(Note: I've posted this previously at Debunking Christianity.

You can go to the archives at that site and read the follow-up of objections

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On the Possibility of a Beginningless Past: A Reply to Craig

William Lane Craig has argued vigorously that, cosmological discoveries aside, it’s reasonable to believe on purely a priori grounds that the set of past events is finite in number.1 He offers two main types of a priori arguments for this claim: (i) that it’s metaphysically impossible for an actually infinite set of concrete things to exist, in which case the set of past events can’t be actually infinite, and (ii) that even if such a set could exist, it’s impossible to traverse it even in principle. Craig doesn’t pursue this claim for it’s own sake, however. Rather, he does so as a means to demonstrating that a theistic god exists. He reasons that if the set of past events is finite, then the universe as a whole had an absolute beginning with the first moment of time2. But since nothing can come into existence without a cause, the universe as a whole has a cause. From here, he goes on to argue that such a cause must be timeless (at least sans creation), immaterial, immensely powerful, and a person of some sort.

I intend to show that one of Craig’s most popular versions of (ii) is unsound. In this essay, I’ll state this argument, prefacing it with an explanation of the concepts crucial to understanding it. Then, I’ll examine a common objection to his argument, along with Craig’s response to it, in order to shed light on an unstated assumption of the argument. Finally, I’ll show that the unstated assumption is false, and how this is fatal to his argument.

I

As I mentioned above, several concepts that are crucial for understanding the argument need clarification.3 First of all, one needs a fairly perspicuous idea of a set and of a proper subset. A set is a collection of entities, called members of the set. The precise number of members contained in a set is its cardinal number. A proper subset is a part of another set, the former lacking at least one member which the latter contains, and which contains no other members (e.g., from a totally distinct set). More formally, a set A is a proper subset of a set B if and only if every member of A is a member of B, and some member of B isn’t a member of A. To illustrate: Suppose you have ten bottlecaps, five of which are from Pepsi bottles and five of which are from Coke bottles. Then we can call this the set of bottlecaps, the cardinal number of which is 10. Let’s call this set, A. Furthermore, the set of Pepsi caps (call it B) is a proper subset of A, since B consists in a collection of members that belong to A, and A has members that B does not (i.e., the Coke caps).

Another concept that plays an important role in the argument is that of a one-to-one-correspondence. This is a concept used to determine whether two sets have the same number of members (or, the same cardinal number). So there is a one-to-one correspondence between two sets, A and B, if and only if each member of A can be paired up with exactly one member of B, and each member of B can be paired up with exactly one member of A. To illustrate this concept, consider our set of bottlecaps. Now suppose that you didn’t know how to count, but you wanted to know if your had just as many Coke caps as you had of Pepsi caps. You could accomplish this task by pairing each Coke cap with each Pepsi cap, and each Pepsi cap with each Coke cap. If this can be accomplished with no remaining bottlecaps, then there is a one-to-one correspondence between the set of Pepsi caps and the set of Coke caps. If follows that the respective sets of bottlecaps have the same cardinal number.

The concept most important for our purposes is that of actual infinity. To obtain a grasp of this concept, consider the set of all the natural numbers (i.e., {1, 2, 3, …}). This set, as well as any set that can be put into a one-to-one correspondence with it, is an actually infinite set (It’s actually the “smallest” of the infinite sets, but we won’t be concerned with “larger” infinites here). An actual infinite has several interesting features. First of all, it is complete, in the sense that it has an infinite number of members; it is not merely increasing in number without limit. Second, any actually infinite set (of the “size” we’re here considering) can be put into a one-to-one correspondence with one of its proper subsets. This can be demonstrated by putting the set of natural numbers in a one-to-one correspondence with its proper subset of even numbers:

1 2 3 4…

2 4 6 8…

This example shows that a part of an actually infinite set can have as many members as the whole set! The cardinal number of an actually infinite set that can be put into a one-to-one correspondence with the natural numbers is called “aleph null” (let’s use ‘A0’ for brevity).

The final concept relevant to our discussion is order-type. I won’t talk at length about this concept here. Rather, I’ll barely do more than mention the order-types of certain sets containing A0 members. Four our purposes, it will suffice to know that sets can be sequentially ordered according to certain patterns or types. The order-type given to the set of natural numbers so ordered that, beginning with 1, each natural number is succeeded by the next largest natural number – i.e., {1, 2, 3, …} – is ‘omega’, or 'w’, and the set of negative integers so ordered that they are sequentially the opposite of w is w* (i.e., {…-3, -2, -1}). Sets with A0 members can have other order-types, however. For example, an A0 set can have the order type w+1 (i.e., {1, 2, 3, ..., 1}), or the order-type w+2 (i.e., {1, 2, 3, …, 1, 2}), etc. In fact, a set with A0 members can have the order-type w+w (i.e., {1, 2, 3, …, 1, 2, 3, …}), or the order-type w+w+w (i.e., {1, 2, 3, …., 1, 2, 3, …, 1, 2, 3, …}), etc.! To see this, recall that any set that can be put into a one-to-one correspondence with the natural numbers has a cardinal number of A0. But sets with the order-types mentioned above can be put into such a correspondence. So, for example, a set with the order-type w+1 can be put into a one-to-one correspondence with the natural numbers as follows:

