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Can two things or x number of things that are not infinite create infinity? Let's assume that x is not infinity. I am thinking that it's not the case, and there's no paradox here, but I just want to make sure. But I am pretty sure you can prove mathematically that this cannot be true.

Mauro ALLEGRANZA
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Sayaman
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    Things are (almost) not mathematical objects. Anyway, a thing is a system, a system is a set of parts. Each part is a system, and so on, recursively/fractally. So, how many systems and subsystems has a grain of sand? Infinite. In case of two things? Certainly, more. – RodolfoAP Jan 27 '22 at 02:38
  • Can a/the universe be infinite? Many arguments say it contains infinities. – J Kusin Jan 27 '22 at 03:02
  • Are you asking if putting together finitely many finite collections is finite? If so, the answer is yes. – Conifold Jan 27 '22 at 04:44
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    Infinity is only a conceptual abstract object not perceived concretely. A single person called George Cantor mentally created different orders of infinities such as countable infinity vs uncountable infinity, and many other aleph infinities, together with certain arithmetic rules for different types of infinities (ordinal and cardinal) resembling but not same as those of finite natural numbers. So mind has this transcendental property to speculate and conceive potential infinity notions... – Double Knot Jan 27 '22 at 06:02
  • Just tossing this out, but we can express the Peano axioms in finitely many symbols, and thereby create an infinite collection of natural number. We can write down the axioms of ZF using finitely many symbols (even though there are infinitely many axioms!) and create all standard mathematical sets such as the real numbers. For that matter, "The natural numbers" is a finite string that invokes or evokes an infinite collection of numbers. – user4894 Jan 27 '22 at 08:06

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Depending on what the question is actually trying to get at, this question technically has a standard, but fully mathematical, in fact precisely set-theoretic, answer. OTOH it might not be a strong enough post for MathSE, regardless. But all that being said, the "standard" answer goes as follows:

Definition. A cardinal κ is inaccessible ("from below") if, for all λ < κ, 2λ < κ, and if cf(κ) = κ.

:::"cf(κ)" is shorthand for the function that takes a cardinal for its input and outputs the "length" of the cardinal summation sequence that yields that cardinal. If the "length" is as long as the cardinal itself is, the cardinal is regular; otherwise, the cardinal is singular.

:::ℵ0 is inaccessible.

:::Therefore, there is no way using a cardinal summation sequence that is not as "long" as the zeroth aleph, to get the zeroth aleph (which is then redundant); and 2 to the power of any of the finite n perforce does not equal ℵ0, so there is no ZFC operation on finite sets as such that gives us infinite sets.

But in fact no mere compounding of the powerset or union operations would work, either. Like, tetrating finite numbers never reaches infinity, neither hexating them or so and on down the hyperoperator line. As long as the index of the hyperoperation on two finite numbers is itself a finite index, we don't get infinity. (I did read an article that talked about something called "omegation," an unwieldy word, but intended to replicate the etymology of the normal hyperoperator names, which said that 2 ↑n 3 tends toward infinity, so that if we set n to the zeroth aleph, the operation would perhaps go to the zeroth aleph, but again we're presupposing the same number as we're trying to get, so we didn't really get it the way we wanted to.) {To be even more technical, this whole phenomena where compounding the basic operations on sets never solves the problem is tantamount to a semiotic condensation of cf(x) issues anyway, but that would take a lot more time to go in to.}

Kristian Berry
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