Infinitesimal Gunk

Abstract

In this paper, I advance an original view of the structure of space called Infinitesimal Gunk. This view says that every region of space can be further divided and some regions have infinitesimal size, where infinitesimals are understood in the framework of Robinson’s (1966) nonstandard analysis. This view, I argue, provides a novel reply to the inconsistency arguments proposed by Arntzenius (2008) and Russell (2008), which have troubled a more familiar gunky approach. Moreover, it has important advantages over the alternative views these authors suggested. Unlike Arntzenius’s proposal, it does not introduce regions with no interior. It also has a much richer measure theory than Russell’s proposal and does not retreat to mere finite additivity.

Appendix: Countable Saturation

The language of standard analysis ${\mathcal{L}}$ that we focus on includes constant symbols for all real numbers and set constructions from real numbers through the iterations of the powerset operation and union. More precisely, let $U_{n}(X)=U_{n-1}(X)\cup \mathcal {P}(U_{n-1}(X))$ and $U(X)=\bigcup _{n=0}^{\infty } U_{n}(X)$. The language includes constants for all members of $U(\mathbb {R})$, which have associated ranks according to the least $U_{n}(\mathbb {R})$ they belong. ${\mathcal{L}}$-terms and ${\mathcal{L}}$-formulas are defined in the usual way [8, 166-7]. We can show that all the familiar functions and relations in standard analysis (such as addition, integration, Lebesgue measure), considered as sets, are members of $U(\mathbb {R})$ and therefore are referred to by ${\mathcal{L}}$-constants [8, 165-6]. Note that these functions and relations are not just defined over real numbers but can have a variety of ranks. The hyperreal system under consideration (along with the set constructions) $U^{\prime }$ is an alternative model for ${\mathcal{L}}$ that is an expansion of $U(\mathbb {R})$. There is a unique transfer map from $U(\mathbb {R})$ to $U^{\prime }$ that preserves all the ${\mathcal{L}}$-truths. Members of the image of the transfer map are called standard. For example, the hyperreal line $*\mathbb {R}$ is the image of $\mathbb {R}$ under the transfer map, and is therefore standard. An entity is internal iff it is a member of a standard set (Goldblatt 1998, 172). Any hyperreal number is internal because it is a member of $*\mathbb {R}$. Any hyperreal interval is internal because it is a member of the set $\{X\mid (\exists a,b\in *\mathbb {R})(\forall x\in X)(a\leq x\leq b)\}$, which is the image of the set of all real intervals. A bijective internal function belongs to a set of functions characterized as “bijective” in ${\mathcal{L}}$ in the usual way. It turns out that all sets in the form of {x∣ϕ(x)}, where ϕ is a formula in the language ${\mathcal{L}}^{\prime }$ that extends ${\mathcal{L}}$ with constants for internal entities, are internal sets [8, 177]. For example, ${\mathcal{L}}^{\prime }$ has constants for all hyperreal numbers. A hyperreal interval {x∣a ≤ x ≤ b} ($a,b\in *\mathbb {R})$ is therefore internal. On the other hand, we can prove that any infinite set of real numbers (e.g., $\mathbb {R},\mathbb {N}$) is not internal [8, 176].

In the hyperreal models we are interested in, the internal sets satisfy the following property:

Theorem A.1 (Countable Saturation)

The intersection of a decreasing sequence of nonempty internal sets $X^{1}\supseteq X^{2}\supseteq ...$is always nonempty [8, 138].

Countable Saturation implies this principle:

Corollary A.2 (Nested Intervals)

For any countable nested sequence of intervals $I_{1}\supseteq I_{2}\supseteq ...,$their intersection is non-empty and includes an (open) interval.

Proof

All hyperreal intervals are internal sets. Thus, according to Countable Saturation, the countable nested sequence of intervals I₁, I₂,... have non-empty intersection. Moreover, the interiors of these intervals also have non-empty intersection. Let x be a point in the intersection of their interiors. Then, the intersection of the parts of the intervals to the right of x is non-empty. Let y be a point in this intersection. Then, [x, y] is included in the intersection of I₁, I₂,.... (Clearly, the intersection also contains the open interval (x, y).) □

With Nested Intervals, we can prove that Infinitesimal Gunk violates Countable Basis:

Theorem A.3

Under Infinitesimal Gunk, the topology of space does not have a countable basis.

