"Chris M. Thomasson" <chris.m.thomasson.1@gmail.com> writes:

For some damn reason when I hear end segments from WM I think of a
tree. Take the following infinite 2-ary tree that holds the positive integers:
___________________________________________
0
/ \
/ \
/ \
/ \
1 2
/ \ / \
/ \ / \
3 4 5 6
/ \ / \ / \ / \
.........................
___________________________________________

this goes on and on for infinity... We all can see how this can go for infinity, right WM? Wrt trees there are only leaves in a finite view of
it. However, the "infinite view" of the tree has no leafs because it never ends... Fair enough? Or too out there?

That's a can of worms in WMaths. WM has written 734,342,120 nonsense
posts about binary trees over the years. It's one of his favourite
examples to use to bamboozle his poor students.

The infinite binary tree -- simply a graph with node set N and edge set
(n, 2n+2) (in your numbering) -- is a particular puzzle for WM because
the node and edge sets are countable but the path set isn't.

Can you see a proof that the infinite rooted paths can be mapped, one to
one, with an uncountable subset of R?

... The infinite one has no leaves.

If you consider graphs in general, they do not have to be infinite to
have no leaves.

--
Ben.

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Ben Bacarisse <ben@bsb.me.uk> writes:

"Chris M. Thomasson" <chris.m.thomasson.1@gmail.com> writes:

For some damn reason when I hear end segments from WM I think of a
tree. Take the following infinite 2-ary tree that holds the positive
integers:
___________________________________________
0
/ \
/ \
/ \
/ \
1 2
/ \ / \
/ \ / \
3 4 5 6
/ \ / \ / \ / \
.........................
___________________________________________

this goes on and on for infinity... We all can see how this can go for
infinity, right WM? Wrt trees there are only leaves in a finite view of
it. However, the "infinite view" of the tree has no leafs because it never >> ends... Fair enough? Or too out there?

That's a can of worms in WMaths. WM has written 734,342,120 nonsense
posts about binary trees over the years. It's one of his favourite
examples to use to bamboozle his poor students.

The infinite binary tree -- simply a graph with node set N and edge set
(n, 2n+2) (in your numbering) -- is a particular puzzle for WM because

Correction, there are two such edges of course: (n, 2n+1) and (n, 2n+2).

the node and edge sets are countable but the path set isn't.

Can you see a proof that the infinite rooted paths can be mapped, one to
one, with an uncountable subset of R?

... The infinite one has no leaves.

If you consider graphs in general, they do not have to be infinite to
have no leaves.

--
Ben.

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Le 22/07/2024 à 01:10, Ben Bacarisse a écrit :

That's a can of worms in WMaths. WM has written 734,342,120 nonsense
posts about binary trees over the years.

It appears so to poors who cannot think straight.

Paths in the Binary Tree can be distinguished by nodes only. There must be
at least as much nodes as paths.

Regards, WM

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WM <wolfgang.mueckenheim@tha.de> wrote:

Le 22/07/2024 à 01:10, Ben Bacarisse a écrit :

That's a can of worms in WMaths. WM has written 734,342,120 nonsense
posts about binary trees over the years.

It appears so to poors who cannot think straight.

Paths in the Binary Tree can be distinguished by nodes only. There must be at least as much nodes as paths.

I explained this too you a while back. Infinity is complicated. What
is true in the infinite case is that there are as many FINITE paths as
nodes, namely a countable infinity of them. However there are an
uncountable infinity of INFINITE paths.

Regards, WM

--
Alan Mackenzie (Nuremberg, Germany).

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Le 22/07/2024 à 17:08, Alan Mackenzie a écrit :

WM <wolfgang.mueckenheim@tha.de> wrote:

Le 22/07/2024 à 01:10, Ben Bacarisse a écrit :

That's a can of worms in WMaths. WM has written 734,342,120 nonsense
posts about binary trees over the years.

It appears so to poors who cannot think straight.

Paths in the Binary Tree can be distinguished by nodes only. There must be >> at least as much nodes as paths.

I explained this too you a while back. Infinity is complicated.

No reason to go without logic.

What
is true in the infinite case is that there are as many FINITE paths as
nodes, namely a countable infinity of them. However there are an
uncountable infinity of INFINITE paths.

