Discussion:
Limits
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c***@gmail.com
2017-08-09 11:46:55 UTC
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The sequence of natural numbers

{1}, {2}, {3}, ...

and the sequence of all (unreduced) positive fractions

{1/1}, {1/2}, {2/1}, {1/3}, {2/2}, {3/1}, {1/4}, {2/3}, {3/2},{4/1}, {1/5}, ...

have empty limits because the enumeration of the (unreduced) positive fractions does not leave a remainder.

So the Sequence

{1/1}, {2/2}, {3/3}, ...

as a subsequence of the latter has the limit empty set too.

But has this subsequence, when evaluated,

{1}, {1}, {1}, ...

also the empty set as its limit?
Me
2017-08-09 21:30:37 UTC
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Post by c***@gmail.com
the sequence of all (unreduced) positive fractions
{1/1}, {1/2}, {2/1}, {1/3}, {2/2}, {3/1}, {1/4}, {2/3}, {3/2},{4/1}, {1/5}, ...
ha[s an] empty limit[.] because [...]
This sequence only has an empty set-theoretic limit if, say, 1/1 =/= 2/2, etc. Since we USUALLY (i.e. in an arithmetic context) would tend to claim 1/1 = 2/2, etc. it would be preferable to write your "unreduced positive fractions" as the usual ordered pairs (that represent "fractions" in classical mathematics/set theory):

{(1,1)}, {(1,2)}, {(2,1)}, {(1,3)}, ...
Post by c***@gmail.com
[Now] the sequence
{1/1}, {2/2}, {3/3}, ...
Actually, the sequence

{(1,1)}, {(2,2)}, {(3,3)}, ...
Post by c***@gmail.com
[...] has the [set-theoetic] limit empty set too.
Right. Hint: All the elements in this sequence are distinct, i. e. (1,1) =/= (2,2), (1,1) =/= (3,3), (2,2) =/= (3,3), etc
Post by c***@gmail.com
But has this subsequence, when evaluated,
{1}, {1}, {1}, ...
also the empty set as its limit?
No. The latter sequence has the set-theoretic limit {1}.
c***@gmail.com
2017-08-10 19:25:50 UTC
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Post by Me
Post by c***@gmail.com
the sequence of all (unreduced) positive fractions
{1/1}, {1/2}, {2/1}, {1/3}, {2/2}, {3/1}, {1/4}, {2/3}, {3/2},{4/1}, {1/5}, ...
ha[s an] empty limit[.] because [...]
This sequence only has an empty set-theoretic limit if, say, 1/1 =/= 2/2, etc.
Wrong. All fractions, whether cancelled or not, can be enumerated.
Post by Me
Since we USUALLY (i.e. in an arithmetic context) would tend to claim 1/1 = 2/2, etc.
Nevertheless 1/1 is not 2/2 when we focus on the written symbols.
Post by Me
{(1,1)}, {(2,2)}, {(3,3)}, ...
Post by c***@gmail.com
[...] has the [set-theoetic] limit empty set too.
Right. Hint: All the elements in this sequence are distinct, i. e. (1,1) =/= (2,2), (1,1) =/= (3,3), (2,2) =/= (3,3), etc
Post by c***@gmail.com
But has this subsequence, when evaluated,
{1}, {1}, {1}, ...
also the empty set as its limit?
No. The latter sequence has the set-theoretic limit {1}.
That's the problem of set theory. When considering the symbols, then
{1/1}, {2/2}, {3/3}, ... has an empty limit.
When cancelling the numbers, then the limit becomes {1}.

In fact set theoretic limits are nonsense. There is never the empty set of sequences like {1}, {2}, {3}, ... because the infinite is not finished.
Me
2017-08-10 21:08:47 UTC
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Post by c***@gmail.com
Nevertheless 1/1 is not 2/2 when we focus on the written symbols.
*lol* You are an idiot, you know. If you WANT to "focus on the written symbols", you should refer to *them*. This could be done by writing

"1/1", "2/2", etc.

(instead of 1/1, 2/2, etc.) for example. Then clearly

"1/1" =/= "2/2",

etc.

Still these expressions have to denote certain SETs in the context of set theory. Actually, "fractions" usually are represented by ordered pairs in classical mathematics/set theory. Hence we might define

"1/1" := (1, 1)
"2/2" := (2, 2)
etc.
Post by c***@gmail.com
When considering the symbols, ...
You should write

{"1/1"}, {"2/2"}, {"3/3"}, ...

instead of

{1/1}, {2/2}, {3/3}, ... ,

ESPECIALLY in an arithmetic context.

Alas, it's pointless to TELL you ANYTHING, you are to dumb for mathematics, Mucke. Just get over it.
c***@gmail.com
2017-08-11 18:52:12 UTC
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Post by Me
If you WANT to "focus on the written symbols", you should refer to *them*. This could be done by writing
"1/1", "2/2", etc.
Quotation marks are not compelling.
Post by Me
Still these expressions have to denote certain SETs in the context of set theory.
They do.

According to set theory, the limit of the sequence {1_1}, {1_2}, {1_3}, ... is empty because "all indices are used up". But set theorists overlook the fact that every index is followed by infinitely many further indices.
FredJeffries
2017-08-11 19:32:42 UTC
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Post by c***@gmail.com
According to set theory, the limit of the sequence {1_1}, {1_2}, {1_3}, ... is empty because "all indices are used up".
No, the indices never get "used up".
Post by c***@gmail.com
But set theorists overlook the fact that every index is followed by infinitely many further indices.
No, they never overlook that fact.

That's why they devised the Axiom of Infinity and Cantor's second method of ordinal generation and limits; about all of which you glory in displaying your total ignorance.
Me
2017-08-11 22:04:35 UTC
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Post by FredJeffries
Post by c***@gmail.com
According to set theory, the limit of the sequence {1_1}, {1_2}, {1_3}, ...
is empty because "all indices are used up". (Mucke)
First of all this claim is nonsense.
Post by FredJeffries
No, the indices never get "used up".
On the other hand, one might claim that ALL natural numbers "are used" as indices. With other words, that a sequence is a function f with dom(f) = IN (if it's an infinite sequence).

So what?
Post by FredJeffries
Post by c***@gmail.com
But set theorists overlook the fact that every index is followed by
infinitely many further indices. (Mucke)
Huh?!

Actually, a rather simple theorem in set theory:

An e IN: card({k e IN : k > n}) = aleph_0 .

@Mucke, hint: For all n e IN: card({1, ..., n} U {n+1, n+2, n+3, ...}) = card(IN) = aleph_0. On the other hand, for all n e IN: card({1, ..., n}) = n e IN. Hence the number of indices (numbers) following n can't be finite (for each and every n e IN).
FredJeffries
2017-08-11 23:01:58 UTC
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Post by FredJeffries
Post by c***@gmail.com
According to set theory, the limit of the sequence {1_1}, {1_2}, {1_3}, ...
is empty because "all indices are used up". (Mucke)
First of all this claim is nonsense.
Post by FredJeffries
No, the indices never get "used up".
On the other hand, one might claim that ALL natural numbers "are used" as indices. With other words, that a sequence is a function f with dom(f) = IN (if it's an infinite sequence).
But our distinguished Professor did NOT say that "each natural number is used". He claimed that "ALL indices are used UP" [emphasis added].

He is using his own special meaning (or, rather, lack of meaning) for "all" and is invoking some type of supertask that "uses up" the indices one-by-one.
Post by Me
So what?
So, you are allowing him to play his shell game -- pretending that when he babbles about his "All Theory" he is talking about what the rest of the world sees as "Set Theory".

But he isn't. All Theory is not Set Theory. He (quite naturally) finds "contradictions" in All Theory (the same contradictions that have been known for millennia). He then announces that he has found a "contradiction in set theory". And you all fall for the shell game.

His (infinite) sequences are NOT "functions with domain IN". They are processes that advance step by step.
c***@gmail.com
2017-08-12 17:02:02 UTC
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Post by FredJeffries
Post by c***@gmail.com
According to set theory, the limit of the sequence {1_1}, {1_2}, {1_3}, ... is empty because "all indices are used up".
No, the indices never get "used up".
Then after indexing infinitely many rational numbers there were natural numbers remaining without partner. That contradicts equinumerosity. Try to understand how Fraenkel explains this by means of Tristram Shandy.
Post by FredJeffries
Post by c***@gmail.com
But set theorists overlook the fact that every index is followed by infinitely many further indices.
No, they never overlook that fact.
That's why they devised the Axiom of Infinity and Cantor's second method of ordinal generation and limits;
The axiom of infinity (1904) allows to exhaust infinite sets. That's just what Fraenkel said and applied in 1928. No day of Shandy's life remains without description. No index remains without being used (up).
FredJeffries
2017-08-12 17:49:39 UTC
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Post by c***@gmail.com
Post by FredJeffries
Post by c***@gmail.com
According to set theory, the limit of the sequence {1_1}, {1_2}, {1_3}, ... is empty because "all indices are used up".
No, the indices never get "used up".
Then after indexing infinitely many rational numbers there were natural numbers remaining without partner. That contradicts equinumerosity. Try to understand how Fraenkel explains this by means of Tristram Shandy.
Post by FredJeffries
Post by c***@gmail.com
But set theorists overlook the fact that every index is followed by infinitely many further indices.
No, they never overlook that fact.
That's why they devised the Axiom of Infinity and Cantor's second method of ordinal generation and limits;
The axiom of infinity (1904) allows to exhaust infinite sets.
No, it doesn't. The axiom of infinity is an entirely different paradigm from your medieval one-by-one examination.
Julio Di Egidio
2017-08-13 06:01:57 UTC
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<snip>
Post by FredJeffries
Post by c***@gmail.com
The axiom of infinity (1904) allows to exhaust infinite sets.
No, it doesn't.
You don't like "exhaust"? Seems innocuous, in fact just fine, to me.
Post by FredJeffries
The axiom of infinity is an entirely different paradigm
from your medieval one-by-one examination.
He denies the existence of infinite sets: is that medieval? OTOH, please
correct me if I am mistaken, but in ZF-Inf indeed there are no infinite sets.