1 1 2 3…

1 2 3 4…

Similarly, a set with the order-type w+w+w can be put into a one-to-one correspondence with the natural numbers as follows:

1 2 3…1 2 3…1 2 3…

1 4 7…2 5 8…3 6 9…

Therefore, since sets with such order-types can be put into a one-to-one correspondence with the natural numbers, it follows that their cardinal number is A0.

At this point, an interesting feature of certain sets with A0 members emerges. For consider any A0 set with an order-type other than w. For example, consider a set of A0 offramps on an infinitely long freeway, such that a distance of one mile separates each offramp from its predecessor and successor (except, of course, the first offramp, since it has no predecessor). Suppose further that the order-type of the offramps is w+1 ({1, 2, 3, …, 1}). The offramp assigned the first 1 would seem to be infinitely distant from the offramp assigned the second 1. Such a set has the interesting feature of being non-traversable in principle – it cannot, even in principle, be exhaustively counted through one offramp at a time. This is because it is logically impossible to count to a number that has no immediate predecessor. But the offramp assigned the second 1 has no immediate predecessor. Therefore, a driver on such a freeway could never reach the offramp assigned the second 1. Call this particular logical ban on traversing sets with w+1 or “higher” order-types ‘LB’.

Now it may be tempting to think that this consideration is decisive for the view that the past must be finite, since any set with A0 members can be ordered according to the order-type w+1. In this way, one might think, LB infects all actually infinite sets, and thus no set with A0 members is traversable. This reasoning can be expressed as follows:

1. A set is LB non-traversable if and only if it contains at least one member A0 distant from at least one of the other of its members.

2. Any set with the order-type w+1 is such that it contains a member A0 distant from at least one of the other of its members.

3. Therefore, any set with the order-type w+1 is LB non-traversable.

4. Any set with A0 members can be assigned the order-type w+1.

5. Therefore, any set with A0 members is LB non-traversable.

But this would be rash. For the inference from (3) and (4) to (5) is a non sequitur. For consider a set with A0 members that is assigned the order-type w. A set with this order-type is such that (i) no member is infinitely distant from any other member, and (ii) each member does have an immediate predecessor, as so is immune to LB. But any set with A0 members can be assigned the order-type w. So if the inference from (3) and (4) to (5) were valid, then by similar reasoning the following inference should go through as well:

3’. Any set with the order-type w is not LB non-traversable.

4’. Any set with A0 members can be assigned the order-type w.

5’. Therefore, any set with A0 members is not LB non-traversable.

But (5’) contradicts (5). Therefore, since the same pattern of reasoning yields contradictory results, it’s faulty. So, just because an A0 set can be assigned a non-traversable order-type, it doesn’t follow that such a set is non-traversable. What really follow from (3) and (4) is rather

5’’. Any set with A0 members can be assigned an LB non-traversable order-type.

Which, needless to say, doesn’t help to establish the finitude of the set of past events.

The arguments above suffer from another problem as well. For (2) is clearly false. To see this, consider again our infinitely long freeway. No suppose that it only has one lane, and that it has an infinitely long traffic jam. Finally, suppose that each car in the jam is assigned a number from the order-type w+1. Does it follow that there is a car A0 distant from any other car? No. For the car assigned the second 1 could be immediately in front of the first car. This illustration shows that the order-type assigned to a set of objects doesn’t necessarily affect the distances between its members. To drive this implication home, suppose that the cars of the traffic jam were assigned the order-type w. Would it follow that no car is A0 distant from any other car? Not in the least. For the first and second cars may be infinitely distant from one another. One might reply that we could just stipulate that the distances between the cars and the order-type assigned to the cars correspond. In such a case, each car would only be finitely distant from every other car when assigned the order-type w (e.g., the second car is 2 meters from the end of the traffic jam, the third car is 3 meters from the end of the jam, etc. [these are small cars!]). One could then reassign the cars with the order-type w+1, but then the correspondence between the order-type and the distances of the cars would break down. This is because no car assigned a number from the w order-type in our scenario is infinitely distant from any other car. Therefore, the second 1 of the newly assigned w+1 reordering would be assigned to a car that is only finitely distant from any other car. These illustrations show that (i) some sets assigned the order-type w+1 are such that no member is infinitely distant from any other, and (ii) we can know a priori than an A0 set of concrete objects cannot be reordered from w to w+1 in such a way that the distances between the members of such a set correspond to their order-type. Therefore, if a set of objects has A0 members (arranged linearly), it does not follow from this that it has members infinitely distant (whether in time or in space) from other members (and is therefore LB non-traversable).4 Thus, to show that an A0 past is non-traversable, Craig must show that no A0 set with either the order-type w or w* (and is such that no member is infinitely distant from any other) is traversable. Let’s consider one of Craig’s main attempts to do this.