Only a matheologian fixed in his views can claim that after knowing my
game

Conquer the Binary Tree

Here is a variant of the construction by infinite paths, a game that only
can be lost if set theory is true: You start with one cent. For a cent you
can buy an infinite path of your choice in the Binary Tree. For every node covered by this path you will get a cent. For every cent you can buy
another path of your choice. For every node covered by this path (and not
yet covered by previously chosen paths) you will get a cent. For every
cent you can buy another path. And so on. Since there are only countably
many nodes yielding as many cents but uncountably many paths requiring as
many cents, the player will get bankrupt before all paths are conquered.
If no player gets bankrupt, the number of paths cannot surpass the number
of nodes. [Hippasos: "What can we learn from the new game CTBT that I
devised for my students?", MathOverflow (2 Jul 2010). W. Mückenheim:
"History of the infinite", HI12.PPT, current lecture]

But there are more nutcakes like you.

"You seem to be ignoring the fact that, after you have colored a countable family of pathes, say P0, P1, ..., Pn, ..., there may be other paths Q
that are not on this countable list but have, nevertheless, had all their
nodes and edges colored. Perhaps the first node and edge of Q were also in
P1, the second node and edge of Q were in P2, etc. [...] by choosing the sequence of Pn's intelligently, you can, in fact, ensure that this sort of thing happens for every path Q." [Andreas Blass, loc cit] My reply: If the second node is in P2 then also the first node is in P2, and so on for all
n   – for every path Q of the Binary Tree. No way to get rid of
already coloured paths by choosing "intelligently"! Here not even the antidiagonal is constructed.

Regards, WM

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WM <wolfgang.mueckenheim@tha.de> wrote:

Le 22/07/2024 à 17:08, Alan Mackenzie a écrit :

WM <wolfgang.mueckenheim@tha.de> wrote:

Le 22/07/2024 à 01:10, Ben Bacarisse a écrit :

That's a can of worms in WMaths. WM has written 734,342,120 nonsense
posts about binary trees over the years.

It appears so to poors who cannot think straight.

Paths in the Binary Tree can be distinguished by nodes only. There
must be at least as much nodes as paths.

I explained this too you a while back. Infinity is complicated.

No reason to go without logic.

What
is true in the infinite case is that there are as many FINITE paths as
nodes, namely a countable infinity of them. However there are an
uncountable infinity of INFINITE paths.

Only a matheologian fixed in his views can claim that after knowing my
game

We've known your game for years; it is to obfuscate, confuse, and lie.

Conquer the Binary Tree

Here is a variant of the construction by infinite paths, a game that only can be lost if set theory is true: You start with one cent. For a cent you can buy an infinite path of your choice in the Binary Tree. For every node covered by this path you will get a cent. For every cent you can buy
another path of your choice. For every node covered by this path (and not yet covered by previously chosen paths) you will get a cent. For every
cent you can buy another path. And so on. Since there are only countably many nodes yielding as many cents but uncountably many paths requiring as many cents, the player will get bankrupt before all paths are conquered.
If no player gets bankrupt, the number of paths cannot surpass the number
of nodes. [Hippasos: "What can we learn from the new game CTBT that I devised for my students?", MathOverflow (2 Jul 2010). W. Mückenheim: "History of the infinite", HI12.PPT, current lecture]

But there are more nutcakes like you.

That is really uncalled for.

"You seem to be ignoring the fact that, after you have colored a countable family of pathes, say P0, P1, ..., Pn, ..., there may be other paths Q
that are not on this countable list but have, nevertheless, had all their nodes and edges colored. Perhaps the first node and edge of Q were also in P1, the second node and edge of Q were in P2, etc. [...] by choosing the sequence of Pn's intelligently, you can, in fact, ensure that this sort of thing happens for every path Q." [Andreas Blass, loc cit]

It can happen for every FINITE path Q.

My reply: If the second node is in P2 then also the first node is in
P2, and so on for all n   – for every path Q of the Binary Tree. No way to get rid of already coloured paths by choosing "intelligently"!
Here not even the antidiagonal is constructed.

An infinite path in an infinite binary tree can be coded as an infinite sequence of Ls and Rs, corresponding to whether at the next node one goes
left or right. So, for example, the very first path might be
LLLLLLLL.....

But, supposing these infinite paths can be mapped to the integers, what
is the second path? And the third one? There is no systematic way of numbering these paths.

It is clear that the number of such paths is the same as the power set of
the natural numbers. There are more elements in any power set than in
the original set. So there are more infinite paths than can be indexed
by the natural numbers.

