Controversial concepts in mathematics

What’s it to you?

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. :smiley:

And from what I have seen of many of your posts, “high-school” would be a step up for you.

insanity is not a joke…and you are clearly a kook. take care of yourself.

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.

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

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).

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

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.)

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 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?

“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.

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.

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.

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”).

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? - [size=85]probably not[/size] :confused:

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

Here:
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).)

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.)

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”.

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.

Not really.

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.

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

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.

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.

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

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.

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”.

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).

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.

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

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?

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.

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”.

Well there you go.

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.

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…

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 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.

I think I understood it well enough.

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”.

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)?

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:

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.

(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).

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?

“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.

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.

Does that mean that the codomain of the function that is infinity is the set of real numbers plus “I don’t know”?

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).)

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

I thought you didn’t like people correcting the words you use. :smiley:

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.)

I agree.

You’re absolutely right of course. My apologies.

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

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.

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.

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.

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.

I had to read back to see where this discussion of (\infty ) being an operator came from. I think I can see that it is just another language/communication problem where something has been shortened to the point of being confusing.

In this notation -
(\sum_{i=1}^{5}a_i)
What is meant and I think originally stated is -
(\sum_{i=1}^{i=5}a_i)
Meaning to iterate the summation operation until i=5.

And in this notation -
(\sum_{i=1}^{\infty}a_i)
What is meant is -
(\sum_{i=1}^{i=\infty}a_i)
Meaning to iterate the summation operation until i= {the highest possible number} -even though it doesn’t exist.
Or -
(\sum_{i=1}^{i=Highest Possible Number}a_i)

And from something I recently read in order to make that more logically rational the notation became -
(\sum_{i=1}^{i=>\infty}a_i)
Implying “toward” the highest possible number so as to avoid the issue of i becoming a non-existent number.

When those details were left out, the idea of what was meant by -
(\sum_{i=1}^{\infty}a_i)
became ambiguous.

I just had this very same conversation with Certainly real concerning other word usage shortcuts that cause ambiguity and confusion.

So in maths, (\infty) is actually referring to “Highest Possible Number”.
And in that summation notation (\sum_{}^{}a_i) it is referring to a “direction toward the highest possible number” - as the DESTINATION of the Summation Operation.

At no time is (\infty) meant as an operator - always a destination of “highest possible number” even though it is known there is no highest to be reached.

It is not clear whether it means “the highest possible number” or “non-specific number greater than every integer”. I am inclined to believe it refers to the latter.

No need to apologize, but since you already did, let me accept it.

Right. But if someone gives you something you didn’t really ask for, even if its partly or entirely your fault, you have to note it, right?

I am not sure I am following you. The proof merely shows that “2 = 1” not that “anything equals anything”. And while we both agree that its conclusion is false, the fact that it is does not tell us where the flaw lies. (The flaw can’t be in the conclusion.) And since this is a deductive argument involving definitional premises, the flaw must be in the logic since definitional premises cannot be false. So where exactly is the flaw?

But which part of the proof is wrong? Perhaps the very first line? Infinity plus infinity isn’t equal to infinity?

And here is what I think:

The problem with the proof is that (\frac{\infty}{\infty}) does not equal to (1).

This is based on the premise that (\infty + \infty = \infty) stands for “When you take a number greater than every integer and add to it a number greater than every integer, you always get a number greater than every integer”. And that is true even if “a number greater than every integer” is a contradiction in terms. But when you take a number greater than every integer and divide it by a number greater than every integer, you do not always get (1).

I think you misunderstood me. When I said “It doesn’t tell me much”, I meant “It doesn’t tell me what I want to know” and what I want to know is how many arguments it accepts, what kind of arguments and what kind of value it outputs.

Right. So I made several assumptions:

  1. You think infinity is a function. (Turns out to be correct.)

  2. You think it is a function that can accept single argument. (Turns out to be correct.)

  3. You think it’s a function that can accept an integer. (Turns out to be incorrect.)

What I know now is that you think that infinity is a function that takes a single argument of type “function” making it a higher-order function.

What I don’t know is:

  1. What kind of function does it take? Any or some specific?

  2. What kind of value does it output?

  3. How does it map its input to its output?

  4. Who else defines infinity this way?

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Sock-puppets… there are numerous and many.