First Known Solution to Thomson’s Lamp – Update

My complete paper is now available as a draft for review, but not for distribution. I hope to get it published in an academic journal of philosophy. The main proof concerning Thomson’s Lamp has not changed, but I expanded it to anticipate what I think will be a common objection. That proof is reproduced below, and the draft paper is attached.

This post is an update of which was published on June 7th, 2014, though I already had the solution sketched on May 13, 2014 (when I started working on the rest of the paper). To my knowledge this is the first known publication of a proof for Thomson’s Lamp.

bobo Turning Out The Light on Thomsons Lamp Paper DRAFT


Solving Thomson’s Lamp

Can we establish that there exists a supertask that is logically possible? We maintain the following original solution constitutes the first known solution to Thomson’s Lamp ‘on the merits.’ Whereas Benacerraf’s solution involved pointing out a contradiction in the particular proof Thomson attempted to construct after assuming the existence of his lamp, our solution takes as given the logically defensible portion of the original thought experiment and proceeds to answer the nexus question: What state is the lamp in after two minutes of infinite button presses have elapsed?


As we have shown, there are compelling reasons to continue to ask this original question Thomson posed, and Thomson himself corroborates the possibility of a proof’s appearance, writing “I am now inclined to think that there are no simple knock-down arguments to show that the notion of a completed ω-task is self-contradictory” (ZP, 131). Given the original experimental setup, and ignoring Thomson’s original and invalid claim that no solution is possible due to contradiction, what state is the lamp in after two minutes of infinite button switches?


As Thomson notes, an even number of switches will maintain the lamp’s state while an odd number will change it (5). The only question that requires answering is, “Is infinity even?” It is, when we use the set-theoretic definition of even to mean capable of subdivision into two disjoint subsets of equal cardinality, or size. As we will show, this set-theoretic definition of even is equivalent to the numerical, “divisible by 2” definition to which we are accustomed.


It is trivial to show that a countably infinite set can be divided into two disjoint subsets of equal cardinality. Consider the set ℕ of natural numbers. From the definitions for even and odd we know that if a given integer j is even, j+1 is odd. From the definition of ℕ we know that for all j, j+1 exists. Accordingly we can subdivide ℕ into two disjoint subsets of evens and odds respectively, both of which have the same cardinality as each other.[1]


One might object that the set-theoretic definition of even may not apply, since Thomson was asking a question about the number of tasks performed, and of what number (1 or 0) would represent the lamp’s final set. However, the two definitions are equivalent, once the construction ‘number’ is understood set-theoretically. We traditionally define even and odd as follows.


Eqn 1.1 An even integer n = 2k where k is some other integer

Eqn 1.2 An odd integer m = 2k + 1 where k is some other integer


We can rewrite these definitions in terms of the set-theoretic definition given above. Let us define a natural number r as a set s that contains all natural numbers that precede r, starting from and inclusive of 0. So again we have the integer 5 defined as the set {0, 1, 2, 3, 4}. Note that the cardinality of s is equivalent to r. Let us use the set-theoretic definition of odd and even to define the oddness or evenness of set s. If set s can be divided into two disjoint sets (whose union contains all its elements) that are equinumerous (whose cardinalities are equivalent), then we say set s is even. If not, set s is odd.


Let us denote the sets u and v as the two disjoint subsets into which we divide s. Let us denote the cardinality of u and v as w and x respectively. If and only if w = x, can we say that the set s has successfully been divided into two disjoint subsets of equinumerality. However, if w = x, then we can express r, the cardinality of set s, as 2 * w. If not, we express r as 2 * w + 1, since at most the count can be off by one (since we have been choosing from among two subsets, and thus if we had two extra items left in our parent set, then each subset would have received one, etc.). Thus we have arrived at our so-called ‘numerical’ definition of even from the set-theoretic.


We can see that the two disjoint subsets into which we divide the button presses of the Thomson Lamp can be none other than the sets of ‘switch to on’ and ‘switch to off,’ just as they appeared in Thomson’s use of the Grandi sequence[2] (6). What is novel is the knowledge that the termination of the supertask depends, in this case, on the initial configuration. If the lamp was on, it remains on at the conclusion. Else, it remains off.