Julio
Me
2017-08-13 10:33:41 UTC
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Post by Julio Di Egidio
Post by FredJeffries
The axiom of infinity is an entirely different paradigm
from your medieval one-by-one examination.
He denies the existence of infinite sets: is that medieval?
No, he denies the *possibility* to allow for infinite sets in mathematics.

Actually, he *claims* that allowing for infnite sets leads to "contradictions" in mathematics.
Post by Julio Di Egidio
OTOH, please correct me if I am mistaken, but in ZF-Inf indeed there are no
infinite sets.
Indeed, but in ZF(C) they are, aren't they?
FredJeffries
2017-08-13 16:07:26 UTC
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Post by Julio Di Egidio
Post by FredJeffries
The axiom of infinity is an entirely different paradigm
from your medieval one-by-one examination.
He denies the existence of infinite sets: is that medieval?
Just the opposite. It is his one-by-one examination of infinite collections that is medieval: That the sum of an infinite series is calculated by calculation infinitely many binary sums, that the union of an infinite family of sets is calculated by infinitely many pairwise unions, that one can somehow perform infinitely many translations or infinitely many distinguishings, ...
Post by Julio Di Egidio
OTOH, please correct me if I am mistaken, but in ZF-Inf indeed there are no infinite sets.
You are mistaken. From the axioms of ZF-Inf one cannot PROVE the existence of an infinite set, but every model of ZF is likewise a model of ZF-Inf.
Julio Di Egidio
2017-08-14 10:28:11 UTC
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Post by FredJeffries
Post by Julio Di Egidio
Post by FredJeffries
The axiom of infinity is an entirely different paradigm
from your medieval one-by-one examination.
He denies the existence of infinite sets: is that medieval?
Just the opposite. It is his one-by-one examination of infinite
collections that is medieval: That the sum of an infinite series is
calculated by calculation infinitely many binary sums, that the union of
an infinite family of sets is calculated by infinitely many pairwise
unions, that one can somehow perform infinitely many translations or
infinitely many distinguishings, ...
He just denies the existence of infinite anything, the rest is simply a
consequence of that: so it's not medieval in that sense, and it'd not be
medieval anyway as rather in the middle ages there was a strong sense
of the absolute. OTOH, back to mathematics, I do find merit in remembering
that "limits" of any sort are NOT in fact arbitrary constructions: there are
structural rules of "continuity" to be satisfied, such as conservativeness
of extensions or similar (and maybe see "structural induction" for a
stricter presentation of the requirements), so that on this line I find
your point more misleading than his is imprecise. FWIW.
Post by FredJeffries
Post by Julio Di Egidio
OTOH, please correct me if I am mistaken, but in ZF-Inf indeed there are no infinite sets.
You are mistaken. From the axioms of ZF-Inf one cannot PROVE the existence
of an infinite set,
Right.
Post by FredJeffries
but every model of ZF is likewise a model of ZF-Inf.
Hm, thanks but that's the pill I still can't swallow: sure, there are even
unicorns out there, but how's that relevant to the theory, in this case,
ZF-Inf and to the mathematics we can do with it?

Julio
Me
2017-08-14 11:49:30 UTC
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He just denies the existence of infinite [sets] ...
No, he denies the *possibility* to allow for infinite sets in mathematics.

Actually, he *claims* that allowing for infinite sets leads to "contradictions" in mathematics.
FredJeffries
2017-08-14 14:22:04 UTC
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Post by Julio Di Egidio
of the absolute. OTOH, back to mathematics, I do find merit in remembering
that "limits" of any sort are NOT in fact arbitrary constructions: there are
structural rules of "continuity" to be satisfied, such as conservativeness
of extensions or similar (and maybe see "structural induction" for a
stricter presentation of the requirements)
That is correct. Objects such as limits, order types, cardinality, ... are properties of (structured) sets which cannot be reduced to properties of the sets' elements or relationships between pairs of elements.

The limit of a sequence is a property of the sequence as a whole, a thing-in-itself. You can even throw out an element of a sequence without altering its limit.
Julio Di Egidio
2017-08-15 09:14:49 UTC
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Post by FredJeffries
Post by Julio Di Egidio
of the absolute. OTOH, back to mathematics, I do find merit in remembering
that "limits" of any sort are NOT in fact arbitrary constructions: there are
structural rules of "continuity" to be satisfied, such as conservativeness
of extensions or similar (and maybe see "structural induction" for a
stricter presentation of the requirements)
That is correct. Objects such as limits, order types, cardinality, ... are
properties of (structured) sets which cannot be reduced to properties of
the sets' elements
Of course.
Post by FredJeffries
or relationships between pairs of elements.
No, I think that's not correct, there is nothing more (to a set) than a
relationship between the elements to each other: there are no handles
to that bag...

Julio
FredJeffries
2017-08-14 14:27:50 UTC
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Post by FredJeffries
Post by Julio Di Egidio
OTOH, please correct me if I am mistaken, but in ZF-Inf indeed there are no infinite sets.
You are mistaken. From the axioms of ZF-Inf one cannot PROVE the existence
of an infinite set,
Right.
Post by FredJeffries
but every model of ZF is likewise a model of ZF-Inf.
Hm, thanks but that's the pill I still can't swallow: sure, there are even
unicorns out there, but how's that relevant to the theory, in this case,
ZF-Inf and to the mathematics we can do with it?
For one thing, it indicates that the axiom set ZF-Inf is inadequate do do the mathematics that you want to do.

If you want to bar infinite sets, ZF-Inf is not the way to do it-- you have chosen the wrong tool for the job.

At that point, you can either whine about and blame the inadequate tool, or you can try to devise another tool.
Julio Di Egidio
2017-08-15 09:26:19 UTC
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Post by FredJeffries
Post by Me
Post by FredJeffries
Post by Julio Di Egidio
OTOH, please correct me if I am mistaken, but in ZF-Inf indeed there are no infinite sets.
You are mistaken. From the axioms of ZF-Inf one cannot PROVE the existence
of an infinite set,
Right.
Post by FredJeffries
but every model of ZF is likewise a model of ZF-Inf.
Hm, thanks but that's the pill I still can't swallow: sure, there are even
unicorns out there, but how's that relevant to the theory, in this case,
ZF-Inf and to the mathematics we can do with it?
For one thing, it indicates that the axiom set ZF-Inf is inadequate do do
the mathematics that you want to do.
That *you* want to do, you are the one claiming that there are infinite
sets in ZF-Inf, I am the one objecting to that.
Post by FredJeffries
If you want to bar infinite sets, ZF-Inf is not the way to do it-- you
have chosen the wrong tool for the job.
Funny, as my goal is the opposite of that, as you should know, while you are
the finitist... Never mind: just please enlighten us and tell us how one
goes about banning infinite sets (correctly) if one wants, as I am still
not sure WTF you are even talking about, honestly.

