The following series of posts deal with the dichotomies of choice and determinism in math, as well as the corresponding theological dichotomy of free will and divine providence. How does each field deal with the dichotomy? Is there even a real dichotomy between the two?
The Axiom of Choice
When I was an undergrad, my math program required us to write an undergrad thesis in order to graduate. It could be some original research or a more in depth exploration of a mathematical topic. At the time, I had no idea what to research, but I hoped to become a logician. My very cool (and unknowingly prescient) advisor suggested an old school theorem which incorporates ideas from both Logic and Analysis (my eventual field), as well as a smattering of Group theory, known as the Banach-Tarski theorem. I'm glad I did it, as it was quite helpful in learning both how to read academic papers and texts at a higher level, as well as introducing me to some basic concepts in graduate level Mathematics, especially as my own undergraduate program was rather small.
To put it in layman's terms, the Banach-Tarski theorem states that the unit ball in three dimensional Cartesian space $\mathbb R^3$ can be partitioned into finitely many (very weird) sets (as small as 5 sets), and by rotating and shifting these sets alone, reconstruct two unit balls out of that one. It sounds pretty magical, but what is at work is how these sets are constructed to begin with... or rather not. In fact, they can't be "constructed" in any "humanly conceivable" way. The key principle in finding such sets is the Axiom of choice. One formulation of it goes something like this: Suppose you have a set $A$ whose elements are non-empty disjoint sets. $(A_i)_i$ (just think of the $i$'s as elements in some indexing set $I$). Then there is a set that contains exactly one element in each of the $A_i$'s in $A$.
Seems plausible, right? But there is something sneaky about this. First of all, when you employ the Axiom of Choice at its fullest potential, this set $A$ can be of arbitrary size. If $A$ is finite, you don't even have to use choice, but if it's infinite, even countably so, then some amount of choice is may be necessary. Furthermore, such a set created by choice is what mathematicians folklorically call "non-constructive." We are just told that such a set exists, but we don't really know how to build it. Intuitively, it's because we as humans tend to reason in discrete steps, but the construction can take uncountably many steps if the indexing set $I$ mentioned above is uncountable.
Here a nuance must be made clear, not all representative sets require choice. For example, suppose that for my collection $(A_i)_i$, every set $A_i$ is a subset of the natural numbers indexed by $i$. Then I can easily just pick the smallest number. But this no longer requires choice, as the "smallest number" in a set of natural numbers is something one can theoretically distinguish beforehand.
Oddities of Choice: non-measurable sets
So when does the axiom of choice non-trivially appear? A common example where a student might first witness a funny outcome of choice is in the construction of non-measurable sets. But to understand this, we must first understand what a measure is. To illustrate the properties of a measure, let's work on the unit interval $[0,\infty]$. In set theory, we might consider the cardinality of a set to be its size, but $[0,1]$, and any other interval inside it, has uncountably many points. Yet for us, the interval is still "small", geometrically speaking. If we measure it, we get, well a line of length 1. More generally, a measure is a more "geometric" way to assign size to subsets of a set. Formally speaking, it is a (partial) function $\mu$ whose inputs are (as we will see, certain) subsets of the unit interval, and whose output is some nonnegative number.
On the onset, we want this measure $\mu$ to behave in intuitively nice ways for us. For instance, if we have an interval $[a,b]$, we want $\mu$ to correspond to the length of the interval: $\mu([a,b]) = b - a$. It's also ok for non-empty sets to have measure 0. This also works with our intuitions; after all, what is the length of a single point in the real line? We also can work beyond the unit interval, picking a larger set such as $\mathbb R$ or $\mathbb R^n$.
Now, we want some ground rules for how such a measure works.
1. Monotonicity: if $A$ and $B$ are measurable sets with $A \subseteq B$, then $\mu(A) \leq \mu(B)$.
2. Countable additivity: if $(A_i)_i$ is a sequence of mutually disjoint sets, then
\[ \mu \bigg(\bigcup_i A_i \bigg) = \sum_i \mu(A_i) \]
(Note: you can have sets of infinite measure. In such a case for some measurable set $A$, we just let $\mu(A) = \infty, $ but don't think about set theoretic sizes of infinity)
3. Closed under countable unions, intersections, and complements.
With those requirements above, we've got a pretty good rigorization of size in a geometric, and not merely set theoretic sense. Observe also that we don't care much about countably many points. In our particular case, the measure of an open interval $(a,b)$ is the same as that of $[a,b]$. Also, any countable set has measure 0. This can get pretty strange for some people, since it implies, for example, that the set of rational numbers in $[0,1]$ has measure 0. It helps to think of such sets as "dust" in comparison to the intervals, however, even though the dust seems to be everywhere.
But is every set measurable?
