Controversial concepts in mathematics

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Re: Controversial concepts in mathematics

Postby Magnus Anderson » Sun Jan 17, 2021 10:58 am

obsrvr524 wrote: \(0 + 0 + 0 + \dots\) - that - is an infinite SERIES. It has a final sum which is the "sum of an infinite series".

Some people might say, "infinite sum" to mean the sum of an infinite series. It's shorter. But it is misleading and technically wrong.


That might be true but it doesn't seem to be a relevant correction (it's more of an effort to help me better express myself.) The point is that what we're dealing with here is an expression involving an infinite addition of terms that exists outside of time rather than inside of time. What I am asking is "What does it mean for such an expression to never stop?" To say that such an expression stops or that it never stops is a figurative use of the word "stop" since the word "stop" can be literally applied only to things that exist in time. Since such an expression can be represented as a sequence of terms (not merely as a set of terms), the question can be reframed as "What does it mean for a sequence to have no end?" That's the question I asked you, right? And it is better to ask Silhouette the same question if it's possible to do so. So that's what I did.

Magnus wrote:Let's say that the word "infinity" represents the number of natural numbers.


Observer wrote:[wrong word but ok]


Actually, the truth value of that statement can be barely contested (: That's because I provided a PROVISIONAL definition -- basically my own -- to be used for the duration of that particular argument I presented. I did NOT claim that's the official definition. (The official definition, I believe, is more general than that.)

No. When you added the "a", you started a new sequence.


We didn't add \(a\) to a different sequence (one that was moreover empty.) Rather, we added it to the existing sequence. Thus, we didn't really start a new one.

"a" does not represent the next higher number


\(a\) is not a number at all, it's a letter. It's a letter that we added to a sequence of natural numbers. The index of the place it occupies in the sequence, however, is a number and that number is basically the number of natural numbers. (The only thing I am not entirely sure about is whether that number is the number of natural numbers OR the number of natural numbers plus one. The reason being that the index of the first place in the sequence is \(1\) rather than \(0\). In fact, I am far more inclined to believe its actual index is infinity plus one.)

By the way, assuming that this can more easily change your mind, James had no problem using infinity as an index. (There was a discussion between him and Carleas on whether or not the set of even natural numbers is in one-to-one correspondence with the set of natural numbers -- or something like that.)
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Re: Controversial concepts in mathematics

Postby Magnus Anderson » Sun Jan 17, 2021 11:55 am

One has to choose the method of indexing.

There are two methods that immediately come to mind:

1) the method where the index of any element is a number that is equal to the number of elements that come before that element

2) the method where the index of any element is a number that is equal to the number of elements that come before that element plus one

If we choose the first method, the index of the first element would be \(0\). The index of \(a\) in \((1, 2, 3, \dotso, a)\) would then be the number of natural numbers (we can denote this with \(\infty\)).

If we choose the second method, the index of the first element would be \(1\). The index of \(a\) in \((1, 2, 3, \dotso, a)\) would then be the number of natural numbers plus one (we can denote this with \(\infty + 1\)).
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Re: Controversial concepts in mathematics

Postby obsrvr524 » Sun Jan 17, 2021 1:54 pm

Magnus Anderson wrote:The point is that what we're dealing with here is an expression involving an infinite addition of terms that exists outside of time rather than inside of time. What I am asking is "What does it mean for such an expression to never stop?" To say that such an expression stops or that it never stops is a figurative use of the word "stop" since the word "stop" can be literally applied only to things that exist in time. Since such an expression can be represented as a sequence of terms (not merely as a set of terms), the question can be reframed as "What does it mean for a sequence to have no end?" That's the question I asked you, right? And it is better to ask Silhouette the same question if it's possible to do so. So that's what I did.

Technically the word "stop" does imply time so a process stops or comes to an end - ends. But if you are talking strictly about the concept, void of time issues, the end merely means a highest number or position in the sequence above which there are none. Obviously that doesn't apply to an infinite set or series because it doesn't have an end point above which there is none.

I don't understand why that is not clear.

Magnus Anderson wrote:
Magnus wrote:Let's say that the word "infinity" represents the number of natural numbers.

Observer wrote:[wrong word but ok]

Actually, the truth value of that statement can be barely contested (: That's because I provided a PROVISIONAL definition -- basically my own -- to be used for the duration of that particular argument I presented. I did NOT claim that's the official definition. (The official definition, I believe, is more general than that.)

And I "barely" did contested it. :D
I said, "but ok".
And if you had said, "For this discussion let's say that infinity refers to the number of naturals" I would have only said "ok". But now we are clear. :D

And since we both know what "infA" means - why not just used that.

Magnus Anderson wrote:
No. When you added the "a", you started a new sequence.

We didn't add \(a\) to a different sequence (one that was moreover empty.) Rather, we added it to the existing sequence. Thus, we didn't really start a new one.

If you are using the 1to1 correspondence, you have to use a different sequence because all of your naturals are used up.
Magnus Anderson wrote:
"a" does not represent the next higher number


\(a\) is not a number at all, it's a letter. It's a letter that we added to a sequence of natural numbers. The index of the place it occupies in the sequence, however, is a number and that number is basically the number of natural numbers. (The only thing I am not entirely sure about is whether that number is the number of natural numbers OR the number of natural numbers plus one. The reason being that the index of the first place in the sequence is \(1\) rather than \(0\). In fact, I am far more inclined to believe its actual index is infinity plus one.)

If you are free to change your index system starting at zero then you can change it to start at -2 and also add "b" and "c". It is the same difference - use a1, a2, a3,... and b1, b2, b3... or use 1, 2, 3,... and -1, -2, -3. It doesn't matter. All I said was that you have to modify your index method - and you did.

Magnus Anderson wrote:By the way, assuming that this can more easily change your mind, James had no problem using infinity as an index. (There was a discussion between him and Carleas on whether or not the set of even natural numbers is in one-to-one correspondence with the set of natural numbers -- or something like that.)

Yes, I read that. He also had a long discussion on a different board concerning hyperreals and that 1=0.999... issue.

I don't have an issue with the infA+1 notation. My issue is that when defining the terms you asked about like "largest number", "infA+1" does not represent a natural number. It is a greater quantity than merely infA but there is no natural number that can represent it. That is why it has its own word.

James specifically stated that you have the freedom to set a chosen standard and must do so if you are going to add or do maths with infinity. I can see that makes sense. And as long as you make it clear that you have, it isn't a problem.

On that other board ("Rational" something) they were telling James that he "doesn't know shit" and that he should just accept what he is told (definitely not James' style - much like Mr Trump in that regard). But then James pointed out that what was told back in 1947 or so by a Hewitt somebody is exactly what he was saying merely in different notation. James was explaining, with many indisputable arguments why he, that Hewitt bloke, and others was right. James was big on the "why" questions - to an extraordinary extreme. If you want to know the "why" behind just about anything - study up on James - maybe get that book Mithus published.