II

Craig advances an argument for the proposition that one cannot traverse a beginningless past and end at the present moment.5 To do this, Craig assumes, for the sake of argument, that there could be a beginningless past, conceived as a set of events with the cardinal number A0 and the order-type w* (i.e., {…,. -3, -2, -1}), where each negative integer represents an event of the past. He then argues,

“…suppose we meet a man who claims to have been counting down from eternity and who is now finishing:…,-3, -2, -1, 0. We could ask, why didn’t he finish counting yesterday or the day before or the year before? By then an infinite amount of time had already elapsed, so that he should already have finished. Thus, at no point in the infinite past could we ever find the man finishing his countdown, for by that point he should already be done! In fact, no matter how far back into the past we go, we can never find the man counting at all, for at any point we reach he will already be finished. But if at no point in the past do we find him counting, this contradicts the hypothesis that he has been counting down from eternity.”6

Craig’s argument is a reductio ad absurdum, where we suppose that a beginningless past is possible in order to show that it entails a contradiction. The argument can be expressed as follows, with (1) as the premise set up for reduction:

1. The past is beginningless (conceived as a set of events with the cardinality A0, and the order-type w*).

2. If the past is beginningless, then there could have been an immortal counter who counts down from such a past at the rate of one negative integer per day.

3. The immortal counter will finish counting if and only if he has an infinite number of days in which to count them.

4. If the past is beginningless, then there are an infinite number of days before every day.

5. Therefore, the immortal counter will have finished counting before every day.

6. If the immortal counter will have finished counting before every day, then he has never counted.

7. Therefore, the immortal counter has both never counted and has been counting down from a beginningless past (contradiction)

8. Therefore, the past is not beginningless (from 1-7, reductio)

In short, Craig argues that the past must -- logically must -- have a beginning. For the very notion of traversing a beginningless past entails a contradiction. Craig’s underlying intuition here is that if the past is beginningless, then it must contain an actually infinite proper subset of events that was not formed by successive addition, and that this is absurd.

Critics typically attack (3), arguing that Craig mistakenly assumes that to count an infinite number of negative integers is to count all of them. However, critics of Craig’s argument point out that one can count an infinite set of numbers without counting them all.7 For example, suppose our eternal counter just finished counting all the negative integers down to -3. Then it would be true that he has counted an infinite number of integers, and yet he has not counted all the integers. This can be demonstrated by the following one-to-one correspondence:

Days counted: -3 -4 -5…

Nat. numbers: 1 2 3…

In this case, the set of days counted has the cardinal number A0, since its members can be put into a one-to-one correspondence with the natural numbers. Yet he clearly hasn’t counted all the negative integers, since he has failed to count -2 and -1. Therefore, since counting an infinite number of things is not synonymous with counting them all, Craig’s (3) is based on an equivocation.

Craig has denied that he is guilty of this charge8:

“I do not think the argument makes this alleged equivocation, and this can be made clear by examining the reason why our eternal counter is supposedly able to complete a count of the negative numbers, ending at zero. In order to justify this intuitively impossible feat, the argument’s opponent appeals to the so-called Principle of Correspondence…On the basis of the principle the objector argues that since the set of past years can be put into a one-to-one correspondence with the set of negative numbers, it follows that by counting one number a year an eternal counter could complete a countdown of the negative numbers by the present year. If we were to ask why the counter would not finish next year or in a hundred years, the objector would respond that prior to the present year an infinite number of years will have elapsed, so that by the Principle of Correspondence, all the numbers should have been counted by now.

But this reasoning backfires on the objector: for on this account the counter should at any point in the past have already finished counting all the numbers, since a one-to-one correspondence exists between the years of the past and the negative numbers.”9

From this passage, we see Craig’s rationale for (3):

(R) The counter will have finished counting all of the negative integers if and only if the years of the past can be put into a one-to-one correspondence with them.

Furthermore, from the passage cited, we see that Craig thinks that the defender of an A0 past agrees with (R). But since the type of correspondence depicted in (R) can be accomplished at the present moment, it follows that the counter should be finished by now. Therefore, Craig’s opponent is committed to a view that entails the absurdity surfaced by the above reductio.