It turns out this is not the case. Going back to the Banach-Tarski theorem, for instance the weird sets partitioning the unit ball have to be non-measurable sets. Here, we will give a different, earlier example of a non-measurable set that has some of the flavors of the Banach Tarski partitions. It is called the Vitali set, after the mathematician who discovered it. It's described in the following way: define the following relation $E$ on $[0,1]$ by
\[ s E t \text{ iff } s - t \text{ is rational } \]
In other words, if the difference between two numbers is rational, we'll say they are related. Now, for a given $s \in [0,1]$, let $E_s$ be the set of all $0 \leq t \leq 1$ so that $s E t$. $E$ is a special type of relation: an equivalence relation, as it is reflexive, transitive, and symmetric. Based on this, we can show that all the distinct $E_s$'s a partition of $[0,1]$, and given any $E_s$ and $E_t$, either they are disjoint or they are the same set (for the mathematically curious who aren't familiar with this construction, try to prove why this is the case using the definition of equivalence relations!). Also, each $E_s$ is countable in size, as there are countably many rational numbers, and $E_s$ is defined to be all the $t$ in the unit interval that are a rational distance away from $s$. Since there are countably many elements in each $E_s$, there are uncountably many distinct $E_s$'s (again, if you haven't seen this before, try to reason why this must be the case).
Here is where we use choice: for each distinct $E_s$, choose exactly one "representative" element $s'$, and put it into a set $V$. it doesn't matter which, any single element will do. It turns out that this $V$ is not measurable. to see why this is the case, we first observe the following: for every rational $q \in [0,1)$, let $V+q = \{ (s +q) \text{ mod 1}: s\in V$. Basically, $V+q$ is a modular translation of $V$: shift all the elements of $V$ by $q$, and subtract 1 from any of the resulting elements greater than 1. Three observations come to mind:
1. All the $V_q$'s are mutually disjoint. This follows from the property that $V$ contains exactly one element from each of the distinct $E_s$'s.
2. The union of all the $V_q$'s is the unit interval: for any $r \in [0,1]$, find $s \in V$ such that $r-s$ is rational. Then $r$ itself is in the set $V+(r-s)$.
3. If $V$ is measurable, then all the $V_q$'s have the same measure. Think of it this way: split $V$ into two sets: $V \cap [0,1]$ and $V \backslash [0,1]$. These sets should then also be measurable. $V+q$ shifts $V\cap [0,1]$ by $q$, and shifts $V\backslash [0,1]$ back by $1-q$. These are just translations, and translations preserve measures.
So if $V$ is measurable, then either $\mu(V) = 0$ or $\mu(V) > 0$. Seems obvious enough. But it can't be either, because of the countable additivity property for measures. If it were, then because the $V_q$'s partition $[0,1]$ the sum of the $\mu(V_q)$'s should be 1. But a series of infinitely many zeros is still zero, so $\mu(V) > 0$. Yet a infinite sum of $\mu(V)$ is also infinite. In other words, assuming that $V$ is measurable leads to a contradiction. So we must conclude that $V$ is not measurable.
But what is Choice anyway?
Now though such a set $V$ may exist, we can only show it by assuming the axiom of choice; in order to construct $V$, we had to choose an arbitrary element from each of the $E_s$'s. Also, we have no idea what this set "looks" like. It's not a constructed set in the sense that there are clear steps on how we distinguish certain representative points. The arbitrary nature of the selection is why the logician Bertrand Russell was said to state about the axiom of choice that you don't need it to select one shoe for each pair when there are infinitely many pairs of shoes (just pick the left one!), but that you need it to select a sock from each pair of infinitely many socks.
One thing that is philosophically, and perhaps ideologically, interesting about this is that the act of calling this procedure "choice" says something about those who came to use or analyze the axiom and call it the Axiom of "Choice".
In a sense, the mathematical notion of "choice" in practice has a very particular feel to it in that choice is associated with arbitrariness. When you can distinguish some object according to unique properties, and these properties are mathematically definable in the appropriate ways, you don't necessarily need "choice" to isolate the object. But if you want to distinguish a point from other points that are not necessarily distinguishable from it, except that they are just not the same points, then the axiom tells you, "just pick one." And on an intuitive level, why would that be a "wrong" mathematical move to make? It's not in itself, if the mathematician is operating under a theory where mathematical choice is allowed, and a lot of fields like analysis assume it. I myself have used the Axiom of Choice or some mathematical equivalent of it several times in my own work, often implicitly, but sometimes more explicitly.
But this notion of choice as unfettered arbitration is found in all sorts of fields, not just in math. For instance, in neuroscience, there is the Libet experiment, which found that for a certain simple tasks like when to push one of two buttons, the time the subject made the decision to push the button was preceded by some milliseconds by neurons firing in certain regions of the brain before the subject was actually conscious of their decision. Such an experiment has been used by pop materialists and determinists to argue that free will does not exist, despite its flawed design and limitations of implication. Here, the experiment's structure itself (pressing one of two buttons without any prior prompting or reasoning) reveals an understanding of choice that, in its rawest and most rudimentary of forms, involves at some level mere arbitration.
However, is this how all fields understand choice and free will? In the next article, let us see what a few theologians have to say about that!
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