia.org/wiki/Cantor's_diagonal_argument#cite_note-11
(Yes, wikipedia has issues, but this does show the common understanding of the paper.) )

 

[phyti]

I have read the jamesmayer paper. His translation is the same.

Then Cantor is filling the list with subsets of M.
By definition/construction of each will not contain E0.
As Cantor states, "then there always exists ... "

"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."

Now he states the contradiction resulting from his cda procedure is somehow related to the cardinality of M. Yet the binary tree and you declare there is no missing element in M. Where is the cause and effect action?
Then it can correspond to N with no problem.
It's his transferring of the failing incomplete lists to the set M that is the problem. They are independent of each other.
Logically a thing can’t be in a container and not in a container simultaneously.
He never decides which is correct.
He can remove the contradiction by undoing the cda formation of E0. That's its origin.

You are still doing it. The set 'N now' is a tree of sequences for the base 10. It is NOT a list!
It is not defined by Cantor.

 

[pzkpfw]

I have read the jamesmayer paper. His translation is the same.

Mostly; he doesn't bother with the "Ding" stuff.

Then Cantor is filling the list with subsets of M.

Yes. Subsets. (As shown.)

By definition/construction of each will not contain E0.
As Cantor states, "then there always exists ... "

"then there always exists ... " was part of the proposition, later proven.
(And to clarify the above: E0 for one subset (or even one ordering of a subset) won't be the same E0 for another subset. E0 from one subset may be in another subset. (We've covered this before in this thread.))

"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 Dong] E0 would be both an element of M, but also not an element of M."

Yes, not all of M is in the list/subset. Shown by E0 not being in the list, but being in M.

Now he states the contradiction resulting from his cda procedure is somehow related to the cardinality of M.

Yes. The list/subset is infinite. Yet there's still more in M. Have you seen the title of this thread?

Though I'd quote:

Note that the term “cardinality” for infinite sets was not in current usage at the time Cantor wrote this paper; he uses the term “Mächtigkeit”, which can have corresponding English meanings such as ‘thickness’, ‘width’, ‘mightiness’, ‘potency’, etc. I have used the term “magnitude” as a suitable translation.

Yet the binary tree and you declare there is no missing element in M. Where is the cause and effect action?

Why is "declare there is no missing element in M" an issue or question? Who said there was any element missing from M?
M is all the possible elements (by definition).
The list is an infinite subset (by proof).
... the element E0 is missing from the list. It is not missing from M.

Same with your binary tree.
The list/subset of elements/sequences starting with 0 is infinite.
The list/subset of elements/sequences starting with 1 is also infinite
... and none of the second lot are in that first lot (and vice versa).

This all tells us that the concept of infinite has (warning, non math word coming ...) "depth".

Then it can correspond to N with no problem.
It's his transferring of the failing incomplete lists to the set M that is the problem. They are independent of each other.

What do you mean by "transferring" here? It sounds like you mean the missing-from-M thing.
But nobody says that.

Logically a thing can’t be in a container and not in a container simultaneously.

Nobody says that.
Cantor showed E0 isn't in the list, but by definition it is in M. At no point is something simultaneously in and not in a container.
M=[ ...E0... List={...} ]
E0 is in M, it is not simultaneously not in M.
E0 is not in the list, it is not simultaneously in the list.

It's actually that potential contradiction that proves the list is a subset of M (even though it's infinite).
If the list were all of M, then showing E0 isn't in the list would mean it isn't in M, but by definition M is all elements.
Yes, that would be a contradiction - but that's the proof that the list isn't all of M.

He never decides which is correct.
He can remove the contradiction by undoing the cda formation of E0. That's its origin.

The diagonal is just a way to select one coordinate (symbol, character) from each element in the list.
There's nothing wrong with it, it's simple and elegant.
You could instead use a random number generator, with a list kept of previous numbers so they don't repeat.
D = { 1st char of row 77, 2nd char of row 99999, 3rd char of row 1, 4th char of row 3645, ... }
(Of course - that's simply the same as randomly ordering the rows, then applying the nice simple diagonal.)
It doesn't really matter.

