Can "Infinity" ever be more than a mathematical abstraction?

[someguy1]

I can't help but point out that Dirac's delta "function" is infinite-valued on an infinitesimal interval. It provides, moreover, a highly simplifying and intuitively satisfying notation for his vastly influential treatment of quantum mechanics.
(most of that last sentence was lifted from this paper--https://arxiv.org/pdf/1608.06878.pdf)

Mathematically, it's not a function, it's a distribution. There are no infinitesimal intervals in the real numbers.

This is merely a heuristic characterization. The Dirac delta is not a function in the traditional sense as no function defined on the real numbers has these properties.[17] The Dirac delta function can be rigorously defined either as a distribution or as a measure.

https://en.wikipedia.org/wiki/Dirac_delta_function#Definitions

Surprised you didn't check on Wiki. It's perfectly obvious that it's not a mathematical function. The only question is the mathematical formalisms used to make it work out. I thought this was fairly well known. Then again, that's because I learned what I know about QM from the math side.

Now I can see you put "function" in quotes so you're indicating you know this. But it doesn't take an infinite value on an infinitesimal interval. That's the physicist's intuition, but it's not the mathematical formalism.

 

[arfa brane]

Mathematically, it's not a function, it's a distribution.

I know that.

There are no infinitesimal intervals in the real numbers.

Then calculus is in big trouble. Can you prove there are no infinitesimal intervals in the real numbers?

Surprised you didn't check on Wiki.

Ah. And yet

Now I can see you put "function" in quotes so you're indicating you know this. But it doesn't take an infinite value on an infinitesimal interval. That's the physicist's intuition, but it's not the mathematical formalism.

If it doesn't take an infinite value on an infinitesimal interval, that's something you could demonstrate mathematically then?

 

[someguy1]

I know that.Then calculus is in big trouble.

I don't think so.

Can you prove there are no infinitesimal intervals in the real numbers?

Of course. If $$[x, y]$$ is an interval of reals with $$x \leq y$$ then if $$x < y$$ the length of the interval is $$y - x > 0$$. And if $$x = y$$ then the length of the interval is zero. So the length of any interval of reals is either a positive real number or zero. That makes sense since there are no infinitesimals in the real numbers.

Ah. And yet If it doesn't take an infinite value on an infinitesimal interval, that's something you could demonstrate mathematically then?

Yes, the Wiki page is pretty clear. The Dirac delta is not a function. But you say you know that. The physicists are free to think of it any way they want.

 

[arfa brane]

There are plenty of online sources from universities which all define the Dirac delta function thus:

$$ \delta(x) = \begin{cases} \infty\; when\; x = 0 \\ 0\; when\; x \ne 0 \end{cases}; \int_{-\infty}^{+\infty} \delta(x) dx = 1 $$
​

It seems to unequivocally define it as being infinite when x = 0, and 0 otherwise. What's going on?

 

[someguy1]

There are plenty of online sources from universities which all define the Dirac delta function thus:

$$ \delta(x) = \begin{cases} \infty\; when\; x = 0 \\ 0\; when\; x \ne 0 \end{cases}; \int_{-\infty}^{+\infty} \delta(x) dx = 1 $$
​

It seems to unequivocally define it as being infinite when x = 0, and 0 otherwise. What's going on?

Hard to say since you didn't link your source. But the Wiki article is pretty clear. This isn't something I'm going to argue with you about.

 

[someguy1]

$$ \delta(x) = \begin{cases} \infty\; when\; x = 0 \\ 0\; when\; x \ne 0 \end{cases}; \int_{-\infty}^{+\infty} \delta(x) dx = 1 $$
​

ps -- How exactly do you evaluate that integral? Can you show me? On the contrary, since the value at a single point never changes the value of either a Riemann or Lebesgue integral, by your definition it must be zero. So what you wrote isn't logically consistent with delta being defined as a function. The only way to make the integral work out is by properly defining delta as a distribution, as is shown in the Wiki article.

 

[arfa brane]

there are no infinitesimals in the real numbers.

Almost. Apparently 0 is the only real infinitesimal.
And from nLab:

The term ‘infinitesimal’ means the same as ‘infinitely small in absolute value’; in Latin, it literally means ‘infinity-eth’ and should be interpreted in the sense of a fraction.

In the ordinary analysis of real numbers, the only infinitesimal number is zero. However, the basic intuitions of calculus since its beginnings have dealt with infinitely small (and sometimes also infinitely large) numbers. There are several different ways to develop a rigorous theory that includes infinitesimal numbers.

