Friday, April 3, 2009

More on the Kalaam argument

Here is another counterexample to the claim that it is impossible to form a concrete and actual infinity through successive addition. Consider a gamma-widget. A gamma-widget is a stochastic critter that, once it is produced, produces at least one offspring. How many offspring it produces and how quickly is random. For every n>0, a gamma-widget has probability 2n of producing exactly 22n offspring in exactly 2n years. The offspring are exact copies of the parent (this is a way in which the gamma-widget differs from the beta-widget). They, too, are gamma-widgets, and hence produce offspring. All the probabilities are independent.

Here is a mathematical fact (at least, it is a fact, if my scribbled calculations are right): There is a non-zero probability that if a gamma-widget comes into existence, it will be an ancestor of infinitely many gamma-widgets a year later. What has non-zero probability is also possible. Therefore, it is possible for this to happen. Here is another fact: The probability that there is a finite number n such that n years later there are infinitely many gamma-widgets is one. What is almost certain (i.e., has probability one) is a fortiori possible. Therefore, it is possible to have an infinite number of gamma-widgets arise through successive addition.

(If one is worried that the addition is not successive because multiple offspring are produced at the same time, we can stagger the productions in some way, and then not count the ones whose productions happen to overlap.)

Does this argument, and that in my previous post, destroy Craig's Kalaam argument? I think not yet. For there two ways that an infinity prima facie can be built up by successive addition. In one way, there is a first addition but no last one. In the other way, there is a last addition but not first one. What my examples show is that the first way is possible—there is a first addition but no last one. But I have not shown that there is no possibility of successive addition of the other sort.

However, the latter possibility follows from a modification of the case. Suppose that each gamma-widget that comes into existence before the year 2020 randomly and uniformly chooses a time during 2020, and completes a mug at that time. Then there is, almost surely, a subset of the mugs that has a last addition but no first one.

Craig can get out of these arguments by supposing that time cannot be infinitely divisible. Or he might just deny the possibility of my widgets. But what grounds would there be for that denial? Consider a variant on the gamma-widget, the delta-widget. It is just like the gamma-widget, except that it has probability 2n of generating n2 offspring in 2n years. The delta-widget is just like the gamma-widget, except it is much less prolific: n2 instead of 22n. I am pretty sure (I haven't checked) that with probability one, delta-widgets do not have an infinite population explosion. So Craig can't object, it seems, to delta-widgets except by denying the infinite divisibility of time. But if delta-widgets are possible, why not the more prolific gamma-widgets?

Maybe my calculations are wrong. But I am pretty sure they're not. If they are wrong, replacing 22n with an even faster growing sequence should fix things.

17 comments:

Alexander R Pruss said...

Oh, and of course none of this does any damage to the a posteriori version of the Kalaam argument.

Doug said...

When you mention the "a posteriori" versions of the argument, are you referring only to the scientific arguments that the universe began to exist?

Tentatively, I think we can form a cogent inductive argument that the universe began to exist. We might say:

1. The things we observe that end also have beginnings.
2. The universe's past has ended in the present.
3. Therefore, the universe probably had a beginning.

(2) is fairly benign, so the bulk of the defense would rest on (1). Imagine I tell you I just finished reading Popper's "Objective Knowledge," and you ask me when I started. If I tell you I never started, you would look at me like I were crazy, or simply jesting. Similar examples can likewise be thought of.

Were you thinking of an a posteriori argument like this one, or would you restrict it to the two scientific arguments Craig appeals to?

Alexander R Pruss said...

That's a clever argument. It only works given an A-theory, though. Otherwise, (2) is merely a trivial indexical fact.

Drew Mazanec said...

I was wondering a bit about the Hilbert Hotel paradox.

Under a B theory of time, wouldn't an infinite future imply the existence of an acutal infinite?

Alexander R Pruss said...

It sure would, assuming there was new stuff coming into being all the time. That's a nice point.

Drew Mazanec said...

Oh, no. Well, the B theory seems unsalvageable, then, and divine timelessness by extension.

How then do you reconcile divine foreknowledge with libertarian free will?

Alexander R Pruss said...