Regards, WM

--
Alan Mackenzie (Nuremberg, Germany).

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Le 22/07/2024 à 21:45, Alan Mackenzie a écrit :

WM <wolfgang.mueckenheim@tha.de> wrote:

Only a matheologian fixed in his views can claim that after knowing my
game

We've known your game for years;

You have not understood it. Otherwise if not agreeing you could show an
error. But you can only curse:

it is to obfuscate, confuse, and lie.

"You seem to be ignoring the fact that, after you have colored a countable >> family of pathes, say P0, P1, ..., Pn, ..., there may be other paths Q
that are not on this countable list but have, nevertheless, had all their
nodes and edges colored. Perhaps the first node and edge of Q were also in >> P1, the second node and edge of Q were in P2, etc. [...] by choosing the
sequence of Pn's intelligently, you can, in fact, ensure that this sort of >> thing happens for every path Q." [Andreas Blass, loc cit]

It can happen for every FINITE path Q.

Not for finite and not for infinite paths. If the second node is in P2,
then also the first node is in P2. That is the principle of the Binary
Tree.

An infinite path in an infinite binary tree can be coded as an infinite sequence of Ls and Rs, corresponding to whether at the next node one goes left or right. So, for example, the very first path might be
LLLLLLLL.....

It is impossible to use infinite sequences of Ls or Rs. What can be used
is a finite abbreviation like "LLLLLLLL.....". But there are only
countably many finite

But, supposing these infinite paths can be mapped to the integers, what
is the second path? And the third one? There is no systematic way of numbering these paths.

There is no way to enumerate the rationals either. See https://osf.io/preprints/osf/tyvnk, 4 pages English or 4 pages German, according to your preference.

It is clear that the number of such paths is the same as the power set of
the natural numbers.

Yes.

There are more elements in any power set than in
the original set.

Yes, but that has not the least to do with countability.

So there are more infinite paths than can be indexed
by the natural numbers.

There are more fractions than can be indexed. Nevertheless my game shows a contradiction. Can you understand that? The "explanation" of Andreas Blass
is absolute nonsense because of the principle of the Binary Tree. Can you understand that?

Regards, WM

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Am 22.07.2024 um 22:47 schrieb Chris M. Thomasson:

It is impossible to use infinite sequences of Ls or Rs. [...]

@WM: No, you fucking asshole full of shit.

We may DEFINE, say, the infinite sequence A = (a_n)_(n e IN) with a_n =
L for all n e IN.

Then it's possible (at least in math) to "use" this sequence (by
referring to it in mathematical arguments).

Your mind cannot fathom infinity of any kind.

Agree.

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WM <wolfgang.mueckenheim@tha.de> wrote:

Le 22/07/2024 à 21:45, Alan Mackenzie a écrit :

WM <wolfgang.mueckenheim@tha.de> wrote:

Only a matheologian fixed in his views can claim that after knowing my
game

We've known your game for years;

You have not understood it. Otherwise if not agreeing you could show an error. But you can only curse:

it is to obfuscate, confuse, and lie.

"You seem to be ignoring the fact that, after you have colored a
countable family of pathes, say P0, P1, ..., Pn, ..., there may be
other paths Q that are not on this countable list but have,
nevertheless, had all their nodes and edges colored. Perhaps the
first node and edge of Q were also in P1, the second node and edge of
Q were in P2, etc. [...] by choosing the sequence of Pn's
intelligently, you can, in fact, ensure that this sort of thing
happens for every path Q." [Andreas Blass, loc cit]

It can happen for every FINITE path Q.

Not for finite and not for infinite paths. If the second node is in P2,
then also the first node is in P2. That is the principle of the Binary
Tree.

An infinite path in an infinite binary tree can be coded as an infinite
sequence of Ls and Rs, corresponding to whether at the next node one goes
left or right. So, for example, the very first path might be
LLLLLLLL.....

It is impossible to use infinite sequences of Ls or Rs. What can be used
is a finite abbreviation like "LLLLLLLL.....". But there are only
countably many finite

I have just used infinite sequences of Ls and Rs. It's clear you do not
have a degree in mathematics.

But, supposing these infinite paths can be mapped to the integers, what
is the second path? And the third one? There is no systematic way of
numbering these paths.

There is no way to enumerate the rationals either. See https://osf.io/preprints/osf/tyvnk, 4 pages English or 4 pages German, according to your preference.

