First Known Solution to Thomson’s Lamp

Attached is a draft academic paper detailing what I believe to be the first known solution to the paradox posed by Thomson’s Lamp. The paper discusses the paradox, Benacerraf’s critique, and the philosophical consequences of being able to prove a supertask possible. It is currently in draft form and will be revised further. The full solution is already contained within, and that section is unlikely to be revised further. I have reproduced the solution below.

This post and the attached paper are copyright Abhishek Bose-Kolanu 2014, all rights reserved.

DRAFT bobo thomsons lamp first publish

Solving Thomson’s Lamp

While specifying the physical possibility of supertasks is outside the focus of this paper, 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” (Thomson, Zeno’s Paradoxes, 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 (e.g. on –> off –> on, 2 switches) 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 the natural number set 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 (see “Numbers, Transfinite and Ordinal” of this paper). 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 an integer r as a set s that contains all integers 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.

 

We can now put to rest Thomson’s anxieties about having a “method for deciding what is done when a super-task is done” (6). However, the generalizability of our results is an open question. “One difficulty, then, about the question of whether ω-tasks can be completed is that there are different kinds of them, and there is no reason to think that in regard to completability they stand or fall together” (Thomson, Zeno’s Paradoxes, 136). Nonetheless, it is a significant result to be able to declare that there exists a supertask for which a logical rendition of the state of the world following its completion may be given. That this supertask happens to be the ur-supertask is a happy coincidence.



[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 assigned the states ‘0’ and ‘1’ to ‘lamp off’ and ‘lamp on,’ and asked what the sum of Grandi’s sequence could 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 divergent sequences favorable to this approach.

 

bobo

 

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