You are still doing it. The set 'N now' is a tree of sequences for the base 10. It is NOT a list!
It is not defined by Cantor.

So you reserve "N" for use in your binary tree? Sure.

In any case, don't fool yourself. The tree is just another representation of the elements/sequences of coordinates/symbols/characters. m/w or 0/1, whatever. Those elements are a list. How you represent them doesn't change that. (The only difference is you make it hard to use the diagonal method, but as above: there are other ways to achieve the same thing.)

If you don't like the word "list", just use "set" or "subset" as appropriate.

 

[James R]

Moderator note: Some off-topic posts about the Monty Hall problem have been split to a separate thread, here:

Monty Hall problem (again)

James R; I'm not ignoring you, just needed some extra time. I started about 2006 at Physforum. Forums seemed like a good idea, sharing information and learning new things. Over the years, I find people who can't comprehend what they read. It's no longer interesting having debates with people...

www.sciforums.com

 

[phyti]

pzkpfw;

Consider a list beginning

fig.1

Cantor's definition of a sequence:
"Namely, let m and n be two different characters, and consider a set [Inbegriff] M of elements
E = (x1, x2, … , xv, …)
which depend on infinitely many coordinates x1, x2, … , xv, …, and where each of the coordinates is either m or w."
They are a continuous succession of symbols 0 &1.

fig.2

The circled elements are the red diagonal elements in fig.1. Examining the binary tree,
they are not a succession of symbols but isolated elements from various subsets. They are discontinuous bits of existing sequences, similar to two straight line segments meeting at an angle, which are not considered one continuous line. D composed of the diagonal 'b' elements of Cantor's example does not qualify as a sequence. Its elements are in fact sequence u3 in the M0 subset. By symmetry its complement E0 is u14 in the M1 subset. The ordering of the tree requires a sequence remains in the subset where it originates.
Cantor excluded the wrong sequence.
Without the diagonal he has no argument.
Focus!

 

[pzkpfw]

pzkpfw;

Consider a list beginning
View attachment 7610
fig.1

Cantor's definition of a sequence:
"Namely, let m and n be two different characters, and consider a set [banana] M of elements
E = (x1, x2, … , xv, …)
which depend on infinitely many coordinates x1, x2, … , xv, …, and where each of the coordinates is either m or w."
They are a continuous succession of symbols 0 &1.
View attachment 7611
fig.2

The circled elements are the red diagonal elements in fig.1. Examining the binary tree,
they are not a succession of symbols but isolated elements from various subsets. They are discontinuous bits of existing sequences, similar to two straight line segments meeting at an angle, which are not considered one continuous line.

Yes and no (see below). This isn't an issue.

I've already suggested in several posts a "D" could be constructed by selecting coordinates from the elements in that list in other ways, e.g. randomly (with a lookup list to prevent repetition). Even when constructing D from Cantors list, I've suggested you could use random selection (e.g. right there two steps up in post #190). The idea that D must be a continuous sequence in the source is nonsense. The diagonal is just being used to select characters to make D then E with.

For example, instead of u=v, we could (I've also suggested this before) use other functions: v=F(u), like with modulus arithmetic (using M/W for width reasons here, and mod 3):

MMWWMWMW...
WMWMWWMM...
MMMWWMMW...
WMWWWMMM...
MMMWMWMW...
WWMWWMMW...
...
D=MMWWMM...
E=WWMMWW...

This example is less elegant than the diagonal as you need to see 3 rows at a time to construct D (unless you imagine leaving gaps that get back filled), but that doesn't change the outcome: an E which differs from every row in the list by at least one character. (Also note order of rows as previously posted.)

D was constructed by choosing the circled red digits in your fig 2., but the constructed D is just 0010... a sequence just like any other in your tree.

You continue to way over-think what the diagonal is. It's simply a nice easy way to select characters to construct a new sequence with.

D composed of the diagonal 'b' elements of Cantor's example does not qualify as a sequence.

Of course it does, you even marked "D" in your figure 2! That it was constructed by the circled red characters isn't relevant.