Strawman. There is no way to enumerate infinite paths in that binary
tree. As for the rationals, you're just being an idiot, as it is common knowledge how to enumerate the rationals.

It is clear that the number of such paths is the same as the power set of
the natural numbers.

Yes.

There are more elements in any power set than in
the original set.

Yes, but that has not the least to do with countability.

Wrong.

So there are more infinite paths than can be indexed by the natural
numbers.

There are more fractions than can be indexed.

Wrong.

Nevertheless my game shows a contradiction. Can you understand that?
The "explanation" of Andreas Blass is absolute nonsense because of the principle of the Binary Tree. Can you understand that?

I'm sure I could, if I could be bothered. Most of your contradiction
games come from not understanding the differences between the finite and
the infinite. This one is likely another following this pattern.

Regards, WM

--
Alan Mackenzie (Nuremberg, Germany).

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Le 23/07/2024 à 10:49, Alan Mackenzie a écrit :

WM <wolfgang.mueckenheim@tha.de> wrote:

Le 22/07/2024 à 21:45, Alan Mackenzie a écrit :

WM <wolfgang.mueckenheim@tha.de> wrote:

Only a matheologian fixed in his views can claim that after knowing my >>>> game

We've known your game for years;

You have not understood it. Otherwise if not agreeing you could show an
error. But you can only curse:

it is to obfuscate, confuse, and lie.

No comment.

An infinite path in an infinite binary tree can be coded as an infinite
sequence of Ls and Rs, corresponding to whether at the next node one goes >>> left or right. So, for example, the very first path might be
LLLLLLLL.....

It is impossible to use infinite sequences of Ls or Rs. What can be used
is a finite abbreviation like "LLLLLLLL.....". But there are only
countably many finite

I have just used infinite sequences of Ls and Rs. It's clear you do not
have a degree in mathematics.

It is clear that you don't know mathematics. "LLLLLLLL....." is a finite expression, a formula *producing* every desired term of an infinite
sequence. Infinite sequences cannot be written.

Nevertheless my game shows a contradiction. Can you understand that?
The "explanation" of Andreas Blass is absolute nonsense because of the
principle of the Binary Tree. Can you understand that?

I'm sure I could, if I could be bothered.

I doubt that. My students understand it within less than 5 minutes.

Regards, WM

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Am 23.07.2024 um 21:57 schrieb Chris M. Thomasson:

On 7/23/2024 9:49 AM, WM wrote:

"LLLLLLLL....." is a finite expression,

WM is right here.

"LLLLLLLL....." is an infinite expression, indeed....

Nope.

Hint: print(len("LLLLLLLL....."))

But by convention (we think -except WM that is- that) "LLLLLLLL....."
DENOTES (refers to) an infinite sequence of "L"s. :-P

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Am 23.07.2024 um 22:03 schrieb Chris M. Thomasson:

On 7/23/2024 12:57 PM, Chris M. Thomasson wrote:

"LLLLLLLL....." is an infinite expression, indeed....

No, it's not.

Hint: It consists of 8 "L"s and 5 "."s, in total of 13 characters. No?

(A programmer really should know this! :-)

So, it can be reduced to:

(L) = LLL... ? Fair enough?

More math like, defining it as an infinite sequence:

(c_n)_(n e IN) with c_n = "L" for all n e IN. :-P

We'd usually write (a term referring to) this sequence the following way:

("L", "L", "L", ...) .

:-P

Of course, most mathematicains would just write

(L, L, L, ...)

when it's clear that "L" is not some sort of variable or arbitrary
constant but should refer to the letter "L". :-P

Hint: WM does not even know how to write terms for (i.e. referring to)
infinite sequences properly.

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Am 23.07.2024 um 22:28 schrieb Chris M. Thomasson:

On 7/23/2024 1:17 PM, Moebius wrote:

Am 23.07.2024 um 21:57 schrieb Chris M. Thomasson:

On 7/23/2024 9:49 AM, WM wrote:

"LLLLLLLL....." is a finite expression,

WM is right here.

"LLLLLLLL....." is an infinite expression, indeed....

Nope.

Hint: print(len("LLLLLLLL....."))

But by convention (we think -except WM that is- that) "LLLLLLLL....."
DENOTES (refers to) an infinite sequence of "L"s. :-P

Well, I was thinking of an endless sequence of L's

Yes, *I JUST SAID THAT*!

not a literal string.

