Between Atomism and Superatomism

Abstract

There are at least three vaguely atomistic principles that have come up in the literature, two explicitly and one implicitly. First, standard atomism is the claim that everything is composed of atoms, and is very often how atomism is characterized in the literature. Second, superatomism is the claim that parthood is well-founded, which implies that every proper parthood chain terminates, and has been discussed as a stronger alternative to standard atomism. Third, there is a principle that lies between these two theses in terms of its relative strength: strong atomism, the claim that every maximal proper parthood chain terminates. Although strong atomism is equivalent to superatomism in classical extensional mereology, it is strictly weaker than it in strictly weaker systems in which parthood is a partial order. And it is strictly stronger than standard atomism in classical extensional mereology and, given the axiom of choice, in such strictly weaker systems as well. Though strong atomism has not, to my knowledge, been explicitly identified, Shiver appears to have it in mind, though it is unclear whether he recognizes that it is not equivalent to standard atomism in each of the mereologies he considers. I prove these logical relationships which hold amongst these three atomistic principles, and argue that, whether one adopts classical extensional mereology or a system strictly weaker than it in which parthood is a partial order, standard atomism is a more defensible addition to one’s mereology than either of the other two principles, and it should be regarded as the best formulation of the atomistic thesis.

Appendix

Proof Proof of Proposition 6

that every maximal proper parthood chain terminates and consider an arbitrary a. I set out to show that a has an atomic part. Suppose for reductio that a does not have an atomic part and define Z so that, for any x, x ≼Z iff $x \lhd a$. Since every atom has itself as an atomic part, a must be composite; hence it has a proper part, and so Z exist.

• (1) First I show that, for any x ≼Z, there is a y ≼Z such that $y \lhd x$. Consider an arbitrary b ≼Z. b isn’t atomic, since, if it were, a would have an atomic part, contrary to the initial reductio assumption. Hence b is composite, and so has a proper part c. By the transitivity of proper parthood, $c \lhd a$. But then by the characterization of Z, c ≼Z. So there is a y ≼Z such that $y \lhd b$.

• (2) Next I show that there is a maximal chain among Z. Consider again an arbitrary b ≼Z. From the previous result, b has a proper part c ≼Z. So there is a chain formed by b,c = Z_C, where Z_C ≼Z. Since $\lhd$ partially orders its domain, the Hausdorff maximal principle guarantees that there is a maximal chain formed by some M such that Z_C ≼M.

• (3) I now set out to show that the chain formed by M has as a final segment a chain formed by objects among Z that is both maximal and does not terminate. Define N so that, for any x, x ≼N iff x = b or both x ≼M and $x \lhd b$. Since b exists, N exist.

• (a) I now show that N form a final segment of the chain formed by M.

  • (i) I first show that N ≼M. Consider an arbitrary d ≼N. Whether d = b or $d \lhd b$, by the characterization of N, d ≼M. Since d is arbitrary, for every x ≼N, x ≼M.

  • (ii) I now show that every x ≼N is a proper part of every y ≼M such that $y \npreceq \text {N}$. Consider arbitrary d ≼N and e ≼M where $e \npreceq \text {N}$. (If there is no such e, the goal follows trivially.) By the characterization of N, either d = b or both d ≼M and $d \lhd b$. Since d,e ≼M and M form a chain, either $d \lhd e$, $e \lhd d$, or d = e. If $e \lhd d$ then, if d = b, it immediately follows that $e \lhd b$. And if $d \lhd b$, the transitivity of proper parthood guarantees that $e \lhd b$. And since e ≼M, the characterization of N guarantees that e ≼N, contradicting the assumption that $e \npreceq \text {N}$. If e = d, it also follows that e ≼N, since it was assumed that d ≼N, again contradicting the assumption that $e \npreceq \text {N}$. So $d \lhd e$.

• (b) I now show that N ≼Z. Consider an arbitrary d ≼N. By the characterization of N, d = b or $d \lhd b$. If d = b, then, since b ≼Z, the characterization of Z guarantees that $b \lhd a$, and so $d \lhd a$. If $d \lhd b$, then, since, for the same reason, $b \lhd a$, the transitivity of proper parthood guarantees that $d \lhd a$. Either way, $d \lhd a$. So by the characterization of Z, d ≼Z. Since d is arbitrary, for every x ≼N, x ≼Z.

• (c) I now show that N form a chain.

  • (i) I first show that there are x,y ≼N such that $x \lhd y$. By the characterization of N, b,c ≼N. It is also known that $c \lhd b$ (see step (2)).

  • (ii) I now show that, for any x,y ≼N, either $x \lhd y$, $y \lhd x$, or x = y. Consider arbitrary d,e ≼N. Since, by result (3)(a)(i) above, N ≼M, the transitivity of inclusion guarantees that d,e ≼M. Since M is a chain, either $d \lhd e$, $e \lhd d$, or d = e.

• (d) I show next that the chain formed by N is maximal. Suppose for reductio that it is not. Then every x ≼N shares a proper part d. (Remember that the definition of maximal chains presupposes the transitivity of proper parthood.) Consider an arbitrary e ≼N. Then $d \lhd e$. Now consider an arbitrary f ≼M. Since M form a chain and, by (3)(a)(i), e ≼M, either (i) $e \lhd f$, (ii) $f \lhd e$, or (iii) e = f.

  • (i) Suppose $e \lhd f$. The transitivity of proper parthood guarantees that $d \lhd f$.

  • (ii) Suppose $f \lhd e$. Since e ≼N, the characterization of N guarantees that e = b or $e \lhd b$. If e = b, it immediately follows that $f \lhd b$. And if $e \lhd b$, the transitivity of proper parthood guarantees that $f \lhd b$. Since f ≼M, the characterization of N guarantees that f ≼N. And since d is a proper part of every x ≼N, $d \lhd f$.

  • (iii) Suppose e = f. Then since $d \lhd e$, $d \lhd f$.

  Any way, $d \lhd f$. And since f is arbitrary, d is a proper part of every x ≼M. So some y is a proper part of every x ≼M. It now follows that the chain formed by M is not maximal, contradicting the result of the above application of the Hausdorff maximal principle (see step (2)), which ensures that it is.

• (e) I now show that the chain formed by N does not terminate. Suppose for reductio that it does. Then there is a d ≼N such that, for every x ≼N, if x≠d then $d \lhd x$. Since, by result (3)(b), N ≼Z, d ≼Z. Then, by result (1), d has a proper part e ≼Z. Now consider an arbitrary f ≼N. Either f = d or f≠d.

  • (i) Suppose f = d. Then, since $e \lhd d$, $e \lhd f$.

  • (ii) Suppose f≠d. Then, by the reductio assumption, $d \lhd f$. But since $e \lhd d$, the transitivity of proper parthood guarantees that $e \lhd f$.

Either way, $e \lhd f$. Since f is arbitrary, every x ≼N has e as a proper part. But by result (3)(d), N is maximal, and so there is no y that is a proper part of every x ≼N. Contradiction. So the chain formed by N does not terminate.

But since, by result (3)(d), the chain formed by N is maximal, the initial assumption guarantees that it terminates. Contradiction. Hence a has an atomic part. And since a is arbitrary, everything has an atomic part. □