One might object that even though the cardinalities of the two subsets are equivalent, both are ℵ0, which is the same cardinality as our original set ω. Hence, it might be impossible to determine which of our ℵ0 subsets terminates ‘after’ the other, voiding our application of evenness to the proof. However, such worries are easily put to rest once we view the evenness proof as a restatement of 2ω = ω. That 2ω = ω is clearly understood once order-types are applied rigorously. Since ω is an ordinal we know it is specifying a positioning, or sequencing of elements. It is the order-type ω. Hence, to say that 2ω = ω is simply to say that ω pairs in sequence possess the same order-type as ω singletons in sequence.


At first this explanation does not seem to extricate us from our quandary, since the pairs are obviously (switch, switch) pairs of button presses (Thomson’s claim of never once turning the lamp on without turning it off and vice versa). However, we can use our initial knowledge about the lamp’s state at the beginning of the experiment to fix the state of the lamp upon completion. Since the ω pairs of (switch, switch) will take place regardless of what state the lamp is in at the beginning, and since we know that an even number of state transformations preserves state, we can see that the initial state of the lamp can be removed without affecting the even number of state transformations that take place. And in fact, it is removed, since the initial state does not count as a button press. Hence, the initial state remains the final state as previously described.[3]


A particularly vicious reader might inquire if 3ω = ω, thereby demonstrating an arbitrary decision to subdivide into two disjoint subsets instead of three. Perhaps this argument points to slippage between the set-theoretic and numerical definitions of parity, which earlier were claimed to be equivalent. No such slippage is evident. That a number (set) may also be divided into thirds does not mean it cannot also be even.


That many possibilities for subdivision exist should not dissuade us from maintaining the possibility for subdivision into two disjoint subsets of equal cardinality. All that remains is to show that the possibility for a quantity to be subdivided in more than one way does not void the property of subdivision into two in which we are interested.


Division into singletons requires division into ℵ0 many subsets. Division into pairs also requires ℵ0 many subsets. So does division into thirds, or fourths, or fifths…. Division into one subset of size ℵ0 is the case where the subset is equivalent to the parent set, and no division needs take place. Strictly speaking, division into equinumerous subsets of any size < ℵ0 requires ℵ0 many subsets.


What this means is that our first inclination, to hold on to the evenness of ω despite the fact that it is also divisible into subsets of many other sizes (indeed, infinitely many), is correct. If this poses a problem for the reader it is due only to the limitations of language and imagination, rather than to some logical inconsistency. In the first case, with the challenge of dividing into a sequence of triples instead of pairs, we can easily imagine a number that could be divided either way, but which nonetheless possesses the property of evenness in which we are interested, e.g. six. Were six button presses to take place, we could order them as {(switch, switch, switch), (switch, switch, switch)}, or as {(switch, switch), (switch, switch), (switch, switch)}. In either case, the fact that six is capable of subdivision into two disjoint subsets with equivalent cardinality is sufficient to guarantee that the final state of the lamp remains the same as its initial. We are also protected from an instance of a subdivision into triples that is not also possible of subdivision into pairs, since we know that there are infinitely many button presses, and thus for any button press i that happens before the two minutes is up, i+1 exists (because ω – 1 = ω). Or put another way, for any set of j triples, we know that j+1 triples exists, and thus there is no division into subsets of an odd size that cannot also be divided into pairs. It is for this reason that ω, like all limit ordinals, is even.

[1] They happen to also have the same cardinality as the original set ℕ, since all three sets are countably infinite. This quality is at the root of Galileo’s famous perfect squares paradox.

[2] The sequence is 1 – 1 + 1 – 1…. Thomson assigns the states ‘0’ and ‘1’ to ‘lamp off’ and ‘lamp on,’ and asks what the sum of Grandi’s sequence can tell us about the lamp’s final state. In our opinion the question is misguided, as the sum of this divergent series is given as ½, which is not a reason for supertasks being internally self-contradictory, as Thomson supposes, but instead is due to the particular definition of a Cesàro sum, whereby a second converging series is constructed by the partial means in order to find a consistent value to assign as the sum for certain divergent sequences.

[3] For another way of thinking through this proof, we might consider what was at stake when Cesàro sums became the accepted way of fixing the sum of infinite divergent series. The need for a way to fix the sum was precisely because alternative methods, all of which seemed reasonable, gave inconsistent results. If one took Grandi’s sum as 1 + (-1 + 1) + (-1 + 1)… then the answer was 1. If one grouped the terms as (1 – 1 ) + (1 – 1)… then the answer was 0. We can view the initial state of the lamp as an instruction on how to group the terms.




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