Julio
FredJeffries
2017-08-15 18:25:43 UTC
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Post by Julio Di Egidio
Post by FredJeffries
Post by Me
Post by FredJeffries
Post by Julio Di Egidio
OTOH, please correct me if I am mistaken, but in ZF-Inf indeed there are no infinite sets.
You are mistaken. From the axioms of ZF-Inf one cannot PROVE the existence
of an infinite set,
Right.
Post by FredJeffries
but every model of ZF is likewise a model of ZF-Inf.
Hm, thanks but that's the pill I still can't swallow: sure, there are even
unicorns out there, but how's that relevant to the theory, in this case,
ZF-Inf and to the mathematics we can do with it?
For one thing, it indicates that the axiom set ZF-Inf is inadequate do do
the mathematics that you want to do.
That *you* want to do, you are the one claiming that there are infinite
sets in ZF-Inf, I am the one objecting to that.
Post by FredJeffries
If you want to bar infinite sets, ZF-Inf is not the way to do it-- you
have chosen the wrong tool for the job.
Funny, as my goal is the opposite of that, as you should know, while you are
the finitist... Never mind: just please enlighten us and tell us how one
goes about banning infinite sets (correctly) if one wants, as I am still
not sure WTF you are even talking about, honestly.
The system being discussed in another thread, ZF-Inf+~Inf, for example.

https://en.wikipedia.org/wiki/Hereditarily_finite_set#Discussion
Julio Di Egidio
2017-08-16 02:40:55 UTC
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Post by FredJeffries
Post by Julio Di Egidio
Post by FredJeffries
Post by Me
Post by FredJeffries
Post by Julio Di Egidio
OTOH, please correct me if I am mistaken, but in ZF-Inf indeed there are no infinite sets.
You are mistaken. From the axioms of ZF-Inf one cannot PROVE the existence
of an infinite set,
Right.
Post by FredJeffries
but every model of ZF is likewise a model of ZF-Inf.
Hm, thanks but that's the pill I still can't swallow: sure, there are even
unicorns out there, but how's that relevant to the theory, in this case,
ZF-Inf and to the mathematics we can do with it?
For one thing, it indicates that the axiom set ZF-Inf is inadequate do do
the mathematics that you want to do.
That *you* want to do, you are the one claiming that there are infinite
sets in ZF-Inf, I am the one objecting to that.
Post by FredJeffries
If you want to bar infinite sets, ZF-Inf is not the way to do it-- you
have chosen the wrong tool for the job.
Funny, as my goal is the opposite of that, as you should know, while you are
the finitist... Never mind: just please enlighten us and tell us how one
goes about banning infinite sets (correctly) if one wants, as I am still
not sure WTF you are even talking about, honestly.
The system being discussed in another thread, ZF-Inf+~Inf, for example.
That's still nonsense: if, as you claim, there are infinite sets in ZF-Inf,
then there will be infinite sets in ZF-Inf+~Inf, too.
Post by FredJeffries
https://en.wikipedia.org/wiki/Hereditarily_finite_set#Discussion
Standard (Cantorian) mathematics: ex falso quod libet...

Julio
Alan Smaill
2017-08-16 11:33:27 UTC
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Post by Julio Di Egidio
Post by FredJeffries
Post by Julio Di Egidio
Post by FredJeffries
Post by Me
Post by FredJeffries
Post by Julio Di Egidio
OTOH, please correct me if I am mistaken, but in ZF-Inf
indeed there are no infinite sets.
You are mistaken. From the axioms of ZF-Inf one cannot PROVE
the existence of an infinite set,
Right.
Post by FredJeffries
but every model of ZF is likewise a model of ZF-Inf.
Hm, thanks but that's the pill I still can't swallow: sure,
there are even unicorns out there, but how's that relevant to
the theory, in this case, ZF-Inf and to the mathematics we can
do with it?
For one thing, it indicates that the axiom set ZF-Inf is
inadequate do do the mathematics that you want to do.
That *you* want to do, you are the one claiming that there are
infinite sets in ZF-Inf, I am the one objecting to that.
Post by FredJeffries
If you want to bar infinite sets, ZF-Inf is not the way to do it-- you
have chosen the wrong tool for the job.
Funny, as my goal is the opposite of that, as you should know,
while you are the finitist... Never mind: just please enlighten us
and tell us how one goes about banning infinite sets (correctly) if
one wants, as I am still not sure WTF you are even talking about,
honestly.
The system being discussed in another thread, ZF-Inf+~Inf, for example.
That's still nonsense: if, as you claim, there are infinite sets in
ZF-Inf, then there will be infinite sets in ZF-Inf+~Inf, too.
No. There are infinite sets in *some* models of ZF-Inf,
but not in all models.

If you take anything that happens to be a model of ZF, all the axioms of
ZF are true in the model; a fortiori all the axioms of ZF-Inf are true
(there are fewer axioms). There are also models of ZF-Inf with no
infinite sets (assuming it's consistent).

On the other hand, a model of ZF-Inf+~Inf must satisfy the new axiom
~Inf, which says that there are no infinite sets.
Post by Julio Di Egidio
Julio
--
Alan Smaill
Julio Di Egidio
2017-08-16 14:29:46 UTC
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Post by Julio Di Egidio
Post by FredJeffries
Post by Julio Di Egidio
Post by FredJeffries
Post by Me
Post by FredJeffries
Post by Julio Di Egidio
OTOH, please correct me if I am mistaken, but in ZF-Inf
indeed there are no infinite sets.
You are mistaken. From the axioms of ZF-Inf one cannot PROVE
the existence of an infinite set,
Right.
Post by FredJeffries
but every model of ZF is likewise a model of ZF-Inf.
Hm, thanks but that's the pill I still can't swallow: sure,
there are even unicorns out there, but how's that relevant to
the theory, in this case, ZF-Inf and to the mathematics we can
do with it?
For one thing, it indicates that the axiom set ZF-Inf is
inadequate do do the mathematics that you want to do.
That *you* want to do, you are the one claiming that there are
infinite sets in ZF-Inf, I am the one objecting to that.
Post by FredJeffries
If you want to bar infinite sets, ZF-Inf is not the way to do it-- you
have chosen the wrong tool for the job.
Funny, as my goal is the opposite of that, as you should know,
while you are the finitist... Never mind: just please enlighten us
and tell us how one goes about banning infinite sets (correctly) if
one wants, as I am still not sure WTF you are even talking about,
honestly.
The system being discussed in another thread, ZF-Inf+~Inf, for example.
That's still nonsense: if, as you claim, there are infinite sets in
ZF-Inf, then there will be infinite sets in ZF-Inf+~Inf, too.
No. There are infinite sets in *some* models of ZF-Inf,
but not in all models.
If you take anything that happens to be a model of ZF, all the axioms of
ZF are true in the model; a fortiori all the axioms of ZF-Inf are true
(there are fewer axioms). There are also models of ZF-Inf with no
infinite sets (assuming it's consistent).
You still do not answer the question: sure,
there are even unicorns out there, but how's that relevant to
the theory, in this case, ZF-Inf and to the mathematics we can
do with it? Namely, can any putatively infinite sets enter an
argument in ZF-Inf in any useful way at all? For example, can
we do infinite unions in ZF-Inf unless we introduce some infinite
index set anyway? And similar considerations.
Post by Alan Smaill
On the other hand, a model of ZF-Inf+~Inf must satisfy the new axiom
~Inf, which says that there are no infinite sets.
No, it says that "there is no set such that zero is an element of it and the
successor of every element of it is an element of it". That does not imply
that "there are no infinite sets".

Julio
Me
2017-08-16 14:33:38 UTC
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Post by Julio Di Egidio
Post by Alan Smaill
On the other hand, a model of ZF-Inf+~Inf must satisfy the new axiom
~Inf, which says that there are no infinite sets.
No, it says that "there is no set such that zero is an element of it and the
successor of every element of it is an element of it". That does not imply
that "there are no infinite sets".
Obviously it does:

Found the following Q & A at math.stackexchange:

Q: "ZFC without infinity" would mean ZFC with the infinity axiom removed. If you attach the negation of the infinity axiom, you are assuming that there are no inductive sets. (I wonder if that is enough to prove all sets are finite?)

A: If we assume that there is no inductive set but have Choice, we can still prove that every set is equinumerous to an ordinal. Since ω is assumed not to exist, this means that every set must be finite. I think mathematical induction arguments could still be carried out, in the guise of "transfinite" induction, so I would expect one could prove that every set is Dedekind-finite too.
Alan Smaill
2017-08-16 16:03:16 UTC
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[...]
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Post by Julio Di Egidio
Post by FredJeffries
Post by Julio Di Egidio
Post by FredJeffries
If you want to bar infinite sets, ZF-Inf is not the way to do
it-- you have chosen the wrong tool for the job.
Funny, as my goal is the opposite of that, as you should know,
while you are the finitist... Never mind: just please enlighten us
and tell us how one goes about banning infinite sets (correctly) if
one wants, as I am still not sure WTF you are even talking about,
honestly.
The system being discussed in another thread, ZF-Inf+~Inf, for example.
That's still nonsense: if, as you claim, there are infinite sets in
ZF-Inf, then there will be infinite sets in ZF-Inf+~Inf, too.
No. There are infinite sets in *some* models of ZF-Inf,
but not in all models.
If you take anything that happens to be a model of ZF, all the axioms of
ZF are true in the model; a fortiori all the axioms of ZF-Inf are true
(there are fewer axioms). There are also models of ZF-Inf with no
infinite sets (assuming it's consistent).
You still do not answer the question: sure,
there are even unicorns out there, but how's that relevant to
the theory, in this case, ZF-Inf and to the mathematics we can
do with it? Namely, can any putatively infinite sets enter an
argument in ZF-Inf in any useful way at all?
I have no idea how useful ZF-Inf is as a theory,
I'm not aware of anyone exploring what can be done.