James understood, as do I now, that ALL of these confusions are merely about the words not being carefully defined. Once anyone carefully defines/explains the words all of these issues go away. The same is true concerning laws and politics but no one seems to want to fix that either (you can imagine why).
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Re: Controversial concepts in mathematics

Postby Magnus Anderson » Sun Jan 17, 2021 3:20 pm

obsrvr524 wrote:Technically the word "stop" does imply time so a process stops or comes to an end - ends.


Precisely (:

But if you are talking strictly about the concept, void of time issues, the end merely means a highest number or position in the sequence above which there are none. Obviously that doesn't apply to an infinite set or series because it doesn't have an end point above which there is none.


Right. So when talking about sequences, which do not exist in time, the word "end" refers to an element that comes after all other elements. (It can also be used to refer to an element that comes before all other elements but we can ignore that.) Thus, a sequence without an end (which is what an infinite sequence is) is a sequence that has no element that comes after all other elements. But does that imply that there are no elements OUTSIDE of that sequence? Not really. And does it imply that elements that are outside of that sequence cannot come after all of the elements of that sequence? Again, not really. Thus, if you have two sequences, say \(A = (1, 3, 5, \dotso)\) and \(B = (2, 4, 6, \dotso)\), and you join them conceptually into a super-sequence \(C = (1, 3, 5, \dotso, 2, 4, 6, \dotso)\), then the index of the place occupied by \(2\) would be equal to the number of elements in \(A\) plus one which is half the number of natural numbers plus one which, in James's terms, equals \(\frac{infA}{2} + 1\). If you remove \(1\) from \(C\), the index of \(2\) would then be \(\frac{infA}{2}\). And if you add to it another \(\frac{infA}{2}\) elements of your choice, the index of \(2\) would be \(infA\) i.e. it would be located at "the point of infA". In all three cases, \(2\) is located at "the point of infinity" because in all three cases its index is greater than every integer.

I don't understand why that is not clear.


That's clear and I actually agree with it. What's not clear is how you derive "There is no point at infinity" from it. (Also, what you mean by "point at infinity" is not completely clear.)

And I "barely" did contested it. :D
I said, "but ok".


Actually, you said "wrong word" which means you didn't really contest it. Instead, you merely said that my use of words is wrong. I misread (I thought you said "wrong wrong" which would imply that what I'm saying is wrong.) Nonetheless, I am really only interested in hearing what's wrong with what I am saying. I am not so interested in what's wrong with what I am doing and especially not with how I use my words.

And if you had said, "For this discussion let's say that infinity refers to the number of naturals" I would have only said "ok". But now we are clear. :D


I can understand that (:

And since we both know what "infA" means - why not just used that.


I generally don't use it because it's specifically James's term. As far as I am concerned, I am perfectly comfortable using the word "infinity" to mean two different things. Sometimes, I use it the way most people do, which is as a quantity greater than every integer; sometimes, I use it in a narrow sense to refer to some specific quantity greater than every integer (which is similar, but not necessarily the same, as James's \(infA\), since James's \(infA\) refers specifically to the number of natural numbers.)

But yes, I could have used that term.

If you are using the 1to1 correspondence, you have to use a different sequence because all of your naturals are used up.


True. But I am not using the set of natural numbers to count the resulting set.You can't do that -- the set of natural numbers is smaller. What I'm doing is I am using a set that contains every natural number plus \(infA + 1\).

Yes, I read that. He also had a long discussion on a different board concerning hyperreals and that 1=0.999... issue.

I don't have an issue with the infA+1 notation. My issue is that when defining the terms you asked about like "largest number", "infA+1" does not represent a natural number. It is a greater quantity than merely infA but there is no natural number that can represent it. That is why it has its own word.

James specifically stated that you have the freedom to set a chosen standard and must do so if you are going to add or do maths with infinity. I can see that makes sense. And as long as you make it clear that you have, it isn't a problem.

On that other board ("Rational" something) they were telling James that he "doesn't know shit" and that he should just accept what he is told. But then James pointed out that what was told back in 1947 or so by a Hewitt somebody is exactly what he was saying merely in different notation. James was explaining, with many indisputable arguments why he, that Hewitt bloke and others was right.

James understood, as do I now, that ALL of these confusions are merely about the words not being carefully defined. Once anyone carefully defines/explains the words all of these issues go away. The same is true concerning laws and politics but no one seems to want to fix that either (you can imagine why).


That's pretty much it.
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Re: Controversial concepts in mathematics

Postby obsrvr524 » Sun Jan 17, 2021 3:43 pm

Everything you said seemed right to me until you got to this conclusion -
Magnus Anderson wrote:it would be located at "the point of infA". In all three cases, \(2\) is located at "the point of infinity" because in all three cases its index is greater than every integer.

Neither infA nor infA+1 is a "point" (as in a location on a number line). It is an amount - equal to or greater than infA (the standard infinite set of natural numbers). But it is not a number and it is not a point or location in a sequence. It implies a different sequence than the natural numbers alone. So it cannot be "the highest number" being referenced. That "plus 1" is just another item in an infinite list - no hierarchy can be applied. It is not the highest or lowest. It is just another item beyond our ability to count with numbers.

Magnus Anderson wrote:I generally don't use it because it's specifically James's term. As far as I am concerned, I am perfectly comfortable using the word "infinity" to mean two different things. Sometimes, I use it the way most people do, which is as a quantity greater than every integer; sometimes, I use it in a narrow sense to refer to some specific quantity greater than every integer (which is similar, but not necessarily the same, as James's \(infA\), since James's \(infA\) refers specifically to the number of natural numbers.)

But that ambiguity is what is perpetuating the confusion. I don't think James held any ownership to the term "infA" as long as anyone understands it. And it allows for an "infB" or "infC" to maintain distinctions while talking about varied infinities.

If you are not helping - you are probably hurting - or maybe just wasting time and effort. :D
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Re: Controversial concepts in mathematics

Postby Magnus Anderson » Sun Jan 17, 2021 5:13 pm

"Point" is another word for "position" or "place". Thus, "point at \(infA\)" merely means "place whose index is \(infA\)". In the case of \((1, 2, 3, \dotso, a)\), \(a\) is located at the point of \(infA + 1\) which means no more than that it occupies place whose index is \(infA + 1\). Thus, if you agree that it occupies place whose index is \(infA + 1\), you also have to agree that it is located at the point of \(infA + 1\). (And if it occupies place at \(infA + 1\), you will also have to agree that it occupies place at infinity since the word "infinity" means "a number greater than every integer" and that's precisely what \(infA + 1\) is.)

