Actually, the demon only switches the light on and off a countable number of times in this example.
Which premise specifies that? I thought he always hits the switch halfway to the two minute mark.
I’m not sure exactly what you’re asking, zwaldd, but the demon hits the switch for the nth time at time t = sum(1/2[sup]k[/sup], 0 < k < n - 1). So as close as you get to two minutes, the demon has only hit the switch a finite number of times. Does that make sense?
It’s a countable process embedded in an uncountable field; if the field was countable, there would be a ‘most recent event’ and we wouldn’t still be here rattling on about it 
Shade hit the nail bang on the head - we cannot determine what happens at 2 minutes because there is no most recent event from which to draw our conclusion. Say the lamp is on at 1min 59.99999999999 seconds - the next time he hits the switch it will be off, and then on, and then on…there is no point where “if he hits the switch one more time then he will get to two minutes”, and so, for the last time, we cannot determine what happens at 2 minutes!
And zwaldd, I should have been more specific. In the context of the problem, his speed tends to infinity as time tends to 2 minutes (as we don’t know what happens AT two minutes we can’t specify his speed there; for all we know, he takes a coffee break! Mmmmm, coffee…)
His speed is finite and calculable at each stage of the problem, but it tends asymptotically to infinity. The ‘to do infinite things in a finite time you need to move at an infinite speed’ comment was just supposed to illustrate my point, but I admit that wasn’t very clear - thanks for the oppourtunity to clear that up.
I’m not sure exactly what you’re asking, zwaldd, but the demon hits the switch for the nth time at time t = sum(1/2[sup]k[/sup], 0 < k < n - 1). So as close as you get to two minutes, the demon has only hit the switch a finite number of times. Does that make sense?
Actually, that’s a larger cardinality than that of the natural numbers, but not necessarily the next largest. The next largest is omega-1 (or aleph-1); in ZFC, it is consistent that the reals have cardinality omega-1 (the continuum hypothesis), but it’s also consistent for the reals to have just about any larger cardinality, as well. The issue is independent of ZFC.
That’s somewhat debatable, depending on what you require. For something like the OP, we define the “times of the events” by taking a continuous, order preserving function f:omega -> [0,2] (omega = the natural numbers). So f(1) is the time of the first event, f(2) the time of the second, and so on.
If we require a similar setup for an uncountable number of events, such that the times of the events are well ordered, and are given by some continuous, order preserving function g:A -> [0,2], then it can’t be done. If we take the smallest uncountable ordinal, omega-1, then there is no continuous, order preserving function g:omega-1 -> [0,2]. If this function is continuous, then it must be eventually constant, so it can’t be order preserving (or even 1-1).
So, subject to those restrictions, if we have an infinite number of actions performed in any length of time (not necessarily finite, either), the number of actions must be countable.
Here is a link to a page on supertasks, which has some discussion on Thomson’s lamp in particular (see sections 2.3 and 3.1). In the later section, it discusses “physical” models of the lamp, and shows two, one where the “model” predicts on at the end, and a second where the “model” predicts off at the end.
Makes sense, but his task is not complete after a finite number of switches. The only way he can stop hitting the switch while staying true to his task is if he finds an interval where the halfway point to two minutes is exactly the same as the entire way to two minutes. There is only one theoretical moment where this occurs, and that moment is infinity, because half of infinity is infinity.
The OP asks whether the light is on or off at the end of two minutes, not at a certain point close to two minutes. I agree that at any given point prior to two minutes, the demon has only hit the switch a finite number of times. At two minutes however, he has hit the switch an infinite number of times. Infinite is not countable.
Not necessarily. Since the demon only acts at rational times, we can ignore the irrationals, but the rationals are still self-dense.
Quoting myself:
I forgot to include “for some well ordered set A”, but that’s probably clear from context.
Getting back to my original point…the only way there could be an interval of infinite duration, where halfway to the next interval = the entire way to the next interval, is if the passage of time has completely stopped, i.e., the next interval will never arrive. This is the only condition under which the demon can stop hitting the switch and still be on task.
zwaldd:
Oh. As it turns out, you’re using a definition of countable that’s entirely different from what I and the others are using. You’re taking “countable” to mean “finite”, while in mathematics, “countable” means “finite, or having cardinality equal to that of the counting numbers”.
So, for example, the infinite set of integers is countable.
I stand corrected.
Without using the Continuum Hypothesis, is it even possible to prove that Aleph[sub]1[/sub] exists? That is to say: Aleph[sub]1[/sub] is defined as the lowest cardinal which is strictly greater than Aleph[sub]0[/sub], correct? How do we know that there exists any single cardinal which is the minimum of that set?
You do not need the Continuum Hypothesis to prove Aleph[sub]1[/sub] exists. I think you need the Axiom of Choice – in the form of the Well Ordering Principle.
Drifting away from the point a little here guys…not that a little revision of last year’s calculus class isn’t helpful (and needed :()
ultrafilter, point taken about the irrationals. Hey, I had to slip up somewhere, thought I did pretty well back there on page 1 
On.
Definitely on.
Actually, you don’t need the axiom of choice, either. You can construct the ordinals (and cardinals) without the axiom of choice. You do need the axiom of choice to well order an arbitrary set, and to define the cardinality of an arbitrary set.
Thanks Cabbage, I wasn’t sure.