I accept the B-theory. :-)

Moreover, I see no reason why there should be a great difficulty about foreknowledge, even in an A-theory, as long as one isn't a presentist and is willing to allow that God is causally affected by future free actions. (If one accepts an A-theory, one is anyway going to give up on God's immutability. That's pretty bad. But I don't know that it's much worse then to give up on God's not being causally affected.) There is no really good argument against the possibility of backwards causation, I think.

Drew Mazanec said...

I, too, find the B-theory far more intuitively obvious. I just cannot reconcile it with the paradox of Hilbert's hotel.

I assume that if there is an infinite future (infinite number of future events) then any non-presentist view of time involves actual infinites and crashes into Hilbert's hotel. Fortunately for Hilbert, if the crash damages room 1, he can just move the guest into room 2, and the room 2 guest into room 3...

Is this right, or am I missing something?

Alexander R Pruss said...

I just don't see what is paradoxical about Hilbert's Hotel. I have never seen a valid proof starting with the assumption that Hilbert's Hotel exists, and with other non-question-begging assumptions, that arrives at a contradiction.

Drew Mazanec said...

So that's why you have been after al-Ghazali's arguments for Kalam. If true, it throws you on the horns of a nasty dilemma regarding divine sovereignty with libertarian free will (between Calvinism and Open Theism). Craig uses Molinism to get between the horns, but I take it you reject Molinism?

Alexander R Pruss said...

To be honest, this connection has never occurred to me. I just like to refute arguments that I think don't work. Also, as a mathematician, I find myself with no sympathy for the Hilbert's Hotel style anti-infinity arguments. They seem to come down to "infinity minus one equals infinity". Sure--that's the definition of infinity. :-)

Anyway, I find somewhat plausible arguments against backwards-infinite causal sequences. I think some such argument might work even if Craig's arguments don't work. And, if so, that'll be all we need for the Kalaam arguments. Here, the grim reaper situation seems really worth thinking about.

Darrin said...

Bill doesn't hold to continuous time, but, like his atheist friend Quentin Smith, that time is sort of a progression of discrete absolute occurrences, or something like that.

Also:

//1. The things we observe that end also have beginnings.
2. The universe's past has ended in the present.
3. Therefore, the universe probably had a beginning.//

... equivocates on the word "end."

Vlastimil Vohánka said...
This comment has been removed by the author.
Vlastimil Vohánka said...

Alex,

This is the most interesting a priori anti-Kalaam I've seen so far. And I've seen many, indeed. I'm urging you to publish it and/or discuss it with Bill Craig.

Anyway, have you posted somewhere precise calculations of your probabilities? I am esp. interested how you arrived at:

(i) the 100% prob that there is a finite n such that n years after producing the 1st gamma-widget there are infinitely many gamma-widgets, and

(ii) the almost 100% prob that there is a subset of the gamma-widget mugs that has a last addition but no first one.

Truly,

V.

Alexander R Pruss said...

I don't have the calculations any more. I think they were fairly straightforward. I could recover them if I wanted to.

The problem is that now I also have the Grim Reaper arguments in favor of the Kalaam argument. So I need to decide which set of arguments is the stronger. And I think the GR ones win out.

Vlastimil Vohánka said...

"Straightforward" is person-relative. Are you used to forget you have a Ph.D. in math and most people don't? :)

Anyway, what about writing a paper showing in probabilistic terms how your GR Kalaam argument swamps and beats your widget anti-Kalaam argument? That would be a substantial contribution to the debate, in several ways.

Vlastimil Vohánka said...

Ad the related (linked, later) Prosblogion post of yours on 2-widgets

The opponent of supertasks and actual infinities can say:

A first 2-widget is possible with intervention (divine or not) stopping its reproduction before the 1.64... years are up; and impossible without it; so, a 2-widget is, literally, possible.

The intervention would be guaranteed, given a 2-widget. Intervention by who or what? Just someone or something. Maybe by God -- but why God would be necessitated to intervene, if nothing other intervened?, you ask. Reply: Because of His nature, exluding all impossibilities, like squared circles.

Strange, indeed; but not absurd.