--https://ncatlab.org/nlab/show/infinitesimal+number

I can agree with what that quote says, because in calculus you consider dx as an infinitesimal change in the value of x. Plenty of textbooks do this.
What they also do is consider dx to be indistinguishable from 0, although heuristically it's nonzero.

 

[someguy1]

Almost. Apparently 0 is the only real infinitesimal.
And from nLab:--https://ncatlab.org/nlab/show/infinitesimal number

I can agree with what that quote says, because in calculus you consider dx as an infinitesimal change in the value of x. Plenty of textbooks do this.
What they also do is consider dx to be indistinguishable from 0, although heuristically it's nonzero.

arfa brane You quoted me and generated an alert! Thank you!!

That article is crap. 0 is not an infinitesimal. An infinitesimal is a number that is greater than 0 and less than 1/n for every positive integer n. By that definition there are clearly no infinitesimals in the real numbers. The def you quoted is simply wrong. The bit about "infintieth" is garbage.

The site you quoted seems reputable. I can't account for their strange claims. Why are you aiming this stuff at me? I can't do anything about everything on the Internet. It's a category theory based site. Maybe they have a different definition.

 

[someguy1]

ps -- I read farther into the article you quoted. They're not idiots, they are very aware of their choice and they explicitly justify it:


Traditionally, one adds to the definition the requirement that i≠0, but this leads to a less useful notion of the space of all infinitesimals. We allow 0 to be infinitesimal for some of the same reasons that we allow it to be a natural number.

Ok, fine. They acknowledge that their definition is different from everyone else's; and they justify themselves. That's cool. What of it?

And remember these people are doing category theory. This is an entirely different way of thinking about mathematics and mathematical objects than people are used to. There is literally no accounting for how category theorists will express basic notions in order to better fit their framework. There are a lot of oddities. https://en.wikipedia.org/wiki/Category_theory

 

[arfa brane]

Ok, fine. They acknowledge that their definition is different from everyone else's; and they justify themselves.

But "they" are not the only source I've seen that claims 0 is the only infinitesimal real number. Another example is here.
Of course anyone could google "zero and infinitesimal", say.

You seem to be implying that the notion of an infinitesimal depends on how it's defined. Or perhaps the best conclusion a student of calculus can make is that an infinitesimal is only a heuristic? Unless they happen on Robinson's extensions to the reals--nonstandard numbers?

 

[someguy1]

But "they" are not the only source I've seen that claims 0 is the only infinitesimal real number. Another example is here.
Of course anyone could google "zero and infinitesimal", say.

You seem to be implying that the notion of an infinitesimal depends on how it's defined.

Um ... yeah. Doesn't everything in math? The notion of the number 3 depends on how it's defined. Are you some kind of Platonist about infinitesimals? More power to you. Was the quote I bolded above unclear? The authors acknowledge that they changed the usual definition of infinitesimal for their own reasons. Changing the definition changes the notion, or at least provides a variation on the notion.

Or perhaps the best conclusion a student of calculus can make is that an infinitesimal is only a heuristic?

Student of calculus? Pull down the exponent and subtract 1. Chain rule. Product rule. Integration by parts. Related rate problems. And underneath it all, limits. Not infinitesimals. Limits. Epsilons and deltas.

The material you're discussing is not part of the subject of calculus. Philosophy and history of calculus, maybe, but even then you're pretty far afield.

Unless they happen on Robinson's extensions to the reals--nonstandard numbers?

I'm glad you read a Wiki page on those. What do you want to know about them?

 

[arfa brane]

The authors acknowledge that they changed the usual definition of infinitesimal for their own reasons.

See if you can understand the question and the answers at this link: https://math.stackexchange.com/questions/1317987/why-is-zero-the-only-infinitesimal-real-number.

Possibly I need some help with it myself (or maybe I'm pretending, who can say?). Or, maybe math.stackexchange is populated by idiots.

 

[someguy1]

See if you can understand the question and the answers at this link: https://math.stackexchange.com/questions/1317987/why-is-zero-the-only-infinitesimal-real-number.

Ok I stand corrected. Evidently some authors include 0 as an infinitesimal. At least two, evidently. Does this make a difference to anything?

 

[arfa brane]

Does this make a difference to anything?

Maybe. The reals form a ring under addition and multiplication, right? It's a commutative ring. Addition has inverses (negative reals).
$$ \mathbb R $$ also has well-defined relations: <, >, =, ...

Such that (this is pretty obvious) if a and b are real, then if a < b, a - b < 0; if a > b, a - b > 0; if a = b, a - b = 0 (and -b is the additive inverse of a).
Using the relations you can prove 0 is the only solution where -a < x < a for any real a.
The real number line is also called Archimedean, when it's a commutative ring.
So, is it? and why would it make a difference?