Proof

In this proof, we will use this fact: if a set of regions ${\mathcal{B}}$ is a basis for a gunky space, then every region in that space contains some region in ${\mathcal{B}}$. Let $\mathcal {C}$ be any countable set of regions. Take an arbitrary point x on the hyperreal line, and consider the set of all elements in $\mathcal {C}$ that include x in their interiors. Call this set $\mathcal {C}_{x}$. Since $\mathcal {C}$ is countable, $\mathcal {C}_{x}$ is also countable. It follows from Nested Intervals that there exists an infinitesimal neighborhood Δ of x that is included in all elements of $\mathcal {C}_{x}$. Take a closed infinitesimal interval that is strictly included in Δ. This interval does not contain any element of $\mathcal {C}$, so $\mathcal {C}$ is not a basis. Thus, a gunky space does not have a countable basis. □

In Section 5 (p.23), I mentioned that the fusion of any countably infinitely many disjoint measurable regions is not measurable. This claim can be derived from the following corollaries of Countable Saturation:

Corollary A.4

If an internal set X is a countable union of internal sets X₁, X₂,..., then there is a natural number k such that X is the union of X₁,..., X_k [8, 139].

Proof

The proof is adpated from Goldblatt [8, 139-40]. Suppose that for all $k\in \mathbb {N}, X- \bigcup _{n\leq k} X_{n}$ is non-empty. Since $X - \bigcup _{n\leq k}X_{n} = \bigcap _{n\leq k} (X-X_{n})$, we have that $\bigcap _{n\leq k} (X-X_{n})$ is non-empty. Call this set Y^k. Then 〈Y^k〉 is a decreasing sequence of non-empty internal sets. So, by Countable Saturation, there is a point belonging to Y^k for all k, and thus to X − X_k for all k. Therefore, X is not the union of X₁, X₂,.... □

Corollary A.5

For any countably infinitely many disjoint measurable sets, their union is unmeasurable.

Proof

Every measurable set is a union of hyperfinitely many disjoint intervals. This in fact guarantees that it is an internal set. Let A₁, A₂,... be countably infinitely many disjoint measurable sets and let A be their union. Suppose A is measurable. According to Corollary B.4, it follows that A is the union of finitely many A_i. But since A₁, A₂,... are infinitely many and disjoint, their union is not identical to the union of any finitely many A_i. Thus, A is not measurable. □

Corollary A.6

For any countably many disjoint regular closed sets, their union is regular closed.

Proof

Let A₁, A₂,... be countably many disjoint regular closed sets on the hyperreal line. Let A be their union. I will show that A includes all its limit points and is therefore closed. Take any point y outside A. For each regular closed set A_j, there is an open interval that includes y and is disjoint from A_j. According to Nested Intervals, the intersection of these open intervals includes an open interval which includes y and is disjoint from A. Thus y is not a limit point of A. Since y is arbitrarily chosen, no point outside A is a limit point. Therefore, A is closed. □

In Section 6 (p.30), we need to show that there are no “trouble-making” boundaries when it comes to the union of hyperfinitely many regular closed sets.

Theorem A.7

For any hyperfinitely many measurable regular closed sets, their union is regular closed.

Proof Sketch

In standard analysis, we have the following induction principle for the natural numbers: an ${\mathcal{L}}$-formula ϕ with one free variable is satisfied by every natural number (taken as 1,2,...) if (1) ϕ is satisfied by n = 1; (2) if ϕ is satisfied by any natural number n, then it is also satisfied by n + 1. In nonstandard analysis, we have an analogous induction principle for the hypernatural numbers: an ${\mathcal{L}}$-formula ψ with one free variable is satisfied by every hypernatural number if (1) ψ is satisfied by N = 1; (2) if ψ is satisfied by any hypernatural number N, then it is also satisfied by N + 1. Since any set with a hyperfinite cardinality N can be ordered under an internal bijection to {1, 2,..., N}, we will pick any such ordering of the set of hyperfinitely many measurable regular closed sets in question. Now, we can easily confirm the following: (1) the union of the singleton set of a measurable regular closed set is (trivially) regular closed; (2) if the union of the first N measurable regular closed sets is regular closed, then the union of the first N + 1 measurable regular closed sets is also regular closed because the union of two regular closed sets is regular closed. Also, these expressions can indeed be put into ${\mathcal{L}}$-formulas. Then according to the induction principle, for any hypernatural N, the union of the first N measurable regular closed sets is regular closed, which is just what we want. □