Also note that we started with \((1, 2, 3, \dotso)\) which is an infinite sequence that has no last element (= an element that comes after all other elements = an element with the highest index.) But once we add \(a\) to it, and add it after all of its elements, we get \((1, 2, 3, \dotso, a)\) which is an infinite sequence with an end -- it has the last element. \(a\) is now an element with an index that is greater than the index of every other element in the sequence, its index being \(infA + 1\).

That's why it's inaccurate to say that an infinite sequence is a sequence that has no last element. Indeed, it's not even accurate to say that an infinite sequence is a sequence that has no last element and/or no first element. (What's true is that every sequence that has no last element and/or no first element is an infinite one. But a sequence that has both is not necessarily finite.)

The proper definition of the term "infinite sequence" is a sequence whose elements can be put in one-to-one correspondence with one of the infinite sets. (And a set is said to be infinite if its number of elements is greater than every integer. Note that sets have no notion of "end".)
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Re: Controversial concepts in mathematics

Postby Magnus Anderson » Sun Jan 17, 2021 5:28 pm

obsrvr524 wrote:But that ambiguity is what is perpetuating the confusion. I don't think James held any ownership to the term "infA" as long as anyone understands it. And it allows for an "infB" or "infC" to maintain distinctions while talking about varied infinities.

If you are not helping - you are probably hurting - or maybe just wasting time and effort. :D


You may be right that my use of words isn't optimal. The thing is that that's a separate discussion that I am moreover not interested in at the present time.

And while I do welcome feedback of the form "I don't understand what you mean by X", I do not welcome criticism of how I use words nor an advice on how to use them. (Leave these things up to me, please (:)
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Re: Controversial concepts in mathematics

Postby obsrvr524 » Sun Jan 17, 2021 5:48 pm

Magnus Anderson wrote:"Point" is another word for "position" or "place". Thus, "point at \(infA\)" merely means "place whose index is \(infA\)".

But there is no such place. infA is a quality of a set. It is NOT a point at the end of a sequence. There is no "end" to the sequence - no boundary for the set. You cannot ever be at the edge, boundary, or end - no such things exists. So there is no "one position or point beyond it". There is only an addition to the entire infinite set that is not within the set or at an edge of the set or "after" the set but merely appended to it as a new set - infA and also 1 along side or in combination.

Magnus Anderson wrote:That's why it's inaccurate to say that an infinite sequence is a sequence that has no last element. Indeed, it's not even accurate to say that an infinite sequence is a sequence that has no last element and/or no first element. (What's true is that every sequence that has no last element and/or no first element is an infinite one. But a sequence that has both is not necessarily finite.)

Ok. I see what you are saying now. I'm not sure that I erred in that way, but we can assume that back there somewhere I did.

Magnus Anderson wrote:The proper definition of the term "infinite sequence" is a sequence whose elements can be put in one-to-one correspondence with one of the infinite sets. (And a set is said to be infinite if its number of elements is greater than every integer. Note that sets have no notion of "end".)

I can't accept that for similar reason as you just pointed out concerning an infinite sequence not necessarily not having end points. And "infinite" doesn't mean "greater than every integer". The infinite set of every even integer certainly isn't greater than the set of all integers - or "greater than every integer".

A sequence is a progressing pattern. It has a direction of increasing or decreasing. Each element in the pattern might have a 1to1 correspondence with another set. But that is not what defines it as a sequence (infinite or otherwise). And if you add something that is not part of that sequence, it is not appended to "the end". It is merely a part of the total collection. It is not before or after anything because it is not part of any sequence.

Magnus Anderson wrote:And while I do welcome feedback of the form "I don't understand what you mean by X", I do not welcome criticism of how I use words nor an advice on how to use them. (Leave these things up to me, please (:)

That can lead to kickback.
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Re: Controversial concepts in mathematics

Postby polishyouthgotipbanned » Sun Jan 17, 2021 5:49 pm

are you two seriously still getting excited about high-school mathematical induction??????this is what a sectarian does...takes a bit of information out of context out of bible or mathematics or whatever and strings his bizarre theories off it.
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Re: Controversial concepts in mathematics

Postby obsrvr524 » Sun Jan 17, 2021 5:51 pm

polishyouthgotipbanned wrote:are you two seriously still getting excited about high-school mathematical induction??????this is what a sectarian does...takes a bit of information out of context out of bible or mathematics or whatever and strings his bizarre theories off it.

What's it to you?
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Re: Controversial concepts in mathematics

Postby polishyouthgotipbanned » Sun Jan 17, 2021 5:53 pm

its funny watching two insane goons falling over in a well lit room telling eachother they are the only ones that can see. reminds me of my neo-nazi friends.
Kvasirs Mexican gay lover(writing about a need of committing a genocide on 99 percent whites and all non-whites as a Mexican mestizo himself):
And if they are not good, what could be wrong with the genocidal fantasy in my ‘Dies Irae’, published in Day of Wrath, with a vindictive Star Child calling home 500 million Caucasoids (and of course, all non-whites, including Jews) to, ironically, make sure that Dave Lane’s words be fulfilled with the remaining Aryans?.
My step-dads schizophrenic diagnosis:
I see a disease taking over....I will not stay silent. I will do what I can, when I can.
If we do not stand then we shall fall....and the enemy will win.

Remember: the world will end, all there is that is left is to fly between Canada and Greece and have barbecues and drink vodka before you wake up at 40 and remember to have a son who will be 30 when you are 70 and whos children will be 20 when you are 90. paGAYn as fuck.
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Re: Controversial concepts in mathematics

Postby obsrvr524 » Sun Jan 17, 2021 6:17 pm

polishyouthgotipbanned wrote:its funny watching two insane goons falling over in a well lit room telling eachother they are the only ones that can see. reminds me of my neo-nazi friends.

What is more funny is that neither of us was telling eachother that we are the only ones who can see. You seem to be the one doing that. All wise and perceptive are you? - Now THAT would be funny. :D

And from what I have seen of many of your posts, "high-school" would be a step up for you.
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Re: Controversial concepts in mathematics

Postby polishyouthgotipbanned » Sun Jan 17, 2021 6:20 pm

insanity is not a joke...and you are clearly a kook. take care of yourself.
Kvasirs Mexican gay lover(writing about a need of committing a genocide on 99 percent whites and all non-whites as a Mexican mestizo himself):
And if they are not good, what could be wrong with the genocidal fantasy in my ‘Dies Irae’, published in Day of Wrath, with a vindictive Star Child calling home 500 million Caucasoids (and of course, all non-whites, including Jews) to, ironically, make sure that Dave Lane’s words be fulfilled with the remaining Aryans?.
My step-dads schizophrenic diagnosis:
I see a disease taking over....I will not stay silent. I will do what I can, when I can.
If we do not stand then we shall fall....and the enemy will win.