 

[someguy1]

Maybe. The reals form a ring under addition and multiplication, right? It's a commutative ring. Addition has inverses (negative reals).
$$ \mathbb R $$ also has well-defined relations: <, >, =, ...

Such that (this is pretty obvious) if a and b are real, then if a < b, a - b < 0; if a > b, a - b > 0; if a = b, a - b = 0 (and -b is the additive inverse of a).
Using the relations you can prove 0 is the only solution where -a < x < a for any real a.

Making a mountain out of a molehill. It's like arguing whether zero is a natural number. It makes no difference. If you like you could just RENAME 0 to 1 and 1 to 2 and 2 to 3 and so forth and nothing would change. I agree that having a zero is the best thing since sliced bread when it comes to algebra, but it's not nearly good enough. Nobody ever goes, "Oh we can't do algebra in the natural numbers unless we include 0." Because even with 0, the arithmetic of the natural numbers doesn't have all the nice group, ring, and module properties. For that you need the integers, the positive and negative numbers and zero. So the argument over whether to include 0 is TRIVIAL and so is the argument about adding 0 to the infinitesimals.

arfa brane You did show me something I didn't know, which is that SOME authors include 0 with the infinitesimals. And Keisler's name I know well, he wrote the first calculus text based on the hyperreals. So I learned something.

But you are making much more of it than is warranted. [Of course that's what I WOULD say, having just lost the point! LOL]

The real number line is also called Archimedean, when it's a commutative ring.
So, is it? and why would it make a difference?

No model of the real numbers that contains infinitesimals can be Archimedean. I don't have a link to the proof at hand. Maybe I'll find it later. Being a commutative ring has nothing to do with an ordered set being Archimedean. Commutative rings are essentially generalizations of the integers. They're about algebraic properties, not continuous properties or order properties.

ps -- Ah here is the proof that the hyperreals are not Archimedean. https://en.wikipedia.org/wiki/Hyperreal_number

In the hyperreals there is an element $$\omega$$ such that $$1<\omega, \ \ 1 + 1 < \omega, \ \ 1 + 1 + 1 < \omega$$, etc. That violates the definition of an Archimedean order.

 

[arfa brane]

Introduce a relation: ≈, "infinitely close to".

So that if a ≈ b, then a - b ≈ 0. You also have if a ≈ 0 and b ≈ 0, then a + b ≈ 0.
Multiplication means if at least one of a,b ≈ 0, then ab ≈ 0. Also if b ≈ 0, then 1/b is almost infinite. 1/0 is (still) not defined.

This "new" relation changes the structure of the ring somewhat, it redefines addition, multiplication and their inverses.

Just sayin'

 

[someguy1]

Introduce a relation: ≈, "infinitely close to".
So that if a ≈ b, then a - b ≈ 0. You also have if a ≈ 0 and b ≈ 0, then a + b ≈ 0.
Multiplication means if at least one of a,b ≈ 0, then ab ≈ 0. Also if b ≈ 0, then 1/b is almost infinite. 1/0 is (still) not defined.

This "new" relation changes the structure of the ring somewhat, it redefines addition, multiplication and their inverses.

Just sayin'

That's what they do in expositions of the hyperreals, but so what? You're just sayin', but what are you just sayin'?

 

[Write4U]

Introduce a relation: ≈, "infinitely close to".

I love the logic.
Would that be a 2 dimensional linear measurement? If so, is there a vertical version of that symbol indicating "infinitely close" in a 3 dimensional space?

 

[arfa brane]

What am I just sayin'? Maybe nothing so much. But, here we are in a thread about the mathematical abstraction: infinity. And we're talking about physical infinity (whatever that is).

Both of these things have occupied minds for thousands of years. Do we now have all the mathematical tools we need? Is physics done and dusted? Who really knows? I don't think sciforums is going to decide that one . . .

Even in that definition of (what today is called) Dirac's delta function, which isn't a function; look at what might be called the restriction: $$ \int_{-\infty}^{+\infty} \delta(x) dx = 1 $$

The limits of integration are infinite. That is, the integral is over the whole of $$ \mathbb R $$.
I can still remember a question some student asked in 1st year calc: What do the infinite limits mean? The lecturer said something like, in this case you can consider $$ -\infty $$ and $$ +\infty $$ to be equivalent (??). That is, you consider the real line to be extended indefinitely to the left and to the right, such that both ends "go to infinity", i.e. the same place. (like the real line meets itself at infinity, or some such).

In fact, the real line plus a point at infinity is a well-known construction.

 

[someguy1]

In fact, the real line plus a point at infinity is a well-known construction.

You never seem to have a point or a thesis. You communicate like a bot.