III

It isn’t clear that Craig hasn’t made his case, however. For consider the following scenario. Suppose God timelessly numbers the years to come about in a beginningless universe. Suppose further that He assigns the negative integers to the set of events prior to the birth of Christ, and then the positive integers begin at this point. Then the timeline, with its corresponding integer assignment, can be illustrated as follows:

…-3 -2 -1 Birth of Christ 1 2 3…

Suppose yet further that God assigned Ralph, an immortal creature, the task of counting down the negative integers assigned to the years BCE, and stopping at the birth of Christ. Call this task ‘(T)’. With this in mind, suppose now that Ralph has been counting down from eternity past and is now counting the day assigned (by God) the integer -3. In such a case, Ralph has counted a set of years that could be put into a one-to-one correspondence with the set of negative integers, yet he has not finished all the negative integers. This case shows that, while it is a necessary condition for counting all of the events that one is able to put them into a one-to-one correspondence with the natural numbers, this is not sufficient. For if the events that are to be counted have independently “fixed”, or, “designated” integer assignments set out for one to traverse, one must count through these such that, for each event, the number one is counting is the same as the one independently assigned to the event. In the scenario mentioned above, God assigned an integer to each year that will come to pass. In such a case, Ralph must satisfy at least two conditions if he is to accomplish (T): (i) count a set of years that can be put into a one-to-one correspondence with the natural numbers, and (ii) for each year that elapses, count the particular negative integer that God has independently assigned to it. According to Craig’s assumption (R), however, Ralph is supposed to be able to accomplish (T) by satisfying (i) alone. But we have just seen that he must accomplish (ii) as well. Therefore, being able to place the events of the past into a one-to-one correspondence with the natural numbers does not guarantee that the counter has finished the task of counting all the negative integers. In other words, (R) is false. But recall that (R) is Craig’s rationale for (3). Thus, (3) lacks positive support. But more importantly, (3) is false. This is because the scenario above is a counterexample to both (R) and (3). For (3) asserts that it is sufficient for counting down all the negative integers that one has an infinite amount of time in which to count them. But our scenario showed that one could have an infinite amount of time to count, and yet not finish counting all of the negative integers (e.g., one can count down to -3 in an infinite amount of time, and yet have more integers to count).

To sum up: We’ve looked at an argument that Craig repeatedly gives for the impossibility of a beginningless past. We then saw that one of its premises is false, in which case it is unsound. Thus, this argument, at least, cannot be used to offer a priori support for the key premise in his Kalam argument.

1 See, for example, his The Kalam Cosmological Argument (London: Macmillan, 1979); Craig and Quentin Smith, Theism, Atheism, and Big Bang Cosmology (Oxford: Clarendon Press, 1995). See also Craig’s popular-level book, Reasonable Faith (Wheaton: Crossway Books, 1994).

2 I should mention a wrinkle here: the possibility that the universe did not begin to exist with the first event of time, but rather existed eternally in a quiescent, eventless mode of existence “prior” to the first event. Craig addresses this worry in “The Kalam Cosmological Argument and the Hypothesis of a Quiescent Universe”, Faith and Philosophy 8 (1991), pp. 104-8.

3 My discussion of the following set-theoretic concepts is indebted to J.P. Moreland’s Scaling the Secular City (Grand Rapids: Baker, 1987)

4 The points and illustrations are similar to those made and conceded by Craig in “Reply to Smith: On the Finitude of the Past”, International Philosophical Quarterly 33 (1993), pp. 228-9.

5 Actually, he advances two arguments for this proposition. One is a variation on the famous Tristam Shandy Paradox. In Craig’s construal of it, Shandy writes his autobiography from the beginningless past at the rate of one year of writing per day of autobiography. It seems that Shandy would never finish his autobiography, getting farther behind with each passing day. But since one can put the days of his life into a one-to-one correspondence with the set of past years, it (paradoxically) seems that he should have finished his autobiography by now. The other version is virtually the same as the one I consider here. It asserts that Shandy should be finished by now, irrespective of the rate at which he is writing. However, I won’t consider the former version here. See Craig and Smith’s Theism, Atheism, and Big Bang Cosmology, pp. 99-100, and Craig’s “Feature Review of Time, Creation, and the Continuum”, International Philosophical Quarterly 25 (1985), pp. 319-26. For a briefer exposition, see Craig, Reasonable Faith, pp. 98-9.

6 Craig, Reasonable Faith, p. 99.

7 This objection can be found in David A. Conway. “’It Would Have Happened Already’: On One Argument for a First Cause”, Analysis 44 (1984), pp. 159-66; Richard Sorabji. Time, Creation, and the Continuum (Ithaca: Cornell University Press, 1983), pp. 219-24.

8 See, for example, Craig and Smith, Theism, Atheism, and Big Bang Cosmology, pp. 105-6; Craig, “Review of Time, Creation, and the Continuum”, p. 323.