Its elements are in fact sequence u3 in the M0 subset.

Yes, as above. It's also shown in your fig 1. I showed an example of that too in post #48 (second list) when you seemed to think that was an issue.

By symmetry its complement E0 is u14 in the M1 subset.

Sure. Your tree represents M - all of the sequences.
So any D and any E from any subset of M will clearly be in the tree somewhere. No surprise.
But they won't be in all subsets.
* D in your tree isn't in the M1 subset.
* E in your tree isn't in the M0 subset.

This isn't an issue. Despite you writing this several times: Cantor didn't say E wasn't in M, he showed it wasn't in the list.

The ordering of the tree requires a sequence remains in the subset where it originates.

Why? Where did this rule come from?

Sequences are just sequences. That you are choosing to represent the list/set/herd of sequences in M a certain way doesn't change anything about those sequences or where they must "be".

Cantor excluded the wrong sequence.
Without the diagonal he has no argument.
Focus!

Yes, yes. I've seen your schoolboy debate team repetition technique multiple times. Next you'll start a post with "The dictionary defines 'diagonal' as ...".

You keep making claims that are refuted (or if you prefer, denied), then instead of trying to prove your claim you flip to something else. That's why I started writing "focus". One of the main ones was where you claimed Cantor was trying to show E wasn't in M. You've still not backed up that claim.

 

[phyti]

pzkpfw#193;

I am considering the actual method used by Cantor in his cda.
He considered the diagonal elements in his list as a qualified sequence for the purpose of forming E0. Without the geometric binary tree representing M, he wasn't aware of the diagonal 'b' sequence as an existing sequence with its complement E0. If he was, then he already knew the result.
He used a rule of formation (selecting the b's) which is not a random formation,
meaning not using a method or rule of formation.

"But they won't be in all subsets.
* D in your tree isn't in the M1 subset.
* E in your tree isn't in the M0 subset."
"Why? Where did this rule come from?"

That's part of set theory, the ability to sort, order, and manipulate subsets.
Have you studied any set theory?

Cantor
"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.."

He concludes, because he cannot include any E0 in his random lists, the cardinality of M is greater than that of N. The incompleteness of the list does not imply incompleteness of the set M. He attempts to include the b diagonal sequence with the horizontal E0, are impossible on a sheet of paper. The reason for the exclusion of E0 in his list has been shown to be non parallel sequences in a 2D space. In the binary tree, all sequences are parallel, and cannot meet.

Cantor raises the question of E0 being in M or not in M, with his self created contradiction.
He doesn't give an answer.
Why his conclusion in red?

#190
phyti said:
It's his transferring of the failing incomplete lists to the set M that is the problem. They are independent of each other.
you:
"What do you mean by "transferring" here? It sounds like you mean the missing-from-M thing.
But nobody says that"

You can't speak for everyone.
You don't say that, but others can and do.

 

[pzkpfw]

pzkpfw#193;

I am considering the actual method used by Cantor in his cda.

Yes, but you misinterpret things and make bad assumptions, so I try to illustrate.

He considered the diagonal elements in his list as a qualified sequence for the purpose of forming E0.

He constructed a new sequence using a diagonal to select elements from other sequences.
All of your "gasp, it's 2d but the table is 1d" stuff was and continues to be nonsense.

Without the geometric binary tree representing M, he wasn't aware of the diagonal 'b' sequence as an existing sequence with its complement E0.

Wrong!
He didn't need the tree. He already had M. He knew all sequences are in M - by definition.
All elements in the list are in M.
Any D constructed from any list is in M.
Any E made by complement of any D from any list is in M.
No need for the tree to show that.

If he was, then he already knew the result.
He used a rule of formation (selecting the b's) which is not a random formation,
meaning not using a method or rule of formation.


That's part of set theory, the ability to sort, order, and manipulate subsets.
Have you studied any set theory?

This is meaningless word salad, it's your usual thrashing around looking for straws to clutch on to to maintain your intuition.
The diagonal method is very clear and easy to understand. It's been mainstream accepted math for longer than you or I have been alive.
Accepted by people who know more about set theory than you and I combined.