Sure, but THE EXPRESSION (above) is a STRING-LITERAL, isn't it?

A programmer really should know that! ;-P

So if WM claims that "LLLLLLLL....." is a finite expression, he's right.

A parser can interpret [...]

Whatever your parser may be able to do: What he's DEALING with (from
source code or input) is FINITE EXPRESSION consisting of 13 characters.

Please don't do the Mückenheim, here. Ok?

I'm quite sure some CAS will be able to interprete (another finite
expression) correctly:

(L, L, L, ...)

That's how we usually write (a term referring to) an infinite sequence
(where all terms are just "L"), in math.

So if you are thinking of a numb ahem "endless sequence of L's", just write

(L, L, L, ...) .

(when referring to it):-P

Of course, in certain contexts you may abbreviate such expression (for simplicity, and just write

LLL...

instead.

But (then) still LLL... is an infinite sequence, and "LLL..." is a
finite expression (consisting of 6 characters) which DENOTES this
sequence. :-P

Hint: Paris is a town, but "Paris" is a name consisting of 5 characters (denoting the town Paris).

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On 7/23/2024 4:17 PM, Moebius wrote:

Am 23.07.2024 um 21:57 schrieb Chris M. Thomasson:

On 7/23/2024 9:49 AM, WM wrote:

"LLLLLLLL....." is a finite expression,

WM is right here.

"LLLLLLLL....." is an infinite expression, indeed....

Nope.
Hint: print(len("LLLLLLLL....."))
But by convention
(we think -except WM that is- that)
"LLLLLLLL....."
DENOTES (refers to) an infinite sequence of "L"s.
:-P

I agree.

It is a very powerful distinction between
the length of an expression and
the length of what the expression refers to.

Because the expression with all our expressions
is finite,
we can know that they are all true
when they are all not.first.false,
which we can know by looking at the expressions,
without looking at what they refer to.
(AKA logic)

Because what our expressions refer to
can be infinite, and, even though infinite,
we can know what they say is true,
we can learn about infinitely.many
without we ourselves being infinite.

It takes us both parts to explore the infinite.
Finite expressions and infinite domain.
(That seems to be what WM objects to.)

"LLLLLLLL....."
DENOTES (refers to) an infinite sequence of "L"s.

There are many ways to describe an infinite sequence of "L"s.
Pick a description and say that
"LLLLLLLL....." abbreviates that description.

My most.recent favorite.description is
'well.ordered with step.up and non.min.step.down'.
⎛ Each non.{} subset B of "LLLLLLLL....." holds a first.in.B "L"
⎜ Each "L" in "LLLLLLLL....." has a first.after "L"
⎝ Each non.first "L" in "LLLLLLLL....." has a last.before "L"

That doesn't use the word 'infinite'.
It says what the word 'infinite' means (in this case).
That is actually a much more powerful and useful
thing to say.

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Am 23.07.2024 um 23:20 schrieb Chris M. Thomasson:

On 7/23/2024 1:34 PM, Moebius wrote:

Am 23.07.2024 um 22:03 schrieb Chris M. Thomasson:

On 7/23/2024 12:57 PM, Chris M. Thomasson wrote:
;

"LLLLLLLL....." is an infinite expression, indeed....

No, it's not.

So a finite [term] [referring to] the infinite [sequence of Ls]?

Exactly. :-)

So we may say:

LLL... is a sequence consisting of infinitely many Ls.

But on the other hand we must say:

"LLL..." is a expression (string) conisting of 6 characters
(which denotes a sequence consisting of infinitely many Ls).

:-P

Hint: It consists of 8 "L"s and 5 "."s, in total of 13 characters. No?

Sure, [...]

See?! :-?

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Am 23.07.2024 um 23:21 schrieb Chris M. Thomasson:

On 7/23/2024 1:56 PM, Moebius wrote:

Am 23.07.2024 um 22:28 schrieb Chris M. Thomasson:

On 7/23/2024 1:17 PM, Moebius wrote:

Am 23.07.2024 um 21:57 schrieb Chris M. Thomasson:

On 7/23/2024 9:49 AM, WM wrote:

"LLLLLLLL....." is a finite expression,

WM is right here.

"LLLLLLLL....." is an infinite expression, indeed....

Nope.

Hint: print(len("LLLLLLLL....."))