9 Craig, “Review of Time, Creation, and the Continuum”, p. 323.

You can go to the archives at that site and read the follow-up of objections

and replies in the "comments" section below the post)

On the Possibility of a Beginningless Past: A Reply to Craig

William Lane Craig has argued vigorously that, cosmological discoveries aside, it’s reasonable to believe on purely a priori grounds that the set of past events is finite in number.1 He offers two main types of a priori arguments for this claim: (i) that it’s metaphysically impossible for an actually infinite set of concrete things to exist, in which case the set of past events can’t be actually infinite, and (ii) that even if such a set could exist, it’s impossible to traverse it even in principle. Craig doesn’t pursue this claim for it’s own sake, however. Rather, he does so as a means to demonstrating that a theistic god exists. He reasons that if the set of past events is finite, then the universe as a whole had an absolute beginning with the first moment of time2. But since nothing can come into existence without a cause, the universe as a whole has a cause. From here, he goes on to argue that such a cause must be timeless (at least sans creation), immaterial, immensely powerful, and a person of some sort.

I intend to show that one of Craig’s most popular versions of (ii) is unsound. In this essay, I’ll state this argument, prefacing it with an explanation of the concepts crucial to understanding it. Then, I’ll examine a common objection to his argument, along with Craig’s response to it, in order to shed light on an unstated assumption of the argument. Finally, I’ll show that the unstated assumption is false, and how this is fatal to his argument.

I

As I mentioned above, several concepts that are crucial for understanding the argument need clarification.3 First of all, one needs a fairly perspicuous idea of a set and of a proper subset. A set is a collection of entities, called members of the set. The precise number of members contained in a set is its cardinal number. A proper subset is a part of another set, the former lacking at least one member which the latter contains, and which contains no other members (e.g., from a totally distinct set). More formally, a set A is a proper subset of a set B if and only if every member of A is a member of B, and some member of B isn’t a member of A. To illustrate: Suppose you have ten bottlecaps, five of which are from Pepsi bottles and five of which are from Coke bottles. Then we can call this the set of bottlecaps, the cardinal number of which is 10. Let’s call this set, A. Furthermore, the set of Pepsi caps (call it B) is a proper subset of A, since B consists in a collection of members that belong to A, and A has members that B does not (i.e., the Coke caps).

Another concept that plays an important role in the argument is that of a one-to-one-correspondence. This is a concept used to determine whether two sets have the same number of members (or, the same cardinal number). So there is a one-to-one correspondence between two sets, A and B, if and only if each member of A can be paired up with exactly one member of B, and each member of B can be paired up with exactly one member of A. To illustrate this concept, consider our set of bottlecaps. Now suppose that you didn’t know how to count, but you wanted to know if your had just as many Coke caps as you had of Pepsi caps. You could accomplish this task by pairing each Coke cap with each Pepsi cap, and each Pepsi cap with each Coke cap. If this can be accomplished with no remaining bottlecaps, then there is a one-to-one correspondence between the set of Pepsi caps and the set of Coke caps. If follows that the respective sets of bottlecaps have the same cardinal number.

The concept most important for our purposes is that of actual infinity. To obtain a grasp of this concept, consider the set of all the natural numbers (i.e., {1, 2, 3, …}). This set, as well as any set that can be put into a one-to-one correspondence with it, is an actually infinite set (It’s actually the “smallest” of the infinite sets, but we won’t be concerned with “larger” infinites here). An actual infinite has several interesting features. First of all, it is complete, in the sense that it has an infinite number of members; it is not merely increasing in number without limit. Second, any actually infinite set (of the “size” we’re here considering) can be put into a one-to-one correspondence with one of its proper subsets. This can be demonstrated by putting the set of natural numbers in a one-to-one correspondence with its proper subset of even numbers:

1 2 3 4…

2 4 6 8…

This example shows that a part of an actually infinite set can have as many members as the whole set! The cardinal number of an actually infinite set that can be put into a one-to-one correspondence with the natural numbers is called “aleph null” (let’s use ‘A0’ for brevity).

The final concept relevant to our discussion is order-type. I won’t talk at length about this concept here. Rather, I’ll barely do more than mention the order-types of certain sets containing A0 members. Four our purposes, it will suffice to know that sets can be sequentially ordered according to certain patterns or types. The order-type given to the set of natural numbers so ordered that, beginning with 1, each natural number is succeeded by the next largest natural number – i.e., {1, 2, 3, …} – is ‘omega’, or 'w’, and the set of negative integers so ordered that they are sequentially the opposite of w is w* (i.e., {…-3, -2, -1}). Sets with A0 members can have other order-types, however. For example, an A0 set can have the order type w+1 (i.e., {1, 2, 3, ..., 1}), or the order-type w+2 (i.e., {1, 2, 3, …, 1, 2}), etc. In fact, a set with A0 members can have the order-type w+w (i.e., {1, 2, 3, …, 1, 2, 3, …}), or the order-type w+w+w (i.e., {1, 2, 3, …., 1, 2, 3, …, 1, 2, 3, …}), etc.! To see this, recall that any set that can be put into a one-to-one correspondence with the natural numbers has a cardinal number of A0. But sets with the order-types mentioned above can be put into such a correspondence. So, for example, a set with the order-type w+1 can be put into a one-to-one correspondence with the natural numbers as follows:

