dinsdag 21 januari 2014

An new a priori argument for atomism

In what follows I propose a new a priori argument for mereological atomism. Mereological atomism is the thesis that every composite object is ultimately composed of simple objects. Simple objects are objects without proper parts.

Take a formal additive measure of being. This measure measures the amount of being contained in each object. Let O be an object and denote the amount of being contained in O by being(O). Thus, being(O) is zero in case there is no object O. Let the objects {Oi}i compose object O. Hence {Oi}i is a composition of O. The additive nature of the involved measure implies that being({Oi}i) = Σi[being(Oi)].

Strictly speaking the aformentioned formula is not well-formed since being(.) has been defined as a function on objects and not as a function on sets of objects. Yet, this is not a problem. We can extend the domain of being(.) to the collection of all mereological sums. In that case the formula becomes being(sum{Oi}i) = Σi[being(Oi)].

According to the principle of composition-as-identity, object O simply is the objects {Oi}i taken together, that is, object O is nothing above or beyond the objects {Oi}i taken as a totality. From this it follows that being(O) = being(sum{Oi}i) = Σi[being(Oi)].

Next, let O be an object and let Ω and Ω* be two different compositions of O such that every object in Ω* is either equal to or a part of an object in Ω. In that case Ω* is called a refinement of Ω. It follows that being(Ω) = [being(Ω) – being(Ω*)] + being(Ω*). This formula indicates that the amount of being at a certain level of composition is the arithmetical sum of the amount of being at the previous level and the incremental amount between both levels.

Let {Ωn}n be a sequence of compositions of object O such that Ω0 = O and such that for all natural numbers n composition Ω(n + 1) is a refinement of composition Ωn. The sequence {Ωn}n is either finite or infinite. Suppose first that {Ωn}n is finite and let ΩN denote the final composition in the sequence. It follows that being(O) = Σ(n=1 to n=N) [being(Ω(n – 1)) – being(Ωn)] + being(ΩN).

How should this arithmetical formula be adapted to the case that {Ωn}n is infinite? This case is obtained if N proceeds to infinity and the final composition ΩN vanishes from the sequence. Hence, the only natural answer appears to be that in that case one obtains the formula being(O) = Σ(n=1 to n=∞) [being(Ω(n – 1)) – being(Ωn)].

Note that I’m not claiming here that the formula for the infinite case can be mathematically derived from the formula for the finite case. For, such a claim would be clearly ungrounded. The reasoning is qualitative and not quantitative. The structure of the formula for the finite case expresses the insight that the amount of being of some composite is obtained bottom-up in precisely two ways, namely (a) incremental influx of being between the levels of composition and, (b) inheriting the amount of being already available at the lowest level. But then the equivalent structure for the infinite case is just an infinite sum of incremental infusions of being. After all, in the infinite case there is no lowest level and therefore only (a) applies here. This conceptual reasoning should not be taken for a quantitative mathematical derivation of the infinite formula from the finite one.

Now suppose, for reductio, that atomism is false. In that case there is a composite object C that is not composed of simple objects. Due to the principle of supplementation C is composed of two or more other objects. So, there is a composition of C. Since C is not composed of simple objects there is an infinite sequence of compositions {Ωn}n of C such that for every natural number n composition Ω(n + 1) is a refinement of composition Ωn. It thus follows that being(C) = Σ(n=1 to n=∞)[being(Ω(n – 1)) – being(Ωn)].

Further, the principle of composition-as-identity implies that being(C) = being(Ω(n – 1)) and being(C) = being(Ωn). Hence, for all natural numbers n, it follows that being(Ω(n – 1)) – being(Ωn) = 0. This implies that being(C) = Σ(n=1 to n=∞)[being(Ω(n – 1)) – being(Ωn)] = Σ(n=1 to n=∞)[0] = 0. But then being(C) = 0 which implies that there is no object C. This contradicts with the fact that object C does in fact exist. Therefore, the initial assumption that atomism is false needs to be rejected. It thus follows that atomism is true.

This fragment is obtained from my dissertation (pp. 129-131)

5 opmerkingen:

  1. Beste Bert,

    In mijn argument trek ik geen 'instanties van being' van elkaar af. De functie 'being(.)' beeldt zijnden ('beings') af op getallen. Ik trek dus getallen van elkaar af. En voor alle n is being(Ω(n – 1)) – being(Ωn) gelijk aan of groter dan 0. De reden hiervoor is dat Ω(n) een 'refinement' is van Ω(n-1) en compositie nooit resulteert in een afname van de hoeveelheid 'being', zodat being(Ω(n – 1)) altijd gelijk is aan of groter is dan being(Ωn). Bovendien, als voor twee beings x en y geldt dat being(x) - being(y) < 0, dan volgt daar geenszins uit dat being(x) of being(y) negatief is. De functie 'being(.)' heeft immers als bereik de positieve getallen (inclusief 0).

    Groet,
    Emanuel

    BeantwoordenVerwijderen
  2. Beste Bert,

    Het bereik van de functie 'being(.)' hoeft zich helemaal niet te beperken tot de natuurlijke getallen. Je kunt als bereik bijvoorbeeld de positieve rationale getallen (inclusief 0) of de positieve reële getallen (inclusief 0) nemen. Voor alle atomaire beings (atomen) A geldt being(A) > 0. De waarde van being(zijnsgeheel) is dus ook groter dan 0. Er zijn goede argumenten voor de stelling dat één van de atomen geldt als de eerste oorzaak van alle andere atomen. Dit betekent dat er een uniek atoom bestaat dat zelf onveroorzaakt is en dat tevens geldt als de (in)directe oorzaak van alle andere atomen. In mijn proefschrift heb ik een dergelijk argument ontwikkeld. Zie eventueel http://goo.gl/rquCbh voor een verkorte toegankelijke weergave ervan.

    Groet,
    Emanuel

    BeantwoordenVerwijderen
  3. Beste Bert,

    Natuurlijk heb ik Douglas Hofstadters Gödel-Escher-Bach niet gemist. Zie ook "Het Lucas argument tegen het mechanisme" op gjerutten.nl

    Groet,
    Emanuel

    BeantwoordenVerwijderen
  4. Beste Bert,

    Hoe kom je erbij dat Lucas mij overtuigt? In mijn vorige reactie wees ik je op mijn "Het Lucas argument tegen het mechanisme" op gjerutten.nl. Uit dat stuk kun je helemaal niet opmaken dat ik het met Lucas eens ben. Ik zou het maar eens (goed) bekijken. En dan vooral de laatste slide.

    Groet,
    Emanuel

    BeantwoordenVerwijderen
  5. Beste Bert,

    Inderdaad. Ik beantwoordt die vraag op mijn laatste slide in een discussie altijd met 'nee', en leg dan uit waarom.

    Groet,
    Emanuel

    BeantwoordenVerwijderen