Cantor
"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.."

He concludes, because he cannot include any E0 in his random lists, the cardinality of M is greater than that of N.

Yes and No:
E0 generated from a list cannot be in that list. E0 generated from one list might be in a different list. This was noted many posts ago.
Now, since the list is infinite, but there's an element of M not in it - that tells us something about cardinality comparing the list to M.

The incompleteness of the list does not imply incompleteness of the set M.

He never says M is incomplete. Where do you get this from?

He attempts to include the b diagonal sequence with the horizontal E0, are impossible on a sheet of paper. The reason for the exclusion of E0 in his list has been shown to be non parallel sequences in a 2D space. In the binary tree, all sequences are parallel, and cannot meet.

Using a different representation of M that makes it hard to use the diagonal doesn't change what the diagonal used on the list showed us.
Paper has nothing to do with it. For a start, any paper we have is finite!

Here are three sequences.
X = { m, w ,w, w, m, m, w, m, ... }
Y = { w, w ,m, w, m, m, m, m, ... }
Z = { w, m ,m, m, m, m, w, m, ... }
By definition they are all in M (and if I used 0/1 - they'd all be in your tree too). But can you tell:
* which might have been constructed by applying the diagonal to some list from M?
* which might have been constructed by complement of a diagonal from some list?
You cannot.
They are all just elements of M. How they were constructed is not an issue.

Your 2D space stuff continues to be nonsense. The diagonal is just a method to select characters to make sequence D from.

Cantor raises the question of E0 being in M or not in M, with his self created contradiction.
He doesn't give an answer.
Why his conclusion in red?

Ignoring that what you highlighted in red is badly written (as noted above) ...

You don't understand proof by contradiction. He's showing you this:

1. Elements are sequences of characters.
2. M is all of them.
3. An infinite list is made of elements of M.
4. As the list is infinite, your intuition/assumption might be that it contains all of M.
5. He shows you can construct an element that isn't in the list.
6. If the list were all of M, then an element not in the list would not be in M.
7. Line 6. contradicts line 2.
8. The answer is that the intuition/assumption in line 4. is wrong. The list is not all of M. Even though the list is infinite, M still has more elements. (See the orange bit above.)

#190
phyti said:
It's his transferring of the failing incomplete lists to the set M that is the problem. They are independent of each other.
you:
"What do you mean by "transferring" here? It sounds like you mean the missing-from-M thing.
But nobody says that"

You can't speak for everyone.
You don't say that, but others can and do.

You are the only person I've seen in multiple forums over multiple years, arguing that Cantor is wrong.
Also, Cantor simply is mainstream math - as written in text books and taught in Universities.

When I write "nobody says that" I am usually just trying to get you to really understand how out on a limb you are.

In this case though, it's again the bit where you think Cantor was showing something not in M. This is your claim, and it's quite clearly not the normal interpretation of this paper.

 

[phyti]

pkzpfw#195;

Now, since the list is infinite, but there's an element of M not in it - that tells us something about cardinality comparing the list to M

You agreed earlier his list contained subsets of M, which would always have a missing E0.
We know E0 formed using the cda will differ from every element in his list.
That should hold for the instance when the list is all of M.

Since the set M and his list have no last element, how can he compare their cardinality?

The tree in #192 showed both the diagonal elements he chose and the E0 formed from them were already in M. What did he prove?

Cantor’s vision of an infinite set.
“I say of a set that it can be thought of as finished (and call such a set, if it contains infinitely many elements, "transfinite" or "suprafinite") if it is possible without contradiction (as can be done with finite sets) to think of all its elements as existing together, and to to think of the set itself as a compounded thing for itself; or (in other words) if it is possible to imagine the set as actually existing with the totality of its elements”.
Source: Ewald, W., From Kant to Hilbert, Oxford 1996.

 

[pzkpfw]

pkzpfw#195;

You agreed earlier his list contained subsets of M, which would always have a missing E0.
We know E0 formed using the cda will differ from every element in his list.