1 1 2 3…

1 2 3 4…

Similarly, a set with the order-type w+w+w can be put into a one-to-one correspondence with the natural numbers as follows:

1 2 3…1 2 3…1 2 3…

1 4 7…2 5 8…3 6 9…

Therefore, since sets with such order-types can be put into a one-to-one correspondence with the natural numbers, it follows that their cardinal number is A0.

At this point, an interesting feature of certain sets with A0 members emerges. For consider any A0 set with an order-type other than w. For example, consider a set of A0 offramps on an infinitely long freeway, such that a distance of one mile separates each offramp from its predecessor and successor (except, of course, the first offramp, since it has no predecessor). Suppose further that the order-type of the offramps is w+1 ({1, 2, 3, …, 1}). The offramp assigned the first 1 would seem to be infinitely distant from the offramp assigned the second 1. Such a set has the interesting feature of being non-traversable in principle – it cannot, even in principle, be exhaustively counted through one offramp at a time. This is because it is logically impossible to count to a number that has no immediate predecessor. But the offramp assigned the second 1 has no immediate predecessor. Therefore, a driver on such a freeway could never reach the offramp assigned the second 1. Call this particular logical ban on traversing sets with w+1 or “higher” order-types ‘LB’.

Now it may be tempting to think that this consideration is decisive for the view that the past must be finite, since any set with A0 members can be ordered according to the order-type w+1. In this way, one might think, LB infects all actually infinite sets, and thus no set with A0 members is traversable. This reasoning can be expressed as follows:

1. A set is LB non-traversable if and only if it contains at least one member A0 distant from at least one of the other of its members.

2. Any set with the order-type w+1 is such that it contains a member A0 distant from at least one of the other of its members.

3. Therefore, any set with the order-type w+1 is LB non-traversable.

4. Any set with A0 members can be assigned the order-type w+1.

5. Therefore, any set with A0 members is LB non-traversable.

But this would be rash. For the inference from (3) and (4) to (5) is a non sequitur. For consider a set with A0 members that is assigned the order-type w. A set with this order-type is such that (i) no member is infinitely distant from any other member, and (ii) each member does have an immediate predecessor, as so is immune to LB. But any set with A0 members can be assigned the order-type w. So if the inference from (3) and (4) to (5) were valid, then by similar reasoning the following inference should go through as well:

3’. Any set with the order-type w is not LB non-traversable.

4’. Any set with A0 members can be assigned the order-type w.

5’. Therefore, any set with A0 members is not LB non-traversable.

But (5’) contradicts (5). Therefore, since the same pattern of reasoning yields contradictory results, it’s faulty. So, just because an A0 set can be assigned a non-traversable order-type, it doesn’t follow that such a set is non-traversable. What really follow from (3) and (4) is rather

5’’. Any set with A0 members can be assigned an LB non-traversable order-type.

Which, needless to say, doesn’t help to establish the finitude of the set of past events.

The arguments above suffer from another problem as well. For (2) is clearly false. To see this, consider again our infinitely long freeway. No suppose that it only has one lane, and that it has an infinitely long traffic jam. Finally, suppose that each car in the jam is assigned a number from the order-type w+1. Does it follow that there is a car A0 distant from any other car? No. For the car assigned the second 1 could be immediately in front of the first car. This illustration shows that the order-type assigned to a set of objects doesn’t necessarily affect the distances between its members. To drive this implication home, suppose that the cars of the traffic jam were assigned the order-type w. Would it follow that no car is A0 distant from any other car? Not in the least. For the first and second cars may be infinitely distant from one another. One might reply that we could just stipulate that the distances between the cars and the order-type assigned to the cars correspond. In such a case, each car would only be finitely distant from every other car when assigned the order-type w (e.g., the second car is 2 meters from the end of the traffic jam, the third car is 3 meters from the end of the jam, etc. [these are small cars!]). One could then reassign the cars with the order-type w+1, but then the correspondence between the order-type and the distances of the cars would break down. This is because no car assigned a number from the w order-type in our scenario is infinitely distant from any other car. Therefore, the second 1 of the newly assigned w+1 reordering would be assigned to a car that is only finitely distant from any other car. These illustrations show that (i) some sets assigned the order-type w+1 are such that no member is infinitely distant from any other, and (ii) we can know a priori than an A0 set of concrete objects cannot be reordered from w to w+1 in such a way that the distances between the members of such a set correspond to their order-type. Therefore, if a set of objects has A0 members (arranged linearly), it does not follow from this that it has members infinitely distant (whether in time or in space) from other members (and is therefore LB non-traversable).4 Thus, to show that an A0 past is non-traversable, Craig must show that no A0 set with either the order-type w or w* (and is such that no member is infinitely distant from any other) is traversable. Let’s consider one of Craig’s main attempts to do this.