You cannot prove the existence of an infinite set in
ZF-Inf, you cannot disprove the existence of an infinite set
either. If someone is interested in using set theory but is agnostic
on the existence of infinite sets, then ZF-Inf looks appropriate.
You could reason hypothetically about infinite sets without
invoking ex falso quodlibet.
Post by Julio Di Egidio
For example, can
we do infinite unions in ZF-Inf unless we introduce some infinite
index set anyway? And similar considerations.
You already have unions of arbitrary sets of sets inherited
from ZF.
Post by Julio Di Egidio
Post by Alan Smaill
On the other hand, a model of ZF-Inf+~Inf must satisfy the new axiom
~Inf, which says that there are no infinite sets.
No, it says that "there is no set such that zero is an element of it and the
successor of every element of it is an element of it". That does not imply
that "there are no infinite sets".
Suppose there is an infinite set. Now use replacement to show that there
is a set such that zero is an element of it and the successor of every
element of it is an element of it. Contradiction with ~Inf, so
there is no infinite set.

(Jim Burns has been outlining the main part of the proof.)
Post by Julio Di Egidio
Julio
--
Alan Smaill
Jim Burns
2017-08-16 18:02:05 UTC
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Post by Alan Smaill
Post by Julio Di Egidio
You still do not answer the question: sure,
there are even unicorns out there, but how's that
relevant to the theory, in this case, ZF-Inf and to
the mathematics we can do with it? Namely, can any
putatively infinite sets enter an argument in ZF-Inf
in any useful way at all?
I have no idea how useful ZF-Inf is as a theory,
I'm not aware of anyone exploring what can be done.
Whether the following is useful is a matter of taste,
I suppose.

One thing that we don't need either Inf or ~Inf for
is deriving the familiar form of induction
( P(0) & Ax:( P(x) -> P(Sx) ) -> Ay: P(y)
where we refer only to finite ordinals.

Suppose that
0 := {}

Sx := (x U {x})

Nx :=
(x = 0 \/ Ey e x: Sy = x )
& Az e x: ( z = 0 \/ Ev e x: Sv = z )

Nx == "x is a finite ordinal"

A finite ordinal either has an immediate predecessor
or is 0.

With the axiom of foundation, ordinals are well-ordered.
If there is a counter-example to P(x), there is a first
counter-example to P(x).

If P(x) -> P(Sx), then the first counter-example
doesn't have an immediate predecessor. If P(0), then
it isn't 0, either. If both, there is no first
counter-example.

Therefore (for finite ordinals),
( P(0) & Ax:( P(x) -> P(Sx) ) -> Ay: P(y)

[...]
Post by Alan Smaill
Suppose there is an infinite set. Now use replacement to
show that there is a set such that zero is an element of it
and the successor of every element of it is an element of it.
Contradiction with ~Inf, so there is no infinite set.
(Jim Burns has been outlining the main part of the proof.)
It's really very simple. It's just that we only rarely
use the axiom schema of replacement, so it seems odd.

If we can describe the elements of some class (which is
possibly a set) and that class is _not larger than_ some
set, then the axiom of replacement says that class is
also a set.

In this case, if there is an infinite set B, there is a
minimal infinite set B'. The axiom of infinity asserts
the existence of some set which can be as small as the
minimal infinite set B'. We describe the elements
{}, {{}}, {{},{{}}}, ...
and we show that there are not more of them than the
elements of B'. Therefore, they form a set.

"What? Any description at all can form a set?" one may ask.
It's the restriction in size that prevents all that freedom
from erupting into things like the "set" of all sets not
containing themselves. So I'm told, anyway.

I'm not sure how one would go about supporting that
claim with more than "So I'm told", but ZFC does seem
to be successful out here in the real world.
Julio Di Egidio
2017-08-17 08:54:52 UTC
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<snip>
Post by Alan Smaill
Post by Julio Di Egidio
For example, can
we do infinite unions in ZF-Inf unless we introduce some infinite
index set anyway? And similar considerations.
You already have unions of arbitrary sets of sets inherited
from ZF.
But my objection is that the reason why we can do unions there is just
because the sets are all finite so the "preformal counting numbers" can do
without further complication and pretty much implicitly: but for infinite
unions we need an infinite index set (or equivalent) which is just not
there until we establish a set of natural numbers (or equivalent) to
begin with... -- And I think another way to say it is that proofs must
be *finite*, so we need means to treat infinite sets that are not just
going over the elements one by one, namely some "limits". Which again
are not there until there is an infinite set of indexes to begin with.
Post by Alan Smaill
Post by Julio Di Egidio
Post by Alan Smaill
On the other hand, a model of ZF-Inf+~Inf must satisfy the new axiom
~Inf, which says that there are no infinite sets.
No, it says that "there is no set such that zero is an element of it and the
successor of every element of it is an element of it". That does not imply
that "there are no infinite sets".
Suppose there is an infinite set. Now use replacement to show that there
is a set such that zero is an element of it and the successor of every
element of it is an element of it. Contradiction with ~Inf, so
there is no infinite set.
That makes sense, but I still think it cannot be done really, for the
reasons I have mentioned above. Indeed, I think the possibility or not
of doing infinite unions (could be intersections, of course) is exactly
the culprit.

Julio
Alan Smaill
2017-08-17 10:11:34 UTC
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Post by Julio Di Egidio
<snip>
Post by Alan Smaill
Post by Julio Di Egidio
For example, can
we do infinite unions in ZF-Inf unless we introduce some infinite
index set anyway? And similar considerations.
You already have unions of arbitrary sets of sets inherited
from ZF.
But my objection is that the reason why we can do unions there is just
because the sets are all finite so the "preformal counting numbers" can do
without further complication and pretty much implicitly: but for infinite
unions we need an infinite index set (or equivalent) which is just not
there until we establish a set of natural numbers (or equivalent) to
begin with...
Yes, you cannot prove existence of an infinite set.

But you can reason hypothetically by proving statements
like

all x. all y.
((infinite(x) & infinite(y) & x \subset y)
-> Ux \subset Uy)

The assumption that x,y are infinite doesn't introduce
a contradiction, as it would in ZF-Inf+~Inf.

(I'm deliberately not addressing the question of relation
to pre-formal ideas, I'm not in the business of defending
set theory as a foundational theory.)
Post by Julio Di Egidio
-- And I think another way to say it is that proofs must
be *finite*, so we need means to treat infinite sets that are not just
going over the elements one by one, namely some "limits". Which again
are not there until there is an infinite set of indexes to begin with.
See above.
Post by Julio Di Egidio
Post by Alan Smaill
Post by Julio Di Egidio
Post by Alan Smaill
On the other hand, a model of ZF-Inf+~Inf must satisfy the new axiom
~Inf, which says that there are no infinite sets.
No, it says that "there is no set such that zero is an element of
it and the successor of every element of it is an element of it".
That does not imply that "there are no infinite sets".
Suppose there is an infinite set. Now use replacement to show that there
is a set such that zero is an element of it and the successor of every
element of it is an element of it. Contradiction with ~Inf, so
there is no infinite set.
That makes sense, but I still think it cannot be done really,
what cannot be done, in what theory ?
Post by Julio Di Egidio
for the
reasons I have mentioned above. Indeed, I think the possibility or not
of doing infinite unions (could be intersections, of course) is exactly
the culprit.
Julio
--
Alan Smaill
Me
2017-08-16 13:33:43 UTC
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Post by Julio Di Egidio
Post by FredJeffries
Post by Julio Di Egidio
Post by FredJeffries
Post by Me
Post by FredJeffries
Post by Julio Di Egidio
OTOH, please correct me if I am mistaken, but in ZF-Inf indeed there are no infinite sets.
You are mistaken. From the axioms of ZF-Inf one cannot PROVE the existence
of an infinite set,
Right.
Post by FredJeffries
but every model of ZF is likewise a model of ZF-Inf.
Hm, thanks but that's the pill I still can't swallow: sure, there are even
unicorns out there, but how's that relevant to the theory, in this case,
ZF-Inf and to the mathematics we can do with it?
For one thing, it indicates that the axiom set ZF-Inf is inadequate do do
the mathematics that you want to do.
That *you* want to do, you are the one claiming that there are infinite
sets in ZF-Inf, I am the one objecting to that.
Post by FredJeffries
If you want to bar infinite sets, ZF-Inf is not the way to do it-- you
have chosen the wrong tool for the job.
Funny, as my goal is the opposite of that, as you should know, while you are
the finitist... Never mind: just please enlighten us and tell us how one
goes about banning infinite sets (correctly) if one wants, as I am still
not sure WTF you are even talking about, honestly.
The system being discussed in another thread, ZF-Inf+~Inf, for example.
That's still nonsense: if, as you claim, there are infinite sets in ZF-Inf,
or rather a model of ZF\{Inf} might still have infinite sets
Post by Julio Di Egidio
then there will be infinite sets in ZF-Inf+~Inf, too.
Errr.. Why?

I've found the following Q & A at math.stackexchange:

Q: "ZFC without infinity" would mean ZFC with the infinity axiom removed. If you attach the negation of the infinity axiom, you are assuming that there are no inductive sets. (I wonder if that is enough to prove all sets are finite?)