Remember: the world will end, all there is that is left is to fly between Canada and Greece and have barbecues and drink vodka before you wake up at 40 and remember to have a son who will be 30 when you are 70 and whos children will be 20 when you are 90. paGAYn as fuck.
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Re: Controversial concepts in mathematics

Postby Magnus Anderson » Mon Jan 18, 2021 3:09 am

obsrvr524 wrote:But there is no such place.


But I thought that you already agreed that there is such a place, if not explicitly, then at least implicitly (: In the case of \((1, 2, 3, \dotso, a\)), that place is the one occupied by \(a\). Didn't you agree that the index of that place is \(infA + 1\)? If so, that's a "point at \(infA\)".

Let me restate the entire logic:

1. "Point" is another word for "position" or "place"

2. Thus, "point at infinity" is another expression for "place at infinity"

3. "Place at X" is another expression for "place whose index is X"

4. Thus, "place at infinity" is another expression for "place whose index is infinity"

5. \(infA\) is an instance of infinity

6. Thus, "place whose index is \(infA\)" is also "place whose index is infinity"

7. The index of \(a\) in \((1, 2, 3, \dotso, a\)) is \(infA + 1\)

8. Thus, \(a\) occupies "place whose index is \(infA + 1\)" which means "point at \(infA + 1\)" which means "point at infinity"

Please let me know what you disagree with.

infA is a quality of a set.


I don't think that "quality" is an appropriate way to describe what \(infA\) is. You can't even say that \(infA\) specifies that a set has no end because sets have no notion of "end". ("Endless set" is an instance of figurative speech.) What \(infA\) does is it specifies how many elements there are in a set. Specifically, what it does is it states that the number of elements in a set is equal to the number of natural numbers.

It is NOT a point at the end of a sequence.


It is an index of the place (which is a point) occupied by \(a\) in the case of \((2, 3, 4, \dotso, a)\).

There is no "end" to the sequence - no boundary for the set. You cannot ever be at the edge, boundary, or end - no such things exists.


I am not sure I understand what you're saying here. Are you saying that there is no "end" to the sequence that is \((1, 2, 3, \dotso, a)\)? But there is. Remember how we defined the word "end"? It refers to a place that comes after all other places i.e. to a place with the highest index. And that place in the case of \((1, 2, 3, \dotso, a)\) is the one occupied by \(a\).

So there is no "one position or point beyond it".


If this is true then \((1, 2, 3, \dotso, a)\) is an invalid sequence. Do you agree with that?

There is only an addition to the entire infinite set that is not within the set or at an edge of the set or "after" the set but merely appended to it as a new set - infA and also 1 along side or in combination.


I didn't add \(a\) to a new sequence. That's NOT the operation I performed. I added it to THE SAME sequence and I added it AFTER all of its elements. You are insisting that I performed an operation that I did not actually perform. (It's akin to saying I multiplied \(1\) by \(0\) when in fact I divided it by \(0\).) You may want to argue that I'm performing an impossible operation instead (which is false but which is at least based on what I really did rather than something that I didn't do.)

And note that when we say that a sequence has no end, we're merely saying that it has no place with the highest index. This means we can't add an element at the end of that sequence (which means we can't add it to the place with the highest index -- because there is no such place) which does not mean we can't add it AFTER all of its elements (thereby creating a place with the highest index.)

And "infinite" doesn't mean "greater than every integer".


You're probably aware of the fact that I believe that to be the best definition of the word "infinite" out there -- better than "without an end". The reason being very simple: sets have no notion of "end" and a sequence can be infinite even if it has a beginning and an end.

The infinite set of every even integer certainly isn't greater than the set of all integers - or "greater than every integer".


The set of even integers is NOT greater than the set of integers. Nonetheless, the number of even integers is CERTAINLY greater than every integer. Are you saying that the number of even integers is an integer? If so, which one?

A sequence is a progressing pattern. It has a direction of increasing or decreasing. Each element in the pattern might have a 1to1 correspondence with another set. But that is not what defines it as a sequence (infinite or otherwise).


"Progressing pattern" is a figurative description of what a sequence is. It's not a strict mathematical definition. And since sequences do not exist in time, they are neither increasing nor decreasing.

And if you add something that is not part of that sequence, it is not appended to "the end". It is merely a part of the total collection. It is not before or after anything because it is not part of any sequence.


You choose where you're going to insert it. You can insert it between two existing places e.g. you can insert it between the first and second element (thereby becoming second element.) But you can also insert it between one non-existent and one existent place e.g. you can add it between the zero-eth place (which doesn't exist) and first place (thereby becoming first element.) Finally, you can insert it after all other elements (thereby becoming the last element.)

When adding \(a\) to \((1, 2, 3, \dotso)\), I added it after all of its elements. The result is \((1, 2, 3, \dotso, a)\).

And note that I added it to THE SAME sequence. I didn't start a new one. That's NOT the operation I performed.

You may want to argue such an operation is an impossible one rather than saying I added it somewhere I didn't.
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Re: Controversial concepts in mathematics

Postby obsrvr524 » Mon Jan 18, 2021 5:34 am

Magnus Anderson wrote:
obsrvr524 wrote:But there is no such place.


But I thought that you already agreed that there is such a place, if not explicitly, then at least implicitly (: In the case of \((1, 2, 3, \dotso, a\)), that place is the one occupied by \(a\). Didn't you agree that the index of that place is \(infA + 1\)? If so, that's a "point at \(infA\)".

No that isn't what I agreed to. You have changed what I said. That extra "a" that has been included into the set is NOT "after" or "before" the sequence. It can be indexed as the "infA+1" item. But there is no "infA point", position, or place. The infA set has no upper limit. The total relative volume is described as infA - but that is not an index to "the last element" - because there is no last element in that set. infA is larger or smaller than other infinite sets.

Magnus Anderson wrote:Let me restate the entire logic:

1. "Point" is another word for "position" or "place" - ok

2. Thus, "point at infinity" is another expression for "place at infinity" - not ok - there is no place "at infinity"

3. "Place at X" is another expression for "place whose index is X" - ok

4. Thus, "place at infinity" is another expression for "place whose index is infinity" - infA is not an index at infinity. It is an infinite quantity - not a pointer to locations. X cannot be "infinity"In maths nothing goes TO infinity. It only goes TOWARD infinity.


Magnus Anderson wrote:
infA is a quality of a set.


I don't think that "quality" is an appropriate way to describe what \(infA\) is.

You're right. I was thinking of merely an infinite set. InfA is a specific set, so ok now it specifies a relative quantity (actually I think "volume" is a better word. James used "degree").