II

Craig advances an argument for the proposition that one cannot traverse a beginningless past and end at the present moment.5 To do this, Craig assumes, for the sake of argument, that there could be a beginningless past, conceived as a set of events with the cardinal number A0 and the order-type w* (i.e., {…,. -3, -2, -1}), where each negative integer represents an event of the past. He then argues,

“…suppose we meet a man who claims to have been counting down from eternity and who is now finishing:…,-3, -2, -1, 0. We could ask, why didn’t he finish counting yesterday or the day before or the year before? By then an infinite amount of time had already elapsed, so that he should already have finished. Thus, at no point in the infinite past could we ever find the man finishing his countdown, for by that point he should already be done! In fact, no matter how far back into the past we go, we can never find the man counting at all, for at any point we reach he will already be finished. But if at no point in the past do we find him counting, this contradicts the hypothesis that he has been counting down from eternity.”6

Craig’s argument is a reductio ad absurdum, where we suppose that a beginningless past is possible in order to show that it entails a contradiction. The argument can be expressed as follows, with (1) as the premise set up for reduction:

1. The past is beginningless (conceived as a set of events with the cardinality A0, and the order-type w*).

2. If the past is beginningless, then there could have been an immortal counter who counts down from such a past at the rate of one negative integer per day.

3. The immortal counter will finish counting if and only if he has an infinite number of days in which to count them.

4. If the past is beginningless, then there are an infinite number of days before every day.

5. Therefore, the immortal counter will have finished counting before every day.

6. If the immortal counter will have finished counting before every day, then he has never counted.

7. Therefore, the immortal counter has both never counted and has been counting down from a beginningless past (contradiction)

8. Therefore, the past is not beginningless (from 1-7, reductio)

In short, Craig argues that the past must -- logically must -- have a beginning. For the very notion of traversing a beginningless past entails a contradiction. Craig’s underlying intuition here is that if the past is beginningless, then it must contain an actually infinite proper subset of events that was not formed by successive addition, and that this is absurd.

Critics typically attack (3), arguing that Craig mistakenly assumes that to count an infinite number of negative integers is to count all of them. However, critics of Craig’s argument point out that one can count an infinite set of numbers without counting them all.7 For example, suppose our eternal counter just finished counting all the negative integers down to -3. Then it would be true that he has counted an infinite number of integers, and yet he has not counted all the integers. This can be demonstrated by the following one-to-one correspondence:

Days counted: -3 -4 -5…

Nat. numbers: 1 2 3…

In this case, the set of days counted has the cardinal number A0, since its members can be put into a one-to-one correspondence with the natural numbers. Yet he clearly hasn’t counted all the negative integers, since he has failed to count -2 and -1. Therefore, since counting an infinite number of things is not synonymous with counting them all, Craig’s (3) is based on an equivocation.

Craig has denied that he is guilty of this charge8:

“I do not think the argument makes this alleged equivocation, and this can be made clear by examining the reason why our eternal counter is supposedly able to complete a count of the negative numbers, ending at zero. In order to justify this intuitively impossible feat, the argument’s opponent appeals to the so-called Principle of Correspondence…On the basis of the principle the objector argues that since the set of past years can be put into a one-to-one correspondence with the set of negative numbers, it follows that by counting one number a year an eternal counter could complete a countdown of the negative numbers by the present year. If we were to ask why the counter would not finish next year or in a hundred years, the objector would respond that prior to the present year an infinite number of years will have elapsed, so that by the Principle of Correspondence, all the numbers should have been counted by now.

But this reasoning backfires on the objector: for on this account the counter should at any point in the past have already finished counting all the numbers, since a one-to-one correspondence exists between the years of the past and the negative numbers.”9

From this passage, we see Craig’s rationale for (3):

(R) The counter will have finished counting all of the negative integers if and only if the years of the past can be put into a one-to-one correspondence with them.

Furthermore, from the passage cited, we see that Craig thinks that the defender of an A0 past agrees with (R). But since the type of correspondence depicted in (R) can be accomplished at the present moment, it follows that the counter should be finished by now. Therefore, Craig’s opponent is committed to a view that entails the absurdity surfaced by the above reductio.