A: If we assume that there is no inductive set but have Choice, we can still prove that every set is equinumerous to an ordinal. Since ω is assumed not to exist, this means that every set must be finite. I think mathematical induction arguments could still be carried out, in the guise of "transfinite" induction, so I would expect one could prove that every set is Dedekind-finite too.
Post by Julio Di Egidio
Post by FredJeffries
https://en.wikipedia.org/wiki/Hereditarily_finite_set#Discussion
Standard (Cantorian) mathematics: ex falso quod libet...
Huh?! WTF?!
Shobe, Martin
2017-08-13 21:42:28 UTC
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Post by Julio Di Egidio
<snip>
Post by FredJeffries
Post by c***@gmail.com
The axiom of infinity (1904) allows to exhaust infinite sets.
No, it doesn't.
You don't like "exhaust"? Seems innocuous, in fact just fine, to me.
Sometimes it is. In this case, it's being used to muddy the waters as
it's ambiguous. WM relies on natural language ambiguities to make his
"proofs".
Post by Julio Di Egidio
Post by FredJeffries
The axiom of infinity is an entirely different paradigm
from your medieval one-by-one examination.
He denies the existence of infinite sets: is that medieval? OTOH, please
correct me if I am mistaken, but in ZF-Inf indeed there are no infinite sets.
Sure. In ZF-Inf, there may or may not be infinite sets. If you want to
be certain there are no infinite sets you could use ZF-Inf+~Inf.

Martin Shobe
Me
2017-08-13 21:53:42 UTC
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Post by Shobe, Martin
Post by Julio Di Egidio
You don't like "exhaust"? Seems innocuous, in fact just fine, to me.
Sometimes it is. In this case, it's being used to muddy the waters as
it's ambiguous. WM relies on natural language ambiguities to make his
"proofs".
This man, Mucke, has a mission. He's not interested in actually doing mathematics in the least.

If he were he could just stick to some system of "finitistic mathematics" and be quite happy with it.

See:
https://en.wikipedia.org/wiki/Finitism
http://www.springer.com/de/book/9789400713468
Me
2017-08-13 21:59:22 UTC
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... he could just stick to some system of "finitistic mathematics"
and be quite happy with it.
https://en.wikipedia.org/wiki/Finitism
http://www.springer.com/de/book/9789400713468
If anyone is interested:
https://7e663d17-a-62cb3a1a-s-sites.googlegroups.com/site/fengye63/strictlyfinitisticsystemforappliedmath/strictlyFinitisticSystemForAppliedMath.pdf
c***@gmail.com
2017-08-14 13:58:56 UTC
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Post by Shobe, Martin
Post by Julio Di Egidio
<snip>
Post by FredJeffries
Post by c***@gmail.com
The axiom of infinity (1904) allows to exhaust infinite sets.
No, it doesn't.
You don't like "exhaust"? Seems innocuous, in fact just fine, to me.
Sometimes it is. In this case, it's being used to muddy the waters as
it's ambiguous. WM relies on natural language ambiguities to make his
"proofs".
In set theory the sequence {1_1}, {1_2}, {1_3}, ... has empty limit with no doubt.

And the sequence {1}, {1}, {1}, ... has limit {1}.

But if you only *think* of the indices, then the limit is zero.

Note: If you can exhaust aleph_0 natural numbers, then you can also exhaust aleph_0 digits of the form 1.
Me
2017-08-14 15:30:47 UTC
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Post by c***@gmail.com
In set theory the sequence
({1_1}, {1_2}, {1_3}, ...
Post by c***@gmail.com
has empty [set-theoretic] limit with no doubt.
No, idiot. IF, say, 1_n = 1 for all n e IN, THEN the set-theoretic limit of THIS sequence is {1}. IF, on the other hand, for example, 1_n = n for all n e IN, THEN the set-theoretic limit of THIS sequence is {}.

Too dumb to understand basic mathematics/set theory?
Me
2017-08-14 16:38:50 UTC
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Post by Me
Too dumb to understand basic mathematics/set theory?
Hint: If 1_n = 1 for all n e IN then, for example

{1_1, 1_2, 1_3} = {1, 1, 1} = {1} .

With other words,

card({1_1, 1_2, 1_3}) = 1 .

If, on the other handy, say, 1_n = n for all n e IN, then

{1_1, 1_2, 1_3} = {1, 2, 3} .

With other words, NOW,

card({1_1, 1_2, 1_3}) = 3 .

You know, the MEANING of mathematical symbols quite often DEPEND on the context.

It seems to me that you mistake names/terms for the objects/numbers denoted by these names/terms.

Hint: The terms "4", "2 + 2", "1 + 3" (in an arithmetic context) ALL denote the same natural number, namely 4.

With other words:

4 = 2 + 2 = 1 + 3 .

Hence (in this context)

{4, 2 + 2, 1 + 3} = {4} .

If we DEFINE now a sequence (s_n)_(n e IN) with s_n = 4 for all n e IN. We would get

4 = s_1 = s_2 = s_3 = ...

And hence:

{4, 2 + 2, 1 + 3, s_1, s_2, s_3, ...} = {4} .

If you prefer to write

(4_n)_(n e IN)

for such a sequence, things don't change:

{4, 2 + 2, 1 + 3, 4_1, 4_2, 4_3, ...} = {4} .

Now since 4_1 = 4_2 = 4_3 = ... = 4, of course

{4_1} = {4_2} = {4_3} = ... = {4} .

Hence the sequence

({4_1}, {4_2}, {4_3}, ...)

would just be IDENTICAL with the sequence

({4}, {4}, {4}, ...) ,

i.e. they are one and the same sequence.

Hence the set-theoretic limit of "both" sequences would just be {4}.
WM
2017-08-15 13:21:13 UTC
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Post by Me
Post by c***@gmail.com
In set theory the sequence
({1_1}, {1_2}, {1_3}, ...
Post by c***@gmail.com
has empty [set-theoretic] limit with no doubt.
No,
Handwaving is useless. Take the set -theoretic limit You may lwarn it for instance here: https://www.hs-augsburg.de/~mueckenh/Transfinity/Transfinity/pdf
Post by Me
IF, say, 1_n = 1 for all n e IN, THEN the set-theoretic limit of THIS sequence is {1}. IF, on the other hand, for example, 1_n =
is the n-th 1, then the limit is empty, because there are only aleph_0 1's.

Regards, WM
Me
2017-08-15 13:52:38 UTC
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On Tuesday, August 15, 2017 at 3:21:28 PM UTC+2, WM wrote:

Nonsens, as usual.

Just to sum it up, idiot:

IF 1_n = 1 for all n e IN, THEN the set-theoretic limit of the sequence

({1_1}, {1_2}, {1_3}, ...)

is {1}, since the set-theoretic limit of the sequence

({1}, {1}, {1}, ...)

is {1}.

IF 1_n = n for all n e IN, THEN the set-theoretic limit of the sequence

({1_1}, {1_2}, {1_3}, ...)

is { }, since the set-theoretic limit of the sequence

({1}, {2}, {3}, ...)

is { }.

EOD.
Me
2017-08-14 16:57:20 UTC
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Post by c***@gmail.com
The sequence
({1}, {1}, {1}, ...)
has [set-theoretic] limit {1}.
Right. Same with the sequence

({1_1}, {1_2}, {1_3}, ...)

(if we have 1_n = 1 for all n e IN).
Post by c***@gmail.com
But if you only think of the indices, then ...
...USE them, *hell*!

You might for example adopt the identity function (on IN):

id(n) = n for all n e IN

and write

Id_n

instead of id(n). In this case you would actualy refer to the "indices", since

Id_n = n for all ne IN.