Fine enough so far.

That should hold for the instance when the list is all of M.

No.
Because that would be a contradiction. E0 cannot both be in M and also not in M.
By showing E0 is not in the list, but by definition it is in M, it was shown the list - even though it is infinite - is not all of M.

M = [ ... E0 ... List = { ... } ]
It was shown E0 wasn't in this List. But it must be in M.
All of the List is in M. But not all of M is in the List - E0 is the proof of that.
The blue bits are where M is "more" than the List .. even though the List is infinite.

The whole point is showing that not all infinities are the same.
If your intuition tells you they are - it's your intuition that's wrong.

Since the set M and his list have no last element, how can he compare their cardinality?

By doing what he just did!
Yes they both have no last element - that doesn't mean they are the same.

Consider counting Natural numbers. Easy enough: 1, 2, 3, ...
There are infinite of them (with or without zero), but they are easily countable (given infinite time and space).

Now try counting the same (upwards from 1 for no unfair handicap) from the Real numbers ...
You won't even get to 2. Or even 1.5. Or 1.25, Or 1.125, ...
Even if you started just writing some down, which would be the 2nd? Which is the 3rd?

There is "no last element" of both Natural and Real numbers; they are both infinite. But it's not the "same" infinite.

The tree in #192 showed both the diagonal elements he chose and the E0 formed from them were already in M. What did he prove?

He proved M includes things that the List doesn't - even though the List is infinite.

Your own tree sort of shows this too. Are any of the following wrong?

A: the full tree is an infinite set of sequences of 0 and 1.
B: the set "M0" is an infinite set of the sequences - that start with 0.
C: the set "M1" is an infinite set of the sequences - that start with 1.
D: all of M0 and M1 are in the full tree.

So far: we have three things that are all infinite, and all in (or are) the tree.
But note:

E: none of M1 are in M0, and none of M0 are in M1.

So your tree and these subsets are all infinite - but they are not the same. There are sequences (infinite of them!) in your tree that are not in M0, and also sequences (infinite of them!) that are not in M1. Yet the infinite tree contains all of infinite M0 and infinite M1.

Cantor’s vision of an infinite set.
“I say of a set that it can be thought of as finished (and call such a set, if it contains infinitely many elements, "transfinite" or "suprafinite") if it is possible without contradiction (as can be done with finite sets) to think of all its elements as existing together, and to to think of the set itself as a compounded thing for itself; or (in other words) if it is possible to imagine the set as actually existing with the totality of its elements”.
Source: Ewald, W., From Kant to Hilbert, Oxford 1996.

This cherry picked quote out of context does not prove that Cantor thought all infinities are the same.

 

[phyti]

pkzpfw;

Because that would be a contradiction. E0 cannot both be in M and also not in M.

The sequence E0 begins with 0.
If you look in subset Mo, it's there.
If you look in subset M1, it's not there.

 

[pzkpfw]

pkzpfw;

The sequence E0 begins with 0.
If you look in subset Mo, it's there.
If you look in subset M1, it's not there.

And what do you think that shows?

E0 (from any list) is in M.
Your full tree (M0 + M1) represents M, so it's no surprise that E0 would be in there somewhere.
But it won't be in all subsets - even if they are infinite.

The point is that E0 won't be in the list it was generated off.
So - that list is not all of M (or all of the tree).

From your tree:
Any sequence starting with 0 is in M0.
That same sequence will not be in M1.
So ... the subset M1 - even though it is infinite - is not all of the tree.

 

[phyti]

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.

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.

fig.1 fig.2

Fig.1 shows B and its complement E0 are in the list.
Fig.2 shows E0 can't be in the list since it would conflict with any element of B,
or B can't be in the list since it would conflict with all sequences.
Cantor concludes E0 can't be in the list and not in the list, a contradiction.
By definition, fig.1 represents the set M which contains both B and E0, and all sequences and their complements.
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.

If I removed a sequence from the list, it would still be in the set M. Actions applied to the list do not alter the set M. They are two different and independent things.