Magnus Anderson wrote:You can't even say that \(infA\) specifies that a set has no end because sets have no notion of "end". ("Endless set" is an instance of figurative speech.)

Sets have boundaries. Those boundaries are commonly referred to as the set's "ends" - initial and final. But in the case of an infinite set with no end point, the boundary is denoted as "no more than infinite". The boundary is qualitative rather than quantitative (until the set is exactly specified). Once exactly defined or specified the quantity of the set is that of being "endless" - no end - no index pointing to an end - merely a relative amount compared to some other infinite set that also has no end. It is really just a relative size or volume rather than a quantitative amount.

Does that clear anything up? - probably not :-?
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Re: Controversial concepts in mathematics

Postby Magnus Anderson » Mon Jan 18, 2021 6:09 am

Observer wrote:No that isn't what I agreed to.


You did agree that the index of \(a\) in \((1, 2, 3, \dotso, a)\) is \(infA + 1\).

Here:
https://www.ilovephilosophy.com/viewtop ... 5#p2800918

You said "Everything seemed right until" and this "everything" includes me saying that the index of \(a\) is \(infA + 1\).

(And if you say that the index of \(a\) is \(infA + 1\), then you're also saying that there's a place -- which is another word for point -- whose index is \(infA + 1\).)

Magnus wrote:2. Thus, "point at infinity" is another expression for "place at infinity"


Observer wrote:not ok - there is no place "at infinity"


That statement does not state there's such a thing as "point at infinity". It merely states that "point at infinity" and "place at infinity" are two different expressions that have one and the same meaning. And that holds true even if the word "infinity" has no meaning assigned to it. ("Point at sldfhs" and "place at sldfhs" mean the same thing even though "sldfhs" means literally nothing.)

Magnus wrote:4. Thus, "place at infinity" is another expression for "place whose index is infinity"


infA is not an index at infinity. It is an infinite quantity - not a pointer to locations. X cannot be "infinity". In maths nothing goes TO infinity. It only goes TOWARD infinity.


That has nothing to do with that statement. The statement merely states that "place at infinity" means the same as "place whose index is infinity".

Sets have boundaries. Those boundaries are commonly referred to as the set's "ends" - initial and final.


They do not. If they do, what and where exactly are these boundaries? Perhaps by "initial end" you mean "first element" and by "final end" you mean "last element"? But sets are NOT ordered. There is no such things as "first element" and "last element" in a set. The finite set that is \(A = \{1, 2, 3\}\) has no first element, no second element, no third element and no last element. It's not ordered.

Does that clear anything up? - probably not :-?


Not really.
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Re: Controversial concepts in mathematics

Postby obsrvr524 » Mon Jan 18, 2021 12:15 pm

Magnus Anderson wrote:
Observer wrote:No that isn't what I agreed to.


You did agree that the index of \(a\) in \((1, 2, 3, \dotso, a)\) is \(infA + 1\).

Here:
https://www.ilovephilosophy.com/viewtop ... 5#p2800918

You said "Everything seemed right until" and this "everything" includes me saying that the index of \(a\) is \(infA + 1\).

(And if you say that the index of \(a\) is \(infA + 1\), then you're also saying that there's a place -- which is another word for point -- whose index is \(infA + 1\).)

I have no issue with an index of "infA+1". I have issue with an index of "infA".

The "+1" indicates that the indexed address is not within the subset indicated by infA. That makes it easy to locate. But if you said your index was pointing to \(infA - 1\) - you have a problem. There is no location at \(infA - 1\) (or minus 10, 200, 5 million or any other number). You can't use that index to see what is at the location - because there is no such location - invalid index.

Magnus Anderson wrote:
Magnus wrote:2. Thus, "point at infinity" is another expression for "place at infinity"
Observer wrote:not ok - there is no place "at infinity"

That statement does not state there's such a thing as "point at infinity". It merely states that "point at infinity" and "place at infinity" are two different expressions that have one and the same meaning. And that holds true even if the word "infinity" has no meaning assigned to it. ("Point at sldfhs" and "place at sldfhs" mean the same thing even though "sldfhs" means literally nothing.)

Yes but I didn't say that the statement was wrong. I said that it wasn't "ok" - for the reason mentioned above.

Magnus Anderson wrote:
Sets have boundaries. Those boundaries are commonly referred to as the set's "ends" - initial and final.

They do not. If they do, what and where exactly are these boundaries?

Of course they do. In general the boundary is set by the category definition. Anything outside the set/category definition is outside the boundary of the meaning of the set.

When a set is a finite sequence there is an initial point and a final point. If the set is an infinite sequence there is no final point although there might be an initial point. If the set is not a sequence then it is bound only by its categorical definition.
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Re: Controversial concepts in mathematics

Postby Magnus Anderson » Mon Jan 18, 2021 4:24 pm

Observer wrote:Of course they do. In general the boundary is set by the category definition. Anything outside the set/category definition is outside the boundary of the meaning of the set.


It seems like you're saying that a set is said to be bounded if it is known which elements belong to it and which elements do not belong to it. If that's the case, then that means the set of natural numbers is bounded because we known which elements belong to it (e.g. \(1\), \(2\), \(3\) and so on) and which don't (e.g. \(-1\), \(-2\), \(-3\) and so on.) Indeed, it would be impossible to find a set that is not bounded in this sense of the word simply because sets are by definition that way.

When a set is a finite sequence there is an initial point and a final point.


What is the initial point of the set of natural numbers?

If the set is an infinite sequence there is no final point although there might be an initial point. If the set is not a sequence then it is bound only by its categorical definition.


I think you are confusing sets with sequences. They are two different things. No set is a sequence. The set of natural numbers is not a sequence.

Yes but I didn't say that the statement was wrong. I said that it wasn't "ok" - for the reason mentioned above.


Alright. The thing is I am mostly interested in whether you agree or disagree with those statements. "Okay" and "not okay" are of little importance to me if they do not stand for "agree" and "disagree".

I have no issue with an index of "infA+1". I have issue with an index of "infA".

The "+1" indicates that the indexed address is not within the subset indicated by infA. That makes it easy to locate. But if you said your index was pointing to \(infA - 1\) - you have a problem. There is no location at \(infA - 1\) (or minus 10, 200, 5 million or any other number). You can't use that index to see what is at the location - because there is no such location - invalid index.


You said that you agreed with everything I said before I said "it would be located at the point of infA". And what I said immediately before I said that is "the index of \(2\) would be \(infA\)". So of course, I thought you agreed with that. But perhaps I misunderstood.

I don't know what you mean by "the subset indicated by infA". I guess I'll need your help.