III

It isn’t clear that Craig hasn’t made his case, however. For consider the following scenario. Suppose God timelessly numbers the years to come about in a beginningless universe. Suppose further that He assigns the negative integers to the set of events prior to the birth of Christ, and then the positive integers begin at this point. Then the timeline, with its corresponding integer assignment, can be illustrated as follows:

…-3 -2 -1 Birth of Christ 1 2 3…

Suppose yet further that God assigned Ralph, an immortal creature, the task of counting down the negative integers assigned to the years BCE, and stopping at the birth of Christ. Call this task ‘(T)’. With this in mind, suppose now that Ralph has been counting down from eternity past and is now counting the day assigned (by God) the integer -3. In such a case, Ralph has counted a set of years that could be put into a one-to-one correspondence with the set of negative integers, yet he has not finished all the negative integers. This case shows that, while it is a necessary condition for counting all of the events that one is able to put them into a one-to-one correspondence with the natural numbers, this is not sufficient. For if the events that are to be counted have independently “fixed”, or, “designated” integer assignments set out for one to traverse, one must count through these such that, for each event, the number one is counting is the same as the one independently assigned to the event. In the scenario mentioned above, God assigned an integer to each year that will come to pass. In such a case, Ralph must satisfy at least two conditions if he is to accomplish (T): (i) count a set of years that can be put into a one-to-one correspondence with the natural numbers, and (ii) for each year that elapses, count the particular negative integer that God has independently assigned to it. According to Craig’s assumption (R), however, Ralph is supposed to be able to accomplish (T) by satisfying (i) alone. But we have just seen that he must accomplish (ii) as well. Therefore, being able to place the events of the past into a one-to-one correspondence with the natural numbers does not guarantee that the counter has finished the task of counting all the negative integers. In other words, (R) is false. But recall that (R) is Craig’s rationale for (3). Thus, (3) lacks positive support. But more importantly, (3) is false. This is because the scenario above is a counterexample to both (R) and (3). For (3) asserts that it is sufficient for counting down all the negative integers that one has an infinite amount of time in which to count them. But our scenario showed that one could have an infinite amount of time to count, and yet not finish counting all of the negative integers (e.g., one can count down to -3 in an infinite amount of time, and yet have more integers to count).

To sum up: We’ve looked at an argument that Craig repeatedly gives for the impossibility of a beginningless past. We then saw that one of its premises is false, in which case it is unsound. Thus, this argument, at least, cannot be used to offer a priori support for the key premise in his Kalam argument.

1 See, for example, his The Kalam Cosmological Argument (London: Macmillan, 1979); Craig and Quentin Smith, Theism, Atheism, and Big Bang Cosmology (Oxford: Clarendon Press, 1995). See also Craig’s popular-level book, Reasonable Faith (Wheaton: Crossway Books, 1994).

2 I should mention a wrinkle here: the possibility that the universe did not begin to exist with the first event of time, but rather existed eternally in a quiescent, eventless mode of existence “prior” to the first event. Craig addresses this worry in “The Kalam Cosmological Argument and the Hypothesis of a Quiescent Universe”, Faith and Philosophy 8 (1991), pp. 104-8.

3 My discussion of the following set-theoretic concepts is indebted to J.P. Moreland’s Scaling the Secular City (Grand Rapids: Baker, 1987)

4 The points and illustrations are similar to those made and conceded by Craig in “Reply to Smith: On the Finitude of the Past”, International Philosophical Quarterly 33 (1993), pp. 228-9.

5 Actually, he advances two arguments for this proposition. One is a variation on the famous Tristam Shandy Paradox. In Craig’s construal of it, Shandy writes his autobiography from the beginningless past at the rate of one year of writing per day of autobiography. It seems that Shandy would never finish his autobiography, getting farther behind with each passing day. But since one can put the days of his life into a one-to-one correspondence with the set of past years, it (paradoxically) seems that he should have finished his autobiography by now. The other version is virtually the same as the one I consider here. It asserts that Shandy should be finished by now, irrespective of the rate at which he is writing. However, I won’t consider the former version here. See Craig and Smith’s Theism, Atheism, and Big Bang Cosmology, pp. 99-100, and Craig’s “Feature Review of Time, Creation, and the Continuum”, International Philosophical Quarterly 25 (1985), pp. 319-26. For a briefer exposition, see Craig, Reasonable Faith, pp. 98-9.

6 Craig, Reasonable Faith, p. 99.

7 This objection can be found in David A. Conway. “’It Would Have Happened Already’: On One Argument for a First Cause”, Analysis 44 (1984), pp. 159-66; Richard Sorabji. Time, Creation, and the Continuum (Ithaca: Cornell University Press, 1983), pp. 219-24.

8 See, for example, Craig and Smith, Theism, Atheism, and Big Bang Cosmology, pp. 105-6; Craig, “Review of Time, Creation, and the Continuum”, p. 323.

9 Craig, “Review of Time, Creation, and the Continuum”, p. 323.

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