But by convention (we think -except WM that is- that)
"LLLLLLLL....." DENOTES (refers to) an infinite sequence of "L"s. :-P

Well, I was thinking of an endless sequence of L's

Yes, *I JUST SAID THAT*!

not a literal string.

Sure, but THE EXPRESSION (above) is a STRING-LITERAL, isn't it?

A programmer really should know that! ;-P

Of course. No problem with that. I have been programming for many
decades.

Same, same! :-P

I've even once written a compiler/interpreter for a VERY SMALL computer language (I called "SMALL"). So I know one or two things about parsing.

".(9)" is a string literal as well

Exatly. A FINITE "expression", just like "LLLLLLLL.....".

WM is a crank, sure, but even a blind chicken finds a corn once in a while.

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Am 23.07.2024 um 23:53 schrieb Jim Burns:

"LLLLLLLL....." DENOTES (refers to) an infinite sequence of "L"s.

There are many ways to describe an infinite sequence of "L"s.

Sure, but in a math context I personally prefer the usual set theoretic approach.

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On 7/23/2024 6:25 PM, Moebius wrote:

Am 23.07.2024 um 23:53 schrieb Jim Burns:

"LLLLLLLL....." DENOTES (refers to) an infinite sequence of "L"s.

There are many ways to describe an infinite sequence of "L"s.

Sure, but in a math context I personally prefer
the usual set theoretic approach.

I can see some good reasons for your preference.

At the top of the list, I would put
_being accepted by others_ as
a description of what's intended.

And, sure enough, in a (typical) math context,
the usual set theoretic context does best at
bringing that acceptance.

My particular context is not the unusual,
in that the usual does NOT bring acceptance from
my usual sparring.partner.

"Well.ordered with step.up and non.min.step.down"
_appears_ to me to have come closest to
the flavor of acceptance which I actually want.
(I.am.not.a.mind.reader.)
I think that's why it's my favorite.
I guess it's not a terribly complicated reason.

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Am 23.07.2024 um 23:59 schrieb Moebius:

Am 23.07.2024 um 23:20 schrieb Chris M. Thomasson:

On 7/23/2024 1:34 PM, Moebius wrote:

Am 23.07.2024 um 22:03 schrieb Chris M. Thomasson:

On 7/23/2024 12:57 PM, Chris M. Thomasson wrote:
;

"LLLLLLLL....." is an infinite expression, indeed....

No, it's not.

So a finite [term] [referring to] the infinite [sequence of Ls]?

Exactly. :-)

So we may say:

     LLL... is a sequence consisting of infinitely many Ls.

But on the other hand we must say:

     "LLL..." is a expression (string) conisting of 6 characters
     (which denotes a sequence consisting of infinitely many Ls).

:-P

"Objects I can only /name/. Signs represent them. I can only speak /of/
them. I cannot /assert them/."

--L. Wittgenstein

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Le 23/07/2024 à 22:17, Moebius a écrit :

Am 23.07.2024 um 21:57 schrieb Chris M. Thomasson:

On 7/23/2024 9:49 AM, WM wrote:

"LLLLLLLL....." is a finite expression,

WM is right here.

"LLLLLLLL....." is an infinite expression, indeed....

Nope.

Hint: print(len("LLLLLLLL....."))

But by convention (we think -except WM that is- that) "LLLLLLLL....."
DENOTES (refers to) an infinite sequence of "L"s. :-P

We all know that "LLLLLLLL....." refers to an infinite sequence. But
there are only countably finite expressions. More is not available to
refer to infinite sequences.

Regards, WM

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Le 23/07/2024 à 23:53, Jim Burns a écrit :

On 7/23/2024 4:17 PM, Moebius wrote:

Am 23.07.2024 um 21:57 schrieb Chris M. Thomasson:

On 7/23/2024 9:49 AM, WM wrote:

"LLLLLLLL....." is a finite expression,

WM is right here.

"LLLLLLLL....." is an infinite expression, indeed....

Nope.
Hint: print(len("LLLLLLLL....."))
But by convention
(we think -except WM that is- that)
"LLLLLLLL....."
DENOTES (refers to) an infinite sequence of "L"s.
:-P

I agree.

It is a very powerful distinction between
the length of an expression and
the length of what the expression refers to.

There are only countably many finite expressions which can refer to real numbers.