In the case of \((1, 2, 3, \dotso, a)\), there is no place with an index of \(infA\). But that does not mean there are no such places in general. For example, the index of \(a\) in \((2, 3, 4, \dotso, a)\) is \(infA\) quite simply because 1) the index of a place is defined as the number of places that come before that place plus one, and 2) the number of places that come before \(a\) in that sequence is equal to the number of natural numbers minus one -- \(infA - 1\) -- so its index is \(infA - 1 + 1\) or \(infA\).
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Re: Controversial concepts in mathematics

Postby obsrvr524 » Mon Jan 18, 2021 8:29 pm

Magnus Anderson wrote:
Observer wrote:Of course they do. In general the boundary is set by the category definition. Anything outside the set/category definition is outside the boundary of the meaning of the set.


It seems like you're saying that a set is said to be bounded if it is known which elements belong to it and which elements do not belong to it. If that's the case, then that means the set of natural numbers is bounded because we known which elements belong to it (e.g. \(1\), \(2\), \(3\) and so on) and which don't (e.g. \(-1\), \(-2\), \(-3\) and so on.) Indeed, it would be impossible to find a set that is not bounded in this sense of the word simply because sets are by definition that way.

I meant "bound" in the general use of the word, not "said to be bounded" in the tunnel minded technical geek sense. Call it "borders" if you prefer. The series of natural numbers do NOT include the series of letters because the natural numbers exclude everything outside the bounds (borders/category) of numbers.

Magnus Anderson wrote:
When a set is a finite sequence there is an initial point and a final point.

What is the initial point of the set of natural numbers?

When did the natural numbers become a finite sequence/set?

Magnus Anderson wrote:
If the set is an infinite sequence there is no final point although there might be an initial point. If the set is not a sequence then it is bound only by its categorical definition.

I think you are confusing sets with sequences. They are two different things. No set is a sequence. The set of natural numbers is not a sequence.

Where did you get that idea? Every sequence is a set but not every set is a sequence. I'm not a set theorist so is there some technical geeky definition of "set" I don't know about that excludes defined progressive series - sequences?

Magnus Anderson wrote:
Yes but I didn't say that the statement was wrong. I said that it wasn't "ok" - for the reason mentioned above.

Alright. The thing is I am mostly interested in whether you agree or disagree with those statements. "Okay" and "not okay" are of little importance to me if they do not stand for "agree" and "disagree".

Well ambiguity is important to me so I can't commit to agreeing to a statement if it isn't clear what it really means. That doesn't mean that I disagree. I just stop right there until the ambiguity is removed.

Magnus Anderson wrote:I don't know what you mean by "the subset indicated by infA". I guess I'll need your help.

Of course you do. You specified it as infA = natural number set. Then you added into the set the letter "a" making a subset of the infA of natural numbers and a subset of the letter "a".

Magnus Anderson wrote:In the case of \((1, 2, 3, \dotso, a)\), there is no place with an index of \(infA\).

Well there you go.

Magnus Anderson wrote:But that does not mean there are no such places in general. For example, the index of \(a\) in \((2, 3, 4, \dotso, a)\) is \(infA\) quite simply because 1) the index of a place is defined as the number of places that come before that place plus one, and 2) the number of places that come before \(a\) in that sequence is equal to the number of natural numbers minus one -- \(infA - 1\) -- so its index is \(infA - 1 + 1\) or \(infA\).

If you change the definition of infA to exclude the number 1, of course issues will change. But then you are stuck with your new \(infA-1\) having no index location.
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Re: Controversial concepts in mathematics

Postby Silhouette » Tue Jan 19, 2021 1:33 am

Magnus Anderson wrote:Alright, so you think that the symbol of infinity represents a function of some sort. But what kind of function? Any function or some specific function? If it represents some specific function, what kind of specific function? How many arguments does it have, of what type and what kind of value does it output?

Specifically the function that amounts to "keep doing this".
Like addition and subtraction, by itself this means nothing.
Much like you can't merely "1-" without a finite amount to subtract from one (subtraction has arity of two), there's only certain forms where the finite denotation of "infinity" can validly apply. Sum and product are the most common, and as another example it's implied by "the indefinite" integral so doesn't have to be specified but its application would be valid if you did etc. If infinite sums and products followed the same conventions as indefinite integrals, it couldn't need to be specified there either - and I'd argue that not specifying (through some finite symbol) would actually aid in its communication. In these examples, its arity is one, accepting the arguments of sum, product, integral (all functions themselves). This is what I meant by what you kindly linked: "Just to wrap up what is meant when infinity is said to be an operator - it is only validly used in mathematics as part of an operation, more like an operator on an operator. This doesn't mean it's interchangeable with operators like addition and multiplication etc."

As such its output depends on what arguments it accepts, which can either be convergent to an exact finite limit that could not be any other value, or divergent i.e. "I don't/can't know" or "indetermined/indeterminable". I've already answered this but you asked again, so...

Magnus Anderson wrote:I think the word "infinity" means the same thing in both expressions: \(x = \infty\) and \(\sum_{i=1}^{\infty}a_i\). And in both cases it does not mean "I don't know". If \(x = \infty\) means "I don't know what's the value of \(x\)" then that means that \(x\) can be equal to \(5\) and that's certainly not what it means.

Actually in many cases where infinity is involved, you can manipulate the answer to being any number. It's how you can infamously pretend "1=2", and if you can make that true, you'll soon realise you can make anything true, whether "5" or not. This is genuinely what infinity amounts to when you try to use it as an operand.
It's only when you use it as an operation, as above, that it can be used meaningfully - and then only really when its use converges to an exact finite limit that could not be any other value.

Magnus Anderson wrote:
Silhouette wrote:As for infinity specifying "the point where the summation stops", that's a clear contradiction in terms.

I will have to ask you a question similar to the one I asked obsrvr524.

What does the word "stop" mean with respect to infinite sums such as \(a_1 + a_2 + a_3 + \dotso\)?

I mean quite obviously it means the chain of addition ends after a finite number of terms has been reached. I don't mean that you have to time the expression - take as long as you like or do it instantly. The exact finite limit that could not be any other value doesn't care.
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Re: Controversial concepts in mathematics

Postby Magnus Anderson » Tue Jan 19, 2021 3:01 am

obsrvr524 wrote:I meant "bound" in the general use of the word, not "said to be bounded" in the tunnel minded technical geek sense. Call it "borders" if you prefer.


I think I understood it well enough.

The series of natural numbers do NOT include the series of letters because the natural numbers exclude everything outside the bounds (borders/category) of numbers.


The set of natural numbers does not include letters because the set of natural numbers excludes everything outside the bounds of the set of natural numbers (or in plain terms, everything that does not belong to the set of natural numbers.)