Then we would get

LIM ({Id_1}, {Id_2}, {Id_3}, ...) = { } ,

since

({Id_1}, {Id_2}, {Id_3}, ...) = ({1}, {2}, {3}, ...)

and

LIM ({1}, {2}, {3}, ...) = { } .
Shobe, Martin
2017-08-15 23:35:32 UTC
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Post by c***@gmail.com
Post by Shobe, Martin
Post by Julio Di Egidio
<snip>
Post by FredJeffries
Post by c***@gmail.com
The axiom of infinity (1904) allows to exhaust infinite sets.
No, it doesn't.
You don't like "exhaust"? Seems innocuous, in fact just fine, to me.
Sometimes it is. In this case, it's being used to muddy the waters as
it's ambiguous. WM relies on natural language ambiguities to make his
"proofs".
In set theory the sequence {1_1}, {1_2}, {1_3}, ... has empty limit with no doubt.
There's plenty of doubt. In fact, given what you've said elsewhere, I'd
say the limit is {1}.
Post by c***@gmail.com
And the sequence {1}, {1}, {1}, ... has limit {1}.
But if you only *think* of the indices, then the limit is zero.
[snip nonsense about exhaustion]

Martin Shobe
WM
2017-08-16 12:45:51 UTC
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Post by Shobe, Martin
Post by c***@gmail.com
In set theory the sequence {1_1}, {1_2}, {1_3}, ... has empty limit with no doubt.
There's plenty of doubt. In fact, given what you've said elsewhere, I'd
say the limit is {1}.
You can learn to calculate the limit here: Set-theoretical limits of sequences of sets, p. 55 in https://www.hs-augsburg.de/~mueckenh/Transfinity/Transfinity/pdf
Post by Shobe, Martin
Post by c***@gmail.com
And the sequence {1}, {1}, {1}, ... has limit {1}.
But if you only *think* of the indices, then the limit is zero.
Regards, WM
Me
2017-08-16 13:12:16 UTC
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In set theory the sequence ({1_1}, {1_2}, {1_3}, ...)
Is identical with the sequence ({1}, {1}, {1}, ...), if 1_n = 1 for all n e IN, that is.
In fact, given what [WM] said elsewhere, I'd say the limit is {1}.
Right.
You can learn to calculate the limit here [...]
Mücke, *we* do not have to learn that. It's a well known fact that for a (set) sequence (A_n) with A_n c A_(n+1) for all n e IN:

lim A_n = U{A_n : n e IN} (where "lim" refers to the set-theoretic limit)
n->n

See: https://en.wikipedia.org/wiki/Set-theoretic_limit#Monotone_sequences

Hence
the sequence ({1}, {1}, {1}, ...) has limit {1}.
Exactly!
WM
2017-08-16 13:44:04 UTC
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Post by Me
In set theory the sequence ({1_1}, {1_2}, {1_3}, ...)
Is identical with the sequence ({1}, {1}, {1}, ...), if 1_n = 1 for all n e IN, that is.
There is no necessity to fix any numerical value! Here we are in set theory, not in number theory. All kinds of sets are objects of set theory.
Post by Me
You can learn to calculate the limit here https://www.hs-augsburg.de/~mueckenh/Transfinity/Transfinity/pdf
*we* do not have to learn that.
You are wrong. On page 231, loc cit you can see that some users did not know anything about that topic when I brought it up here last year.
Post by Me
lim A_n = U{A_n : n e IN} (where "lim" refers to the set-theoretic limit)
n->n
It is well known that matheologians repeatedly fall victim to this nonsense. They are simply confusing limit and union. And they are too blocked to learn the difference from the simplest examples like this one:

o
oo
ooo
...

The limit has aleph_0 symbols, the union has not.
Post by Me
See: https://en.wikipedia.org/wiki/Set-theoretic_limit#Monotone_sequences
Unfortunately this nonsense is taught also in Wikipedia.
Post by Me
Hence
the sequence ({1}, {1}, {1}, ...) has limit {1}.
Exactly!
Excatly --- and even if we do not attach a numerical value to the symbol.
The sequence §, §, §, ... has limit §.

Regards, WM
Me
2017-08-16 14:06:47 UTC
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Post by WM
Post by Me
In set theory the sequence ({1_1}, {1_2}, {1_3}, ...)
Is identical with the sequence ({1}, {1}, {1}, ...), if 1_n = 1 for all
n e IN, that is.
There is no necessity to fix any numerical value!
It *is*, since the limit (if it exists at all) DEPENDS on that values.

Hint: Let's consider the (set) sequence S = ({a_1}, {a_2}, {a_3}, ...).

IF a_n = 1 for all n e IN, then

lim {a_n} = {1}
n->oo

IF, on the other hand, say, a_n = 2 for all n e IN, then

lim {a_n} = {2}
n->oo

Note that {1} =/= {2}, if 1 =/= 2. Too complicated for you?

Or IF, say, a_n = n for all n e IN, then

lim {a_n} = { } . WOW!
n->oo

Note that in any case, { } =/= {1} and {} =/= {2}.
Post by WM
Post by Me
the sequence ({1}, {1}, {1}, ...) has limit {1}.
Exactly!
Excatly --- and even if we do not attach a numerical value to the symbol.
The sequence §, §, §, ... has limit §.
Huh? What are you talking about, man?

*We* were talking about the SET-THEORETIC limit of (set) sequences, so please don't switch topics, again!

Right, if we consider the (set) sequence

({a}, {a}, {a}, {a}, ...)

its (set-theoretic) limit will be {a}, for ANY (each and every) set a. :-)

BUT you originally DIDN'T consider SUCH a sequence, you CONSIDERED the sequence


({1_1}, {1_2}, {1_3}, ...) ,

remember? :-)
Me
2017-08-16 14:49:01 UTC
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Post by WM
In set theory the sequence ({1_1}, {1_2}, {1_3}, ...)
is identical with the sequence ({1}, {1}, {1}, ...), if 1_n = 1 for all
n e IN, that is.
There is no necessity to fix any numerical value!
*FUCK* I DIND'T claim that, did I?! I JUST stated A FACT.

On the other hand, it *is* necessary to be more specific concerning the "values" of the terms, since the limit (if it exists at all) DEPENDS on that "values".

Hint: Let's consider the (set) sequence S = ({a_1}, {a_2}, {a_3}, ...).

IF a_n = 1 for all n e IN, then

lim {a_n} = {1}
n->oo

IF, on the other hand, say, a_n = 2 for all n e IN, then

lim {a_n} = {2}
n->oo

Note that {1} =/= {2}, if 1 =/= 2. Too complicated for you?

Or IF, say, a_n = n for all n e IN, then

lim {a_n} = { } . WOW!
n->oo

Note that in any case, { } =/= {1} and {} =/= {2}.

Actually, I already considered two "corner cases" before:

1) IF a_n = a_m for all n,m e IN, THEN the set-theoretic limit of THIS sequence is just {a_k} for some k e IN.

2) IF a_n =/= a_m for all n,m e IN, n =/= m, THEN the set-theoretic limit of THIS (different) sequence is just the empty set.
Post by WM
the sequence ({1}, {1}, {1}, ...) has limit {1}.
Exactly!
Excatly --- and even if we do not attach a numerical value to the symbol.
The sequence §, §, §, ... has limit §.
Huh? What are you talking about, man?

*We* were talking about the SET-THEORETIC limit of (set) sequences, so please don't switch topics, again!

Right, if we consider the (set) sequence

({a}, {a}, {a}, {a}, ...)

its (set-theoretic) limit will be {a}, for ANY (each and every) set a. (See case 1) above).

BUT you originally DIDN'T consider SUCH a sequence, you CONSIDERED the sequence

({1_1}, {1_2}, {1_3}, ...) ,

remember? :-)
Me
2017-08-16 15:04:06 UTC
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There is no necessity to fix any numerical value! (Mückenheim)
*FUCK* I DIND'T claim that, did I?! ***I JUST stated A FACT.***
Again, Just to sum it up:

IF 1_n = 1 for all n e IN, THEN the set-theoretic limit of the sequence

({1_1}, {1_2}, {1_3}, ...)

is {1}, since the set-theoretic limit of the sequence

({1}, {1}, {1}, ...)

is {1}.

IF 1_n = n for all n e IN, THEN the set-theoretic limit of the sequence

({1_1}, {1_2}, {1_3}, ...)

is { }, since the set-theoretic limit of the sequence

({1}, {2}, {3}, ...)

is { }.

EOD.
WM
2017-08-17 12:19:52 UTC
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Post by WM
is identical with the sequence ({1}, {1}, {1}, ...), if 1_n = 1 for all
n e IN, that is.
There is no necessity to fix any numerical value!
*FUCK* I DIND'T claim that, did I?! I JUST stated A FACT.
I mentioned only that your statement was irrelevant
Post by Me
On the other hand, it *is* necessary to be more specific concerning the "values" of the terms, since the limit (if it exists at all) DEPENDS on that "values".
You said also that the limit depends on representation. Not sure about your opinion?

In my example the values are always the natural numbers.

one, two, three, ...
eins, zwei, drei, ...
1, 2, 3, ...
I, II, III, ...
{0}, {0, 1}, {0, 1, 2}, ...

Therefore the different limits are contradictory.
Post by Me
Post by WM
Excatly --- and even if we do not attach a numerical value to the symbol.
The sequence §, §, §, ... has limit §.
Huh? What are you talking about, man?
*We* were talking about the SET-THEORETIC limit of (set) sequences, so please don't switch topics, again!
§ is a set. According to set theory the sequence §, §, §, ... has limit §. Do you need some help in deriving that result?

Regards, WM

Shobe, Martin
2017-08-17 03:05:02 UTC
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In set theory the sequence {1_1}, {1_2}, {1_3}, ... has empty limit with no doubt.
There's plenty of doubt. In fact, given what you've said elsewhere, I'd
say the limit is {1}.
You can learn to calculate the limit here: Set-theoretical limits of sequences of sets, p. 55 in https://www.hs-augsburg.de/~mueckenh/Transfinity/Transfinity/pdf
I already know how to calculate the limit. Since you've said elsewhere
that 1 = 1_1 = 1_2 = 1_3 ..., the limit, if you calculate it correctly,
is {1}.

