I want here to introduce some rather strange relation, it's a little bit made up, possibly ad-hoc. However, the point behind it is that it can straight forwardly interpret the membership relation of set theory.

The idea is about "relative part-hood", to be symbolized as: $ y \ P^x x$ to mean that "$y$ is a part of $x$ as viewed by $x$". So it not any part of $x$ that would qualify to be a "part of $x$ as viewed by $x$", it's only those that $x$ views them as parts of. So $P^x$ is different from the usual part-hood relation $P$.

More officially $y \ P^z x$ should be read as "$y$ is a part of $x$ relative to $z$".

In other words it's a kind of conditional part-hood!

But we'd be only interested here when $z=x$; i.e., only of: $y \ P^x x$

For the sake of simplicity I'll denote that kind of self relative part-hood by $\mathcal P$, so:

Define: $y \ \mathcal P \ x \iff y \ P^x \ x $

If we add the following Uncluttering axiom, then matters would look much like $\in$ membership relation:

Ockhams Razor: $ \forall x \forall y [\forall z(z \ \mathcal P x \to z \ \mathcal P y) \to x \ P \ y]$

This would easily prove Extensionality for relation $\mathcal P$, that is:

$\forall x,y [\forall z (z \ \mathcal P x \leftrightarrow z \ \mathcal P y ) \to x=y]$

In English: no distinct objects view their parts in exactly the same manner.

The reason is because $x,y$ would be parts of each other, thus equal! [M2]

Now clearly even if we work in the traditional General Extensional Mereology "GEM", or even in its atomic variant "AGEM", add the above-mentioned relative part-hood relation $\mathcal P$ to it, still even in this milieu the $\mathcal P$ relation is NOT transitive! Thereby getting very close to the set theoretic membership relation $\in$. Now a $set$ can be defined as:

Define (set): $set(x) \iff \exists y (x \ \mathcal P \ y)$

Although this seems to be made up, what's interesting is that this method allows a single comprehension scheme to derive all rules of $ZF-Regularity-Infinity$.

The single axiom scheme is simply Replacement in terms of $\mathcal P$ relation, which in brief is all closures of:

$\exists \text { set } A \ [\forall y (\phi(y) \to \exists x \ P \ A (y=F(x)))] \to \exists \text { set } B \ \forall y (y \ \mathcal P B \leftrightarrow \phi(y)) $;

for every formula $\phi(y)$ in which $B$ is not free, and function $F$ from parts of sets to parts of sets.

So this single axiom schema together with the Razor axiom would prove: Empty, Pairing, Set Union, Power set, all instances of Separation, and of course Replacement with respect to relation $\in$. In other words, all of the comprehension axioms of ZFC!

The idea is to simply define $\in$ as $\mathcal P$

Actually even infinity can be said to be promoted by this method, since if we allow Gunk Mereology, then by Extensionality for $\mathcal P$, there must exist a gunk set, i.e., a gunk object that is a set, then this would easily prove infinity, since the set of all parts of that object that are seen as relative parts of parts of it, would satisfy infinity.

Moreover, its clear that this method encourages the use of Atomless General Extensional Mereology $\tilde{A}GEM$ as the background Mereological language of this theory, since it increase the plasticity of this method. Accordingly it does motivate infinity!

This means that just from the Razor axiom and the Replacement axiom schema [on top of a background of at least $\tilde{A}GEM$] we can derive ALL axioms of ZF-Regularity, and thus interpret all axioms of ZFC!

Its really astonishing that just a minor tweaking of part-hood relation can prove so powerful that it can interpret the whole of Set Theory.

This shows that Mereological thought is potentially very powerful!

Although the concept of relative part-hood as presented here seem artificial, yet its rather the flawless spontaneous interpret-ability of set theory in it that looks very attractive to me!

So has there been a known work along those lines before?

One of the nice features of this method is that it can have a nice account on empty objects? for example the above full Razor axiom, would lead to Mereology with a bottom atom, which is not plausible. If we want to keep in accordance with classical mereological approaches which are bottom-less, then it needs to be modified to:

$ \forall x \forall y [\forall z(z \ \mathcal P x \to z \ \mathcal P y) \wedge (\exists z (z \ \mathcal P x) \leftrightarrow \exists z (z \ \mathcal P y)) \to x \ P \ y]$

This would prove the empty set to be a mereological atom! Also each Quine atom to be a mereological atom.

  • $\begingroup$ Talking from ignorance here: is y P^z x defined for every set z? $\endgroup$ – jose_castro_arnaud Mar 18 at 2:57
  • $\begingroup$ @jose_castro_arnaud, yes its defined. I've added that. $\endgroup$ – Zuhair Mar 18 at 5:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.