Note that the set of natural numbers is NOT the set of ALL numbers. (You seem to imply this.) There are numbers that are not natural numbers e.g. negative numbers, rational numbers such as \(\frac{1}{2}\) and so on.

The point that I am making (and that you're not addresing) is that there is NO set that is NOT bounded in your sense of the word. The set of natural numbers, even though infinite/endless/boundless, is bounded in your sense of the word. Thus, you did NOT really explain what it means for a set to have a kind of end that finite sets do but infinite sets don't. You merely discovered and explained an irrelevant meaning of the word "boundary".

When did the natural numbers become a finite sequence/set?


Alright, so the set of natural numbers does not have "the initial point". What about the finite set that is \(A = \{1, 2, 3\}\)? What is "the initial point" of \(A\)?

Where did you get that idea? Every sequence is a set but not every set is a sequence. I'm not a set theorist so is there some technical geeky definition of "set" I don't know about that excludes defined progressive series - sequences?


When you say "technical geeky definition", you probably mean "mathematical definition". You somehow forgot this thread is about mathematics.

The main difference between sets and sequences is that sets have no order whereas sequences do. Thus, they are DIFFERENT concepts.

See Wikipedia:

Wikipedia wrote:In mathematics, a sequence is an enumerated collection of objects in which repetitions are allowed and order matters. Like a set, it contains members (also called elements, or terms). [..] Unlike a set, the same elements can appear multiple times at different positions in a sequence, and unlike a set, the order does matter.


Observer wrote:Well ambiguity is important to me so I can't commit to agreeing to a statement if it isn't clear what it really means. That doesn't mean that I disagree. I just stop right there until the ambiguity is removed.


I can understand that. In such a case, I expect people to ask me for clarification. That's what I do when I find it difficult to interpret others.

Of course you do. You specified it as infA = natural number set. Then you added into the set the letter "a" making a subset of the infA of natural numbers and a subset of the letter "a".


\(infA\) does not represent the set of natural numbers. Rather, it represents the number of natural numbers (or the size of the set of natural numbers.)

But yes, I did add \(a\) to \((1, 2, 3, \dotso)\) and by doing so ended up with \((1, 2, 3, \dotso, a)\) which can be thought of as consisting of two subsequences, one being the original \((1, 2, 3, \dotso)\) and the other being \((a)\). So perhaps what you mean by "the subset indicated by \(infA\)" is the subsequence \((1, 2, 3, \dotso)\). And yes, within that subsequene, the index of every place is an integer i.e. there are no places whose index is \(infA\), \(infA + 1\) or \(infA - 1\).

Magnus Anderson wrote:But that does not mean there are no such places in general. For example, the index of \(a\) in \((2, 3, 4, \dotso, a)\) is \(infA\) quite simply because 1) the index of a place is defined as the number of places that come before that place plus one, and 2) the number of places that come before \(a\) in that sequence is equal to the number of natural numbers minus one -- \(infA - 1\) -- so its index is \(infA - 1 + 1\) or \(infA\).


If you change the definition of infA to exclude the number 1, of course issues will change. But then you are stuck with your new \(infA-1\) having no index location.


I did not change the definition of \(infA\). The definition of \(infA\) is the same as always. It refers to the number of natural numbers or the size of the set of natural numbers. Rather, I changed the sequence. I went from \((1, 2, 3, \dotso, a)\) to \((2, 3, 4, \dotso, a)\).

And yes, in the case of \((2, 3, 4, \dotso, a)\) there is no place with an index of \(infA - 1\). What's the problem with that?
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Re: Controversial concepts in mathematics

Postby Magnus Anderson » Tue Jan 19, 2021 4:23 am

Silhouette wrote:Specifically the function that amounts to "keep doing this".
Like addition and subtraction, by itself this means nothing.


"Keep doing this" on its own does not tell me much. But I can make a guess anyway. Perhaps what you mean is a function that maps the set of natural numbers to a set that includes no more than \(true\). Such a function can be then used to tell us whether we should sum the next number in the sequence or not. Of course, being the kind of function that it is (constant function), it will always tell us to sum it.

What that implies is that \(\infty 5\) and \(\infty(5)\) are equal to \(true\). But earlier you said that \(\infty 5\) is not a valid expression (suggesting either that the arity of infinity isn't one or that the only argument it accepts is not an integer.)

But let me examine the rest of your post and see if this is truly the case.

Much like you can't merely "1-" without a finite amount to subtract from one (subtraction has arity of two), there's only certain forms where the finite denotation of "infinity" can validly apply. Sum and product are the most common, and as another example it's implied by "the indefinite" integral so doesn't have to be specified but its application would be valid if you did etc. If infinite sums and products followed the same conventions as indefinite integrals, it couldn't need to be specified there either - and I'd argue that not specifying (through some finite symbol) would actually aid in its communication.In these examples, its arity is one, accepting the arguments of sum, product, integral (all functions themselves). This is what I meant by what you kindly linked: "Just to wrap up what is meant when infinity is said to be an operator - it is only validly used in mathematics as part of an operation, more like an operator on an operator. This doesn't mean it's interchangeable with operators like addition and multiplication etc."


I strikethroughed everything I consider to have little to nothing to do with the question that I asked.

What this tells me is that in some cases the arity of the function that is infinity is one i.e. that it accepts exactly one argument.

"In some cases" means that that may not be always the case. In other words, in some cases, the arity of infinity might be other than one. If that' happens to be the case, i.e. if there are cases when its arity isn't one, what that means is that infinity is either a word that has multiple meanings or a word denoting a category of functions rather than some specific function.

As such its output depends on what arguments it accepts, which can either be convergent to an exact finite limit that could not be any other value, or divergent i.e. "I don't/can't know" or "indetermined/indeterminable". I've already answered this but you asked again, so...


Does that mean that the codomain of the function that is infinity is the set of real numbers plus "I don't know"?

Actually in many cases where infinity is involved, you can manipulate the answer to being any number. It's how you can infamously pretend "1=2", and if you can make that true, you'll soon realise you can make anything true, whether "5" or not. This is genuinely what infinity amounts to when you try to use it as an operand.
It's only when you use it as an operation, as above, that it can be used meaningfully - and then only really when its use converges to an exact finite limit that could not be any other value.


I don't think that to be the case. And you probably know that.

If you want to discuss this, you are welcome to present an argument that involves infinity and that concludes that \(1 = 2\).

Here's one:

\(\infty + \infty = \infty\) // let's divide both sides by \(\infty\)
\(\frac{\infty}{\infty} + \frac{\infty}{\infty} = \frac{\infty}{\infty}\)
\(1 + 1 = 1\)
\(2 = 1\)

My position is that this is a fallacious proof. (Which means you CAN'T use infinity to prove that \(1 = 2\).)

