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God as a necessary, maximally great, endless being vs. the challenge to an actual infinity

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In a recent thread, the Kalam Cosmological argument family was challenged on the issue: can an actual infinity exist? If not (presumably due to Hilbert’s Hotel-like absurdities), then God could not be an infinite being as such is impossible of being.

A thread of discussion developed, and I thought a summary intervention may be helpful. On further thought, perhaps it should be headlined:

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KF, 12: >> I think several themes are worth highlighting.

It can be discussed that non-being, true nothingness cannot be a causal source. Were there ever utter nothing, such would therefore forever obtain. There would be no world.But, manifestly, there is a world.

So, we must ponder the logic of being, at least in a nutshell.

Candidates to being may be such that core characteristics central to identity stand in mutual contradiction such as those of a square circle. Such are impossible of being. And, we see principles of distinct identity necessarily embedded from the outset, especially that truths must be all so together so X and Y where Y = ~ X is not a possible state of affairs. If something could exist in a possible world were it actualised as a state of affairs, it is a possible being.

Of these, some Z can be in at least one possible world Wi, but not in another “neighbouring” one, Wj — contingent beings. The difference Wi – Wj will contain some unmet on-off enabling circumstance for Z . . . a necessary causal condition or factor for Z, say C. If Z is a fire, it requires heat, fuel, oxidiser and an uninterfered-with combustion chain reaction (cf. how Halon extinguishers work).

By contrast, we can see a being N that has no dependence on any such C, which will be in any and all possible worlds. That is, N is a necessary being and will be part of the common framework for any world W to exist. For example, distinct identity (A vs ~A) entails that two-ness and so also the endless set of naturals, must exist. [Beyond which lie the transfinites and the surreals as illustrated.] And, without necessary causal factors C, such has no beginning or end. Given that a world exists, at least one being N must be necessary. (Theists, classically hold that things like numbers are eternally contemplated by God, for instance.)

Any given case Ni is eternal, causeless, framework for any possible world, enabling and structuring it in some way. Notice, eternality not infinity, has been asserted, on the strength that for some W to be, some N must be as key to its framework. We readily see this for two-ness etc.

Such is strange to our ears, maybe, but that is a fault of our education not the logic.

Now too, our world is one of finite stage causal succession as we can see from succession of generations. But it is dubious for such to have existed to the infinitely — endlessly — remote past, as to succeed from some stage s_k to s_k+1, s_k+2 etc is equivalent to a counting succession 0,1,2 . . . which succeeds without limit but in an instance will be such that some later s_p to s_p+1, s_p+2 etc can again be matched 1:1 with 0,1,2 . . . thus showing that a transfinite span, credibly, cannot be traversed in finite-stage successive, cumulative steps. Thus if we are at a now, no S_k is transfinitely remote, even beyond say a big bang at 13.8 BYA. Our world W_a is credibly not some Ni, and has a beginning. It has a cause, a capable, sufficient one.

Where, we exist therein as responsibly and rationally free, morally governed creatures. This constrains the N_a that is at the world-root. For, post Hume et al, only at that level can the IS-OUGHT gap be soundly bridged. And we all know that after centuries of debate, only one serious candidate stands — just put up a viable alternative if you think you can. Good luck with that: ______ (Predictably, a fail.)

This is: the inherently good creator God, a necessary and maximally great being; worthy of loyalty and of the reasonable, responsible service of doing the good in accord with our evident nature.

But, what about, God is infinite?

I suggest, this first means that God is not externally limited or weakened so that he can be defeated or utterly frustrated in his purposes. Which, is among other things a way of saying that God is not evil, that being the privation, frustration or perversion of good capabilities out of their proper end.

God is also eternal and indestructible, as he has no dependence on external, on/off enabling causal factors. Thus, his being is without beginning or end, endless. This, being a characteristic of necessary being, which is required once a world is.

So, I think we need to reflect on what sense is meant when it is suggested that an actual infinity is impossible of being, and what are its strengths and limitations. For sure, an endless past of finite stage causal succession seems impossible and a physical, materially based infinite quantity is also dubious. But, the transfinite set of natural numbers and beyond the surreals great and small all seem necessary — framework to any world. Which in turn suggests mind capable of such a contemplation.