Regards, WM

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Am 24.07.2024 um 23:42 schrieb Chris M. Thomasson:

On 7/24/2024 1:09 PM, WM wrote:

We all know that "LLLLLLLL....." refers to an infinite sequence. But
there are only countably finite expressions. More is not available to
refer to infinite sequences.

Completely agree with with you, WM. But we can refer to more infinite
sequences by using variables and quantifiers. :-)

We "know" that there are more infinite sequences than finite expressions.

You know, mathematics is "immaterial".

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Am 24.07.2024 um 23:51 schrieb Chris M. Thomasson:

On 7/23/2024 3:06 PM, Moebius wrote:

I've even once written a compiler/interpreter for a VERY SMALL
computer language (I called "SMALL"). So I know one or two things
about parsing.

Cool! Have you ever got into C and/or C++? [...]

C, but only as a hobby language. Java (and other languages) by profession.

I implemented SMALL in two languages: C and (Turbo)Pascal. :-)

The tiny c compiler (TCC) is an interesting project:

Yeah. :-)

WM is a crank, sure, but even a blind chicken finds a corn once in a
while.

No shit! :^) Hard Core HYPER finite type of mind. Ultra finite? Sure.

Yes. ->Ultrafinitism.

See: https://en.wikipedia.org/wiki/Ultrafinitism

But WM is not a RATIONAL person. Most of his claims are nonsenical or wrong.

https://youtu.be/vNefwlE1oCg

Pure horror!

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Am 25.07.2024 um 01:37 schrieb Chris M. Thomasson:

Humm... Perhaps this song is more up WM alley with the tin foil hat...
DUH. DUH, DUHUDDHDDUDUUUDDDDUUHHHH, shit like that:

https://youtu.be/32ZTjFW2RYo

Beautiful lady. But the "music", well ... (holy shit).

(DUH. DUH, DUHUDDHDDUDUUUDDDDUUHHHH)

How about that?

https://www.youtube.com/watch?v=pwDo0JUeKqM

REAL MUSIC.

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Am 25.07.2024 um 01:58 schrieb Chris M. Thomasson:

On 7/24/2024 4:45 PM, Moebius wrote:

Am 25.07.2024 um 01:37 schrieb Chris M. Thomasson:

Humm... Perhaps this song is more up WM alley with the tin foil
hat... DUH. DUH, DUHUDDHDDUDUUUDDDDUUHHHH, shit like that:

https://youtu.be/32ZTjFW2RYo

Beautiful lady. But the "music", well ... (holy shit).

(DUH. DUH, DUHUDDHDDUDUUUDDDDUUHHHH)

How about that?

https://www.youtube.com/watch?v=pwDo0JUeKqM

REAL MUSIC.

For WM:

https://youtu.be/hAMQIvEtcJM

https://www.youtube.com/watch?v=js1wG_veGEA

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Am 25.07.2024 um 06:41 schrieb Chris M. Thomasson:

On 7/24/2024 3:26 PM, Moebius wrote:

However, if a tree falls

and no one [and no device] is there to hear [record] it, does it make a
sound? :-P

Who knows? :-)

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On 7/25/2024 6:46 PM, Moebius wrote:

Am 25.07.2024 um 06:41 schrieb Chris M. Thomasson:

However, if a tree falls

and no one [and no device] is there
to hear [record] it,
does it make a sound? :-P

Who knows? :-)

If a virtual particle with energy < ∆E
exists for a time < ∆t
_too short to observe_ ∆E⋅∆t < ℏ/2
can we know it exists?

Yes, we know.
Its existence has consequences
to spectral lines and such as.
Some consequences we are clever enough
to identify and to observe,
even if it is impossible to observe the particles. https://en.wikipedia.org/wiki/Lamb_shift
etc.

Our many.times.great.grandparents tracked
prey they hadn't seen
by the observation of a bent branch.
Consequences of the prey passing by.
That worked well enough to wipe out
a lot of delicious species.

A million times we observe
a tree falling in the forest,
a great, crashing roar, and then
a tree lying on the ground.
Consequences.

We embrace the theory that
reality doesn't give a s#!t
whether we observe or we don't observe.
By that theory, the tree made a noise.

Either that's a thing we know,
or we don't know anything,
like we don't know that
Descartes' demon hasn't deceived us
about nearly everything 'known'.

I propose that
life (apparently) is too short, and
we should consider things 'known' to be known,
and fallen trees not.seen.falling
to have made a sound.

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