The sequence that grows remarkably large, then drops to zero! 

we dont remain sane my good friend

Sadly, Dr. Kaczynski is no longer with us.

Their secret is to be slightly off-kilter before they start their careers. And if you are in doubt of this, just check out their fashion sense.

I recall Kurt Gödel starving himself out of paranoia

False premise

But either both sequences terminate, or neither do. And we know Goldstein terminates at 0. So just stating that it terminates is enough, it terminating at 0 is just a consequence

Thanks!

@@DcubedJ what itsimpsosible anyone could guess that number goes to zero from those first few terms? How could they?? I see no evidence it willgo to zero..

@@DcubedJ omega isnt same size as omega plus 1 though since it has one more term..could you clarify?

It did. The views kept increasing, but surprisingly dropped to 0.

why do you belive it only deserves 66k views?

I have a version of the Collatz Conjecture that has the same result, but doesn't subtract 1. The -1 seems to be superfluous.

No, it is not the same. The 1 really has nothing to do with it in either case. In Goodstein's theorem the real reason it goes to zero is because as you map the powers you end up mapping the numbers to lesser and lesser possibilities. That is, there are far fewer numbers with higher hereditary power which is why they have to grow so fast. That is, take any number below 10^10, there will be no numbers in that range with hereditary power 10 except 1 itself(that that is when the -1 comes in to play but it's really not that relevant). So the point is simple, when you arbitrarily convert the powers you are taking numbers that have low hereditary power(which there are a lot) and mapping it in to a space of much higher hereditary powers but only in to a much smaller set. It really doesn't matter about the "magnitude" of the numbers as that relatively meaningless. Collatz is much deeper since it has to do with the intrinsic nature of the integers and their parities. [And note we are not subtracting 1 in Collatz but adding, although it doesn't matter much but it does completely change the results(with 3n-1 the Collatz conjecture is false)] There is a reason why the Collatz conjecture has not been proven and that is because it requires extremely advanced mathematics in the sense that mathematicians have not figured out some deeper nature of the integers and it actually may be that the Collatz conjecture is false or unprovable(which could be a reason no one has made any real progress on it). The Collatz conjecture looks very simple but it is asking very complex questions of a very deceptive process that is fractal in nature. Because of it's "simplicity" it makes it very hard to figure out what really is going on. On the surface it's just a contractive map but there is something else involved that millions of mathematicians can't figure out... which tells you a lot about that "something". Keep in mind that these "mathematical problems" like the arbitrarily modifying the powers doesn't mean much because there are an infinite number of ways to construct such things. There seems to be two types of mathematicians in the world... those that fabricate problems and try to create theorems to get their names in the book and then those few that do real mathematics. It's very easy to prove things when no one else has worked on it. It's very hard to prove things when a million other people have tried. For example, just thinking along the same lines, you could take a base b expansion and convert the base^n to n^(n*base): sum(a_k*b^k) -> sum(a_k*k^(k*b)) to get a new number. If you iterated this process then you get: sum(a_k*(k^b)^(k^b*b)). If you can prove something interesting and non-trivial about it then you could be "famous" and get your names in the history books. Since no one has looked at such a thing(at least very unlikely) it means that anything you find that isn't trivial and you can prove it then you'll have a theorem which you can publish and claim you are special. That is how it works. ("maybe that is how all math works")

To add a nuance to your last point, just because a problem is difficult doesn't mean it is important or interesting, and conversely just because nobody has looked at a problem before doesn't mean it is not. Considering questions nobody has looked at before and trying to answer them is what most mathematicians do. Sometimes it results in an easy theorem, sometimes it results in a hard problem, but either way, it is "real mathematics". Most mathematicians never solve any groundbreaking problem, or get "famous". But they do lay their bricks on the collective construction of mathematical knowledge nonetheless.

The "power of subtracting one" doesn't really exist in this case. What happens is that, eventually, the sequence comes to a point where the Nth term will have no "N"s when written in Hereditary-base N (because N is bigger than it). From that point on, the base-changing operation doesn't do anything, and every new term is just going to be the previous minus 1, until it reaches 0.

​@@MDNQ-ud1tyOH... okay... totally different way of moving through the number space because of the changing bases... which makes Goodstein's work even more fascinating... thanks for clearing all that up!

Yup, you are correct in that what you have described is the mechanism that eventually decreases the sequence values. The proof shown in the video is the rigorous proof that all goodstein sequences will eventually decrease to 0.

Ah, thanks! You've helped me understand why this actually does work instead of just naively accepting that somehow subtracting 1 outpaces the seeming exponential growth of the sequence. More explanations about infinity need a rationalizing explanation like this. So often we're expected to just accept that infinity works strangely and leave it at that, despite it seemingly breaking basic logic. Showing that the basic logic still makes sense is pretty crucial.

@@DcubedJ I don't see whymathematicians oranyone else would concoct this ordiscover it? It doesn't seem smart or necessary to me--so why and how was it done? Thanks for sharing.

@@leif1075, thanks for your question. I am not too sure why or how Goodstein first constructed and proved this theorem. As to why it is important - it was one of the first (wikipedia says 3rd) statements that was shown to be unprovable in Peano arithmetic but is provable in second order arithmetic. For me personally, i found it fascinating that a sequence that can grow larger than Grahams number will always decrease down to 0. Now whether you find the above important/interesting is up to you, and that’s fine :)

Yup makes sense and thanks for your feedback!

I second it! It would be nice to see the end of a sequence and how we reach zero.

This reminds me about PBS infinite series, the video about the Hydra.

G(3) ends very quickly. It’s only 6 terms. The number of terms in G(4) is a 121 million digit number so you’d be waiting a long time for it to reach 0…

@@JohnSmith-nx7zj Obviously, an example does not need to be complete to communicate a concept ;)

In hindsight, I completly agree with you - I think I could have done a better job of introducing the topic. Appreciate your feedback and will keep that in mind for next time.

I completely agree. Once I realized where the video was heading at the end, things started becoming way more interesting than they seemed at the beginning.

i can agree, the beginning of the video was still enjoyable to me but lukewarm, but