Martin Shobe
c***@gmail.com
2017-08-13 13:35:03 UTC
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The axiom of infinity (1904) allows to exhaust infinite sets.
No, it doesn't.
See? That's why so many mediocre mathematicians believe in set theory. They haven't a clue of how it works.
FredJeffries
2017-08-12 18:05:25 UTC
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According to set theory, the limit of the sequence {1_1}, {1_2}, {1_3}, ... is empty because "all indices are used up".
No, the indices never get "used up".
Then after indexing infinitely many rational numbers
We all all still waiting for you to "index infinitely many rationals".
Post by c***@gmail.com
there were natural numbers remaining without partner. That contradicts equinumerosity.
No it doesn't, because that's not the meaning of "equinumerosity"
Post by c***@gmail.com
Try to understand how Fraenkel explains this by means of Tristram Shandy.
You are ignorant what you have yourself (mis)quoted.

Fraenkel makes an explicit distinction between "he will never get ready if continuing that way" (one-by-one) and "if he would live infinitely long" which requires something new. A new paradigm -- taking a limit, the axiom of infinity.
Me
2017-08-11 21:44:40 UTC
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This could be done by writing
"1/1", "2/2", etc.
Quotation marks are not compelling.
I didn't claim, that they are, did I?
Post by c***@gmail.com
Post by Me
Still these expressions have to denote certain SETs in the context of set theory.
Hence we might just use the usual terms from set theory as well.
Post by c***@gmail.com
According to set theory, the [set-theoretic] limit of the sequence
({1_1}, {1_2}, {1_3}, ...)

Non-standard notation again? Will you ever learn, Mucke?
Post by c***@gmail.com
is empty because "all indices are used up".
Huh?! Where did you get this nonsense from? Actually, we can't say IF the set-theoretic limit of the sequence ({1_1}, {1_2}, {1_3}, ...) is empty or not before knowing the meaning of 1_n (n e IN).

Let's just consider two "corner cases".

a) IF 1_n = 1_m for all n,m e IN, THEN the set-theoretic limit of THIS sequence is just {1_k} for some k e IN. You might have had 1_n = 1 for all n e IN in mind. In this case ({1_1}, {1_2}, {1_3}, ...) = ({1}, {1}, {1}, ...), and hence the limit is (would be) {1}.

b) IF 1_n =/= 1_m for all n,m e IN, n =/= m, THEN the set-theoretic limit of THIS (different) sequence is just the empty set.

Of course, we know that you are at odds with proper definitions (which would allow to DECIDE such questions), Mucke.
Post by c***@gmail.com
But set teorists overlook the fact that every index is followed by
infinitely many further indices.
Oh really? Look a pink elephant, Mucke!
c***@gmail.com
2017-08-12 17:12:36 UTC
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Let's just consider two "corner cases".
a) IF 1_n = 1_m for all n,m e IN, THEN the set-theoretic limit of THIS sequence is just {1_k} for some k e IN. You might have had 1_n = 1 for all n e IN in mind. In this case ({1_1}, {1_2}, {1_3}, ...) = ({1}, {1}, {1}, ...), and hence the limit is (would be) {1}.
b) IF 1_n =/= 1_m for all n,m e IN, n =/= m, THEN the set-theoretic limit of THIS (different) sequence is just the empty set.
That means the limit depends on what the reader desires to think.
Post by Me
Post by c***@gmail.com
But set teorists overlook the fact that every index is followed by
infinitely many further indices.
Oh really?
Yes, they are really inclined to conclude from every index is used that they all are used up, without realizing that ewvery index is followed by nearly every index. It is a psychological phenomenon. Most set theorists are not stupid but have a severe block in their brains, which in this respect, makes them less intelligent than 10 year old children. https://www.hs-augsburg.de/~mueckenh/Transfinity/The%20little%20demon.pdf
FredJeffries
2017-08-12 17:53:30 UTC
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But set teorists overlook the fact that every index is followed by
infinitely many further indices.
Oh really?
Yes, they are really inclined to conclude from every index is used that they all are used up
No, they never conclude that. That is just a figment of your misleading (mis)usage of the word "all".
Post by c***@gmail.com
, without realizing that ewvery index is followed by nearly every index.
Ewvery [sic][pun intended] set theorist acknowledges that.
Me
2017-08-12 20:00:30 UTC
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Let's just consider two "corner cases".
a) IF 1_n = 1_m for all n,m e IN, THEN the set-theoretic limit of THIS
sequence is just {1_k} for some k e IN. You might have had 1_n = 1 for all
n e IN in mind. In this case ({1_1}, {1_2}, {1_3}, ...) = ({1}, {1}, {1},
...), and hence the limit is (would be) {1}.
b) IF 1_n =/= 1_m for all n,m e IN, n =/= m, THEN the set-theoretic limit
of THIS (different) sequence is just the empty set.
That means the limit depends on what the reader desires to think.
No, idiot. It depends on the meaning of 1_n (for n e IN) in the context we are talking about.

Since you DIDN'T state a proper definition of 1_n (n e IN) we just don't know if a) or b) or something else is the case or not.
c***@gmail.com
2017-08-13 13:33:59 UTC
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Post by Me
Let's just consider two "corner cases".
a) IF 1_n = 1_m for all n,m e IN, THEN the set-theoretic limit of THIS
sequence is just {1_k} for some k e IN. You might have had 1_n = 1 for all
n e IN in mind. In this case ({1_1}, {1_2}, {1_3}, ...) = ({1}, {1}, {1},
...), and hence the limit is (would be) {1}.
b) IF 1_n =/= 1_m for all n,m e IN, n =/= m, THEN the set-theoretic limit
of THIS (different) sequence is just the empty set.
That means the limit depends on what the reader desires to think.
It depends on the meaning of 1_n (for n e IN) in the context we are talking about.
If the sequence 1, 1, 1, ... has limit 1 and the sequence 1_1, 1_2, 1_3, ... has limit empty set, then there is a contradiction because the numbers in the first sequence are implicitly enumerated too.
Post by Me
Since you DIDN'T state a proper definition of 1_n (n e IN) we just don't know if a) or b) or something else is the case or not.
In mathematics we need not know that. The limit is never empty. In set theory you have to distinguish whether indices are written or even only thought. That proves set theory to be out of mathematics.
Me
2017-08-13 15:20:12 UTC
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Since you DIDN'T state a proper definition of 1_n (n e IN) we just don't
know if a) or b) or something else is the case or not.
In mathematics we need not know that.
Pleeze, idiot, don't bullshiting arround and stop calling your nonsense "mathematics".
Me
2017-08-13 22:15:12 UTC
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If the sequence 1, 1, 1, ... has limit 1 and the sequence 1_1, 1_2, 1_3, ...
has limit ...
Again, WE DON'T KNOW the limit of the sequence

(1_1, 1_2, 1_3, ...)

(if there *is* a limit) without knowing the values 1_1, 1_2, 1_3, ..., etc. (That's why I asked for a DEFINITION of 1_n (n e IN).)
Post by c***@gmail.com
Post by Me
Since you DIDN'T state a proper definition of 1_n (n e IN) we just don't
know if a) or b) or something else is the case or not.
Now, for the sake of the argument, LET'S ASSUME that

1_n = 1 for all n e IN .

In this case, we just have

(1, 1, 1, ...) = (1_1, 1_2, 1_3, ...) ,

hence there is just ONE limit for "both" sequences, namely 1.

Correspondingly, in this case

({1}, {1}, {1}, ... ) = ({1_1}, {1_2}, {1_3}, ...) ,

and the set-theoretic limit for "both" sequences is just {1}.
c***@gmail.com
2017-08-14 13:59:12 UTC
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Post by c***@gmail.com
If the sequence 1, 1, 1, ... has limit 1 and the sequence 1_1, 1_2, 1_3, ...
has limit ...
Again, WE DON'T KNOW the limit of the sequence
(1_1, 1_2, 1_3, ...)
In mathematics it has the limit 1 or 1_oo
Post by Me
Correspondingly, in this case
({1}, {1}, {1}, ... ) = ({1_1}, {1_2}, {1_3}, ...) ,
and the set-theoretic limit for "both" sequences is just {1}.
No, that is wrong. The sequence {1_1}, {1_2}, {1_3}, ... has empty limit. The sequence {1}, {1}, {1}, ... has limit {1}.

But if you only think of the indices, then the limit is zero.

Note: If you can exhaust aleph_0 natural numbers, then you can also exhaust aleph_0 digits of the form 1.
Me
2017-08-14 15:46:26 UTC
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Again, WE DON'T KNOW the limit of the sequence
(1_1, 1_2, 1_3, ...)
(if there *is* a limit) without knowing the values 1_1, 1_2, 1_3, ... etc.
(That's why I asked for a DEFINITION of 1_n (n e IN).)
In mathematics it has the limit 1 or 1_oo
*lol* Fuck off, Mucke. You are dumb like shit.

Seriosly, you really should stop talking nonsense.

I alread tried to explain it to you, but you won't listen.

Again: LET'S ASSUME that

1_n = 1 for all n e IN .

