Are the subdivisions involved in an integral countably infinite or uncountably infinite? If they're countably infinite...

Are the subdivisions involved in an integral countably infinite or uncountably infinite? If they're countably infinite, then could you find a bijection between some any definite integral and some infinite series?

faceplam
>what are Riemann sums

Countably infinite. Yes.

Yes but Im not sure what it would apply to

uncountably infinite by definition.

...

That's only countably many, boyo.

It would allow you to kinda "stretch" a definite integral out over the naturals. Not sure what the purpose would be, but I find it fascinating.

they are finite.

it's a limit n to infinity, where for every n, you have finitely many

just like in

[math] \sum_{n=1}^m \dfrac{1}{n^2} [/math]

you have always finitely many (rational) terms, while

[math] \lim_{m\to \infty} \sum_{n=1}^m \dfrac{1}{n^2} = \dfrac{\pi}{6} [/math]

it's a limit n to infinity, where for every n, you have finitely many

just like in

∑
m
n=1
1
n
2

∑n=1m1n2


you have always finitely many (rational) terms, while

lim
m→∞
∑
m
n=1
1
n
2

=
π
6

limm→∞∑n=1m1n2=π6

do you know what countable means?

nope

what the fuck is going on in this thread ? a riemann sum is a finite sum. the integral is a supremum/infimum of all riemann sums. there are no infinite sums in the integral.

so it's uncountable? You can't write an integral as an infinite sum?

The amount of finite partitiond of a given interval is uncountable infinite.

There are many wayd you can get to a sum as a sucesions of finite sums that converge to the integral. But there isn't just one and the definition doesn't assume it's just one.

you can, but there isn't a cannonical way how to do it and it's not particularly useful. the sum will be countably infinite, because we don't define uncountable sums in the first place (well I suppose it can be done, but I don't know whether there's some theory around it).

Anyone who doesn’t say countably infinite is retarded.
A big giveaway often is the use of n.

/thread

The amount of subdivisiond you can get that converge to an integral is uncountbly infinite.

The Sum has uncountably infinite elements if you actually could get uncountably many slices before the heat death of the universe.

The answer is: You just stop at an n that gives you a result that is exact enough.

Shittily explained example:
e.g. 10-15 unchanging decimal places after the comma. (Or less.)

What if we define Riemann sums but with uncountable subdivisions?

WHAT THE FUCK DOES ""UNCOUNTABLE SUBDIVISION"" EVEN MEAN ??

finally a real answer. You look how the sum behaves for finite subdivisions as subdivision number gets bigger. \thread retards

It would work but you'd have to define and motivate how to calculate the sum over uncountable subdivisions. The brainlet tears would never stop.

>Uncountably many subdivisions
Lol, brainlet, please try to using rationals to destroy our own point yourself.

That's rather wrong in at least one respect, namely x to infinity.

Subdivisions / Riemann sums: countable.
Indefinite integral: uncountable. The reason for this is because the integrals you are speaking of are just mappings of a subset of R onto R. Since any nonsingular subset of R is uncountable, so is the domain of an indefinite integral.

formally you are examining the behavior of a finite sum as the number of terms becomes arbitrarily large, specific infinite cardinalities have no relevance.