I mean quite obviously it means the chain of addition ends after a finite number of terms has been reached. I don't mean that you have to time the expression - take as long as you like or do it instantly. The exact finite limit that could not be any other value doesn't care.


That's not an answer to the question.

And by the way, one should never say something like "It's obvious" when asked to explain something (:
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Re: Controversial concepts in mathematics

Postby obsrvr524 » Tue Jan 19, 2021 5:56 am

_
The symbol for "keep doing this" is - "\(\dotso\)"
Infinity, \(\infty \) , is not a function or process.

Magnus Anderson wrote:And by the way, one should never say something like "It's obvious" when asked to explain something (:

I thought you didn't like people correcting the words you use. :D
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Re: Controversial concepts in mathematics

Postby Magnus Anderson » Tue Jan 19, 2021 6:58 am

obsrvr524 wrote:I thought you didn't like people correcting the words you use. :D


I didn't criticize his use words as much as I criticized how he treats other people. But you're nonetheless right that I disobeyed my own law (: It happens.

The mistake I made is that I gave everyone a general advice (I told everyone how to treat EVERYONE or MOST people) instead of telling Silhouette that he's doing something that I do not find acceptable (something that increases the chances, even if slightly, of losing me as an interlocutor.)

The symbol for "keep doing this" is - "\(\dotso\)"
Infinity, \(\infty \) , is not a function or process.


I agree.
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Re: Controversial concepts in mathematics

Postby Silhouette » Wed Jan 20, 2021 11:32 am

Magnus Anderson wrote:And by the way, one should never say something like "It's obvious" when asked to explain something (:

You're absolutely right of course. My apologies.

Magnus Anderson wrote:
Silhouette wrote:the chain of addition ends after a finite number of terms has been reached. I don't mean that you have to time the expression - take as long as you like or do it instantly. The exact finite limit that could not be any other value doesn't care.

That's not an answer to the question.

I gave what I thought was an answer to the question.

Magnus Anderson wrote:
Silhouette wrote:Actually in many cases where infinity is involved, you can manipulate the answer to being any number. It's how you can infamously pretend "1=2", and if you can make that true, you'll soon realise you can make anything true, whether "5" or not. This is genuinely what infinity amounts to when you try to use it as an operand.
It's only when you use it as an operation, as above, that it can be used meaningfully - and then only really when its use converges to an exact finite limit that could not be any other value.

I don't think that to be the case. And you probably know that.

If you want to discuss this, you are welcome to present an argument that involves infinity and that concludes that \(1 = 2\).

Here's one:

\(\infty + \infty = \infty\) // let's divide both sides by \(\infty\)
\(\frac{\infty}{\infty} + \frac{\infty}{\infty} = \frac{\infty}{\infty}\)
\(1 + 1 = 1\)
\(2 = 1\)

My position is that this is a fallacious proof. (Which means you CAN'T use infinity to prove that \(1 = 2\).)

The proof is only fallacious due to its implications, which are that anything equals anything, which obviously isn't true. So I do think that to be the case, and I do know that.
The reason that this can occur is because you're using infinity as an operand.
That's why operating on infinity is invalid, and why it's avoided.
And since it's not a valid operand, that's when you realise that in practice it only serves validly as an operation.

Magnus Anderson wrote:
Silhouette wrote:Specifically the function that amounts to "keep doing this".
Like addition and subtraction, by itself this means nothing.

"Keep doing this" on its own does not tell me much. But I can make a guess anyway. Perhaps what you mean is a function that maps the set of natural numbers to a set that includes no more than \(true\). Such a function can be then used to tell us whether we should sum the next number in the sequence or not. Of course, being the kind of function that it is (constant function), it will always tell us to sum it.

What that implies is that \(\infty 5\) and \(\infty(5)\) are equal to \(true\). But earlier you said that \(\infty 5\) is not a valid expression (suggesting either that the arity of infinity isn't one or that the only argument it accepts is not an integer.)

Yeah it doesn't tell you much, which is why I said "by itself this means nothing". It has to apply to something. That's a dead giveaway of an operation: it's transitive, not intransitive. Add to that that treating it like an operation invalidly results in "anything equalling anything", and the case is even stronger.
But that doesn't mean you can simply say "\(\infty 5\) and \(\infty(5)\) are equal to \(true\)", which indeed is why I earlier said that they're not valid expressions, and why I said in an earlier post that there are many invalid ways to use any operation. When you use it in a valid way, it does have an arity of one, where it accepts things like sums, products, integrals etc. - making it more like an operation on operations (since sums, products, integrals etc. are operations themselves). Trying to pass an integer as a value is just another way to invalidly use it in its specific capacity as an operator. For other operators you can pass integers - that's not a necessary requirement for an operation.

Magnus Anderson wrote:
Silhouette wrote:Much like you can't merely "1-" without a finite amount to subtract from one (subtraction has arity of two), there's only certain forms where the finite denotation of "infinity" can validly apply. Sum and product are the most common, and as another example it's implied by "the indefinite" integral so doesn't have to be specified but its application would be valid if you did etc. If infinite sums and products followed the same conventions as indefinite integrals, it couldn't need to be specified there either - and I'd argue that not specifying (through some finite symbol) would actually aid in its communication.In these examples, its arity is one, accepting the arguments of sum, product, integral (all functions themselves). This is what I meant by what you kindly linked: "Just to wrap up what is meant when infinity is said to be an operator - it is only validly used in mathematics as part of an operation, more like an operator on an operator. This doesn't mean it's interchangeable with operators like addition and multiplication etc."

I strikethroughed everything I consider to have little to nothing to do with the question that I asked.

What this tells me is that in some cases the arity of the function that is infinity is one i.e. that it accepts exactly one argument.

"In some cases" means that that may not be always the case. In other words, in some cases, the arity of infinity might be other than one. If that' happens to be the case, i.e. if there are cases when its arity isn't one, what that means is that infinity is either a word that has multiple meanings or a word denoting a category of functions rather than some specific function.

You crossed out everything except the fancy math terminology, and yes, your understanding of what it meant is correct.
I've not identified any other cases where its arity is anything other than 1, only invalid cases where its arity is 0.

Magnus Anderson wrote:
Silhouette wrote:As such its output depends on what arguments it accepts, which can either be convergent to an exact finite limit that could not be any other value, or divergent i.e. "I don't/can't know" or "indetermined/indeterminable". I've already answered this but you asked again, so...

Does that mean that the codomain of the function that is infinity is the set of real numbers plus "I don't know"?

No. It means that when you use it like an operand, or an operation with zero arity instead of one, or as an operation that accepts arguments of the type that it does not accept, that it means nothing.
As I explained, you can use operations validly and invalidly, and using them invalidly doesn't invalidate their status as an operation.
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