In this case, we just have

(1, 1, 1, ...) = (1_1, 1_2, 1_3, ...) ,

hence there is just ONE limit for "both" sequences, namely 1.
Post by c***@gmail.com
Post by Me
Correspondingly, in this case
({1}, {1}, {1}, ... ) = ({1_1}, {1_2}, {1_3}, ...) ,
and the set-theoretic limit for "both" sequences is just {1}.
No, that is wrong.
No, idiot, that's NOT wrong.
Post by c***@gmail.com
The sequence ({1_1}, {1_2}, {1_3}, ...) has empty [set-theoretic] limit.
Look, idiot, IF

1_n = 1 for all n e IN

THEN the sequence

({1_1}, {1_2}, {1_3}, ...)

is identical with the sequence

({1}, {1}, {1}, ... ) ,

HENCE "both" sequences have the same (set-theoretic) limit.
Post by c***@gmail.com
The sequence ({1}, {1}, {1}, ...) has limit [set-theoretic] {1}.
Exactly.
Post by c***@gmail.com
But if you only think of the indices, then the limit is zero.
Yeah, in Mückenmatics MATHEMATICAL FACTS depend on our thoughts!

If you only think hard enough,

1 + 2

becomes

2

in Mückenmathics (after all one term plus one term are two terms).
Me
2017-08-14 16:51:28 UTC
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The sequence
({1}, {1}, {1}, ...)
has [set-theoretic] limit {1}.
Right. Same with the sequence

({1_1}, {1_2}, {1_3}, ...)

(if we have 1_n = 1 for all n e IN).
Post by c***@gmail.com
But if you only think of the indices, then ...
...USE them, *hell*!

You might for example adopt the identity function (on IN):

id(n) = n for all n e IN

and write

Id_n

instead of id(n). In this case you would actualy refer to the "indices", since

Id_n = n for all ne IN.

Then we would get

LIM ({Id_1}, {Id_2}, {Id_3}, ...) = {} .
Me
2017-08-13 22:25:24 UTC
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Let's just consider two "corner cases".
a) IF 1_n = 1_m for all n,m e IN, THEN the set-theoretic limit of THIS
sequence is just {1_k} for some k e IN. You might have had 1_n = 1 for all
n e IN in mind. In this case ({1_1}, {1_2}, {1_3}, ...) = ({1}, {1}, {1},
...), and hence the limit is (would be) {1}.
b) IF 1_n =/= 1_m for all n,m e IN, n =/= m, THEN the set-theoretic limit
of THIS (different) sequence is just the empty set.
That means the limit depends on what the reader desires to think.
No, idiot. It depends on how the writer (author) defines 1_n (n e IN).

Actually, something like, say,

1_n = 1 for all n e IN

would do. See "Case a)".
Post by c***@gmail.com
Post by Me
Post by c***@gmail.com
But set teorists overlook the fact that every index is followed by
infinitely many further indices.
Oh really?
Yes, [...] without realizing that every index is followed by nearly every
index.
Huh?! Actually, there's a rather simple theorem in set theory:

An e IN: card({k e IN : k > n}) = aleph_0 .

Hint: For all n e IN: card({1, ..., n} U {n+1, n+2, n+3, ...}) = card(IN) = aleph_0. On the other hand, for all n e IN: card({1, ..., n}) = n e IN. Hence the number of indices (numbers) following n can't be finite (for each and every n e IN).
Simon Roberts
2017-08-12 20:24:37 UTC
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the sequence of all (unreduced) positive fractions
{1/1}, {1/2}, {2/1}, {1/3}, {2/2}, {3/1}, {1/4}, {2/3}, {3/2},{4/1}, {1/5}, ...
ha[s an] empty limit[.] because [...]
{(1,1)}, {(1,2)}, {(2,1)}, {(1,3)}, ...
Post by c***@gmail.com
[Now] the sequence
{1/1}, {2/2}, {3/3}, ...
Actually, the sequence
{(1,1)}, {(2,2)}, {(3,3)}, ...
Post by c***@gmail.com
[...] has the [set-theoetic] limit empty set too.
Right. Hint: All the elements in this sequence are distinct, i. e. (1,1) =/= (2,2), (1,1) =/= (3,3), (2,2) =/= (3,3), etc
wrong, idiot, hint: it's a SEQUENCE
Post by Me
Post by c***@gmail.com
But has this subsequence, when evaluated,
{1}, {1}, {1}, ...
also the empty set as its limit?
No. The latter sequence has the set-theoretic limit {1}.
Simon Roberts
2017-08-12 20:31:57 UTC
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the sequence of all (unreduced) positive fractions
{1/1}, {1/2}, {2/1}, {1/3}, {2/2}, {3/1}, {1/4}, {2/3}, {3/2},{4/1}, {1/5}, ...
ha[s an] empty limit[.] because [...]
{(1,1)}, {(1,2)}, {(2,1)}, {(1,3)}, ...
Post by c***@gmail.com
[Now] the sequence
{1/1}, {2/2}, {3/3}, ...
Actually, the sequence
{(1,1)}, {(2,2)}, {(3,3)}, ...
Post by c***@gmail.com
[...] has the [set-theoetic] limit empty set too.
Right. Hint: All the elements in this sequence are distinct, i. e. (1,1) =/= (2,2), (1,1) =/= (3,3), (2,2) =/= (3,3), etc
wrong, idiot, hint: it's a SEQUENCE
Hey me, are you confused only because of curly brackets? Because you are so f*cking stupid, I will rewrite his SEQUENCE in terms you may possibly comprehend.

1, 1/2, 2 , 1/3, 1 , 3 , 1/4, 2/3, 3/2, 4/1, 1/5, ...

is this preferable to your brain?
Post by Simon Roberts
Post by Me
Post by c***@gmail.com
But has this subsequence, when evaluated,
{1}, {1}, {1}, ...
also the empty set as its limit?
No. The latter sequence has the set-theoretic limit {1}.
Me
2017-08-12 20:33:35 UTC
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Post by Simon Roberts
Post by Me
Actually, the sequence
{(1,1)}, {(2,2)}, {(3,3)}, ...
Post by c***@gmail.com
[...] has the [set-theoetic] limit empty set too.
Right. Hint: All the elements in this sequence are distinct,
i. e. (1,1) =/= (2,2), (1,1) =/= (3,3), (2,2) =/= (3,3), etc
wrong, idiot, hint: it's a SEQUENCE
Huh?! Didn't you take your pills today, asshole?
Simon Roberts
2017-08-12 20:38:05 UTC
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Post by Simon Roberts
Post by Me
Actually, the sequence
{(1,1)}, {(2,2)}, {(3,3)}, ...
Post by c***@gmail.com
[...] has the [set-theoetic] limit empty set too.
Right. Hint: All the elements in this sequence are distinct,
i. e. (1,1) =/= (2,2), (1,1) =/= (3,3), (2,2) =/= (3,3), etc
wrong, idiot, hint: it's a SEQUENCE
Huh?! Didn't you take your pills today, asshole?
You really are stupid.
Jack Campin
2017-08-09 23:32:21 UTC
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***@gmail.com [like fuck he is] wrote: [who cares]

Muck off, you twat.

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mobile 07895 860 060 <http://www.campin.me.uk> Twitter: JackCampin
Simon Roberts
2017-08-12 19:47:25 UTC
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The sequence of natural numbers
{1}, {2}, {3}, ...
and the sequence of all (unreduced) positive fractions
{1/1}, {1/2}, {2/1}, {1/3}, {2/2}, {3/1}, {1/4}, {2/3}, {3/2},{4/1}, {1/5}, ...
have empty limits because the enumeration of the (unreduced) positive fractions does not leave a remainder.
So the Sequence
{1/1}, {2/2}, {3/3}, ...
as a subsequence of the latter has the limit empty set too.
But has this subsequence, when evaluated,
{1}, {1}, {1}, ...
also the empty set as its limit?
Why? Why so much hatred toward you? I have read many of your posts and never an ill word, phrase, nor insult contained in anything i've read. You have show no abuse or abusiveness, inherently within you, as far as I can see.

Bullshit nasty people. Anyway.

I do not to pretend to understand all that you write and the following may not be relevant but it is reposted here:

Given.
x, y, m, n are integers.
n > |x| > m
n > |y| > m
N = number of distinct values of (x / y).
N is the number of ordered ways to choose x and y divided by M.
Is the derivation of M simple?
ALSO,
Given.
p and q are primes.
s > p > t
s > q > t
P = number of distinct values of (p / q).
P is the number of ordered ways to choose p and q divided by Q.
Is the derivation of Q simple?


IOW

x and y are natural numbers.
How many distinct values of x / y given the range or a range of x and y?
It would be the number of ordered ways to pick x and y divided by some M.
Deriving M may be simple or may be very complicated.
I see symmetry in M at the limit.
c***@gmail.com
2017-08-13 13:38:20 UTC
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Why? Why so much hatred toward you?
They have no mathematical arguments but hate to see their pet being destroyed.
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