And more.>>

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Food for thought. END

Comments
DaveS
DaveS: But that doesn’t imply that the iPad missed some positions (at least I haven’t seen any proof of this).
If the iPad always moves to the right, then, given that for every position there are infinitely positions to the left, it is necessarily the case that the iPad missed positions. Why? Since there is no ‘First Left.’ Because there are infinite positions to the left, there are only two options available to anyone who wants to enter them: 1. Go to the left. 2. Go to the right, but depart from the ‘First left.’ (1) Is a journey with no return. (2) Is not possible, since there is no ‘First left’.Origenes
December 29, 2017
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KF,
So, if the actual chain is transfinitely long, it means transfinitely removed members must have existed.
Let's assume this is the case for the moment. Then exactly one of the following must be true: A) Some digits in the decimal expansion of π are infinitely remote from the decimal point. B) The decimal expansion of π has finite length. Which is it?daveS
December 29, 2017
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DS, I have long since shown the problem i/l/o what an infinite actual chain and past of token passing requires. For members must have once held T or have just now received it. So, if the actual chain is transfinitely long, it means transfinitely removed members must have existed. Thence, they would have had to successively pass T +1 at a time to -2, -1, 0, 1, 2 . . . and on to n-1, n. This means some transfinitely remote P passed to P+1, P+2 etc, crossing a transfinite span to reach -2,-1,0 etc. But that means spanning endlessness in steps and we already say that after any finitely large k from P, P+k, we can redefine as K, k+1, K+2 etc and match 1:1 from P on. Or, from a new span, 0,1,2 on. Always, endless succession will still be ahead at any K in finite reach from P. Where of course for all we care there is onward endlessness to the left of P. From P we do not have a powerful enough means to get to -2 etc. So, the issue is, infeasible supertask. Further to this, inferring that we are always after the transfinite has been bridged so we are always within finite reach of -2 etc, begs the question. KFkairosfocus
December 28, 2017
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KF, Ok, I will happily withdraw my assumption which you term "begging the question", and ask you to prove this traversal is impossible directly. I think that has been your strategy all along, anyway. I would ask you to please address the π example first, however, since I believe it is relevant to the argument you would make regarding the traversal itself.daveS
December 28, 2017
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DS, the traversal of a transfinite span is precisely the issue at stake. By assuming that you beg the question. KFkairosfocus
December 28, 2017
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KF, If, by "begging the question" you mean stating in all-caps, flashing neon letters that I'm assuming up front that the traversal has taken place, yes. What about the digits of π illustration? Do you agree that the expansion has infinite length, yet each digit is at finite distance (number of steps) from the decimal point?daveS
December 27, 2017
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DS you full well know that by begging the question of the infinite traverse you leave on the table a finite descent to n. KFkairosfocus
December 27, 2017
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KF, et al, Since my challenges in #128 and #152 didn't lead to much response, I'll post something which I had planned to bring up if that discussion had developed further. We are talking about this infinite chain of people who are passing an iPad toward the person at the end of the chain (John). It has been claimed at various junctures that if this chain is infinite, then there must be people in it who are at infinite remove from John (for example, Ishaan and everyone behind him). Needless to say, I disagree. I hesitate to bring up another model of this scenario since that has led to confusion in the past, but I think this one might be helpful. The people in the chain pair up one-to-one with the digits in the decimal expansion of π. Let's ignore the "3" to the left of the decimal point. John pairs up with the "1" in the tenths position, the next person pairs up with "4" in the hundredths position, and so on. Notice this association gives us an isomorphism of linearly ordered sets in a natural way. A couple of facts/observations: (1) The decimal expansion of π has infinite length, obviously, because π is irrational. Err, right? Does anyone think that the decimal expansion of π is actually finite because the digits are each finitely distant from the decimal point? I sincerely doubt it. Does anyone think that there are digits in the decimal expansion infinitely distant from the decimal point? I doubt this also. (2) I claim it's clear that given any nonempty collection of (the positions of) these digits, there is a single position closest to the decimal point. For example, consider the positions of all the digits which equal zero. The zero closest to the decimal point is in the 32nd position. You can even get fancier and look for the positions of strings of digits. The first occurrence of "12345" begins at position 49702. Interestingly enough, this string of digits occurs 2018 times in the first 200 million digits of pi. This supports my premise (1) in #152.daveS
December 26, 2017
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DS, the assumption begs the question of stepwise traversal of the transfinite. We have put that back on the table and have shown that it is inherently deeply problematic. KFkairosfocus
December 26, 2017
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KF, To your #182: I am assuming such (explicitly, so no question-begging), and asking you to show me how this results in a contradiction. An analogy: I could claim that there are only finitely many primes, and ask you to prove this claim is false by demonstrating that it leads to a contradiction. Note that I am not trying to prove the iPad completed this traversal, so it's impossible for me to beg the question.daveS
December 26, 2017
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Oops---correction to the second sentence: I am assuming it [the iPad] never moves to the left. Rather, it always moves toward the right, toward the (one) end.daveS
December 26, 2017
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DS, your argument keeps implying that at any given stage transfinitely many token-passes have already happened. For instance, see your: "for every position, there are infinitely other positions to the left." This consistently begs the question at stake. KFkairosfocus
December 26, 2017
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Origenes, Yes, for every position, there are infinitely other positions to the left. And I am assuming it never moves to the right. But that doesn't imply that the iPad missed some positions (at least I haven't seen any proof of this). If the iPad is currently 100 steps from the end, then 900 seconds ago, it was 1000 steps from the end, for example.daveS
December 26, 2017
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DaveS: More precisely, I claim that the iPad never entered the chain—it has always been in it. You are right that if I claimed it did enter the chain at some time, then my proposal would be impossible.
Okay, let's suppose that the iPad was always in the chain and, in the past, always moved towards the now. If it ever moved towards the past it would never return. But here is another problem: for any position in the past it goes that there are infinite positions to the 'left'. So, isn't it necessarily the case, that whenever the iPad moves to the right (towards the now), it does not enter positions on the left? IOWs if the iPad was always in the chain and always moved to the 'right', there were always positions on the left it did not enter. Always having been inside the chain, doesn't seem to alter the fact that moving to the right implies always infinite positions on the left. If an infinite can be traversed by incremental steps (which is highly doubtful) then one requirement seems to be that one takes steps towards that infinite and not in the opposite direction. = = = BTW I do think I have provided logically/mathematical prove that the positions constitute a finite set.Origenes
December 24, 2017
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Origenes, More precisely, I claim that the iPad never entered the chain---it has always been in it. You are right that if I claimed it did enter the chain at some time, then my proposal would be impossible. I also don't claim to have an explanation for how this was all accomplished---I'm merely asserting that it did, and asking for anyone to prove that it's logically/mathematically impossible.daveS
December 24, 2017
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KF, The sets {..., -3, -2, -1, 0} and {0, 1, 2, 3, ...} are not isomorphic as ordered sets, agreed? Are we also agreed that for each natural number k, k is finite?daveS
December 24, 2017
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KF: ... supertask: spanning endlessness in successive +1 increments. The means hath not the power as at every k it arrives at, k+1, k+2 etc continue endlessly and in 1:1 correspondence with the counting numbers from 0 on.
Indeed, there is no progress on a relevant level. It is as if incremental steps take place on a micro-level, while infinity takes place on a macro-level, which is insensitive to the first. It does not matter how many steps you take, for any position, between the infinite past and the now, goes: towards the past are infinite positions. How did progress take place? How did the iPad traverse?Origenes
December 24, 2017
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DaveS @75: In the scenario I described, I do not envision the iPad “entering” the infinite regress at all. It’s eternally being passed down through this chain of people, in a “beginningless” process.
I would like Dave to elaborate on his vision, because I don’t understand how it could work. For any position, between the infinite past and the now, goes: towards the past are infinite positions. However, Dave claims that the iPad only goes towards the now only after it entered all positions in the past. But how does the iPad accomplish that? I see two options and both fail: 1) by first going towards the past, but that would be an endless journey. 2) by starting from the first position of the past, but there is no first position of the past.
DaveS: Of course the past is settled and no additional positions can be added to it after the fact.
So, a beginningless infinite process has come to an end. :)Origenes
December 24, 2017
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DS, your challenge is that to have a token T coming from an endless past to now, you have to do the equivalent of spanning and exhausting the set of naturals in a stepwise, +1 per stage, cumulative process. The symmetry between the natural counting numbers and the negative integers -- a simple transform converts the one to the other (and cf. the Surreals as illustrated in the OP) -- implies just that supertask: spanning endlessness in successive +1 increments. The means hath not the power as at every k it arrives at, k+1, k+2 etc continue endlessly and in 1:1 correspondence with the counting numbers from 0 on. We may even view this as an informal mathematical induction . . . though strictly, the succession onwards is implicit in that procedure. The token passed to some p (transfinitely remote) would always have that endlessness still before it. And for the claim that the token has come from an actually transfinite chain, some p that once held the token then passed it +1 to p+1 etc must be transfinitely remote. As a consequence, if a passed token has just arrived at n-1, n from a chain of members that passed it in succession, we may freely infer that the chain is not transfinitely long. KF PS: Transform, in simple representation: {0,1,2 . . . } x(-1) --> {0, -1, -2, . . . } PPS: We must also recall that properties of individual cases and properties of the whole may but need not differ -- for a wall to be a red brick wall, every brick must be red, but the mass of the whole is greater than that of any given brick. Yes, any specific k can be exceeded k+1, k+2 etc. Put that in correspondence with 0, 0+1, 0+2 etc and we see that no particular integer we state has spanned endlessness. The endlessness, however is material. It is a structural component of the set. Extending to Z, this is double-sided. Succession to w and -w is by recognition that countable endlessness has a new type of quantity, symbolised by its order type w.kairosfocus
December 24, 2017
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KF, PS to #173: In my #170, I mean that given any integer k, it lies finitely many steps from 0. I'm not saying you can construct the entire set Z in finitely many steps.daveS
December 23, 2017
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KF, By definition, every natural number can be reached by applying the successor operation a finite number of times to zero. Any number not reachable in this way is not a natural number.daveS
December 23, 2017
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DS, no, we may reach finite values stepwise from 0, but we may not exhaust the naturals that way as for every k or - k we may reach, we may count on endlessly from there. We may place count on from K in correspondence to count from 0, this is structural. Just so, no finite stage cumulative process can span endlessness. KFkairosfocus
December 23, 2017
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PS to #170: I see KF already brought this point up. Of course the past is settled and no additional positions can be added to it after the fact. I don't believe that creates any problems for my position, however.daveS
December 23, 2017
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Origenes, How many elements does the set {..., -3, -2, -1, 0} have? Infinitely many (aleph null). Are any infinitely distant from 0? No, every integer can be obtained by adding 1 (or -1) finitely many times to 0.daveS
December 23, 2017
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KF @168
KF: Origenes, DS’s underlying point is that the naturals, of infinite number, are all finitely removed from 0. This is the context for my pointing out so insistently the endlessness of the set as a whole as a key structural feature.
I believe I do the same when I point out that there are two types of sets, one infinite set and one finite set. Of course, a finite set is a set that has a finite number of elements. I should, perhaps, explicate an assumption that I make during my reasoning: the past is settled. Given that the past is settled, at the moment the iPad arrives in the now, the number of positions that preceded its arrival is a fixed number. Once in the now, that number is not up for change. Given that the past is settled, in the now there can be no extra positions added to that number. Can that number be "infinity"? No, that number is finite, which is guaranteed by the fact that, per DaveS’claim, the iPad in the past is “always a finite distance” from the now and due to fact that no more positions are added, since the past is settled. So, we have a finite set of positions that the iPad entered between the past and the now. I do not see why it is any more complicated than that.Origenes
December 23, 2017
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Origenes, DS's underlying point is that the naturals, of infinite number, are all finitely removed from 0. This is the context for my pointing out so insistently the endlessness of the set as a whole as a key structural feature. Consequently, as each step imposes a +1 increment, the total remove must be equally endless. Where at ANY finite k, no matter how large, you can go on k, k+1, k+2, etc, which can be "slid" to 1:1 correspondence with 0,1,2 etc equally endlessly. Indeed, in previous discussions I used the picture of pink and blue marked tapes endless to the right to make the point. So, there literally is no last natural v such that v, +1 --> w, omega, the first transfinite. Instead, by reckoning endless succession as integral to the naturals, we recognise that w is successor to the endlessness, its order type in more technical terminology. Now, scroll up to the OP and notice the diagram of the surreals. Do you see how the number line is flipped upwards and bent to the right like a bow? The negative integers are a mirror image of the counting numbers. This is where I now point out that to have traversed the chain from a negative transfinite labelled member, a transfinite traverse would have to have been crossed. That cannot be done stepwise, the token just cannot cross endlessness in +1 steps. That it is at n-1, n now means it has only traversed finitely many steps. And indeed you are right to point out that if every negatively labelled member of the chain is finitely remote, the traverse indicated is never infinite. Talk about infinitely many finite and bounded negative integers falls into a mare's nest. Let us trust we can now return from this tangent. KFkairosfocus
December 23, 2017
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DaveS: My assumption is that the positions the iPad has entered is the set {…, -3, -2,-1, 0}, that is all positions with nonpositive index/coordinate. There is no lower bound N on this set of positions. It’s exactly the same as the set of all positions in the chain.
No, that's not possible, because you have stated the following (emphasis added):
DaveS: The iPad’s journey (moving forward in time) has an endpoint, and it’s always a finite distance from that endpoint.
The position of the iPad in the past, relative to the now, is always finite, this means that the set of positions that the iPad entered in the past up to the now can only be a finite set. Never can it be a an infinite set. So, again, we have two sets (see #164), one finite and one infinite — the latter is larger. Unless you are willing to argue that there is no difference between an infinite set and a finite set, your claim is refuted.Origenes
December 23, 2017
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DS, enjoy your Christmas. Meanwhile, the situation still stands that you have not solved the passage from the transfinitely remote problem, and we have reason to hold that no stepwise, finite stage, cumulative process can do so. Saying or suggesting that at any given stage s, the actual transfinite passage has already been done, does not answer the problem. And, I have again pointed out that there is no neighouring finite v in the counting numbers such that v, +1 --> w . . . there is no one step increment of mere degree from finite to transfinite. Endlessness is inherently structural to the naturals [that is a crucial point captured and symbolised by the innocuous-seeming but extremely powerful ellipsis, . . . ], and this is the exact reason why we cannot traverse it stepwise. Remember, N = {0,1,2, . . . } Where w is order type as identifying the scale of the endless succession as a new type of quantity, qualitatively different from finite ones precisely because it embeds endlessness of counting succession. We need a more powerful operation than steps of finite size to bypass a transfinite traverse or to leap over it. Sort of, like taking reciprocals of infinitesimals -- indeed, [0,1] captures a dual of the whole number space including transfinites once we bring in infinitesimals as neighbours of 0 and use reciprocals 1/m --> M, with M > w, where too, trivially any natural value n is dual to 1/n, with w > n. Or in this case, living in a world created by or at least rooted in a necessary being first cause at finite remove. Such a being is necessitated by our being here as a going concern world that is clearly contingent. In such a case, no finite stage stepwise process will be transfinite and there is no need to suggest or imply an infinite past finite stage causally cumulative succession of any kind. KFkairosfocus
December 23, 2017
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All: I'm about to head out for some Christmas season travel, so I'll be semi-offline for a few days. I might be able to sneak in some posts if I get good access to WiFi here and there. Feliz Navidad. Origenes: My assumption is that the positions the iPad has entered is the set {..., -3, -2,-1, 0}, that is all positions with nonpositive index/coordinate. There is no lower bound N on this set of positions. It's exactly the same as the set of all positions in the chain.daveS
December 23, 2017
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DaveS: I’m not seeing any problem yet.
We have two sets: A: All positions of the chain from the infinite past to the now: [infinity ….. -2 , -1 , 0] B: All positions that the iPad entered in the past to the now: [N ….. -2 , -1 , 0] DaveS’ claim: “the iPad entered all positions of the chain from the infinite past to the now.” If the iPad entered all positions from the infinite past to the now, then A – B = 0. However, we know that N is a finite number. And we also know that infinity is larger than any finite number. Because A is infinite and B is finite, it follows: A – B = A Conclusion: DaveS' claim is false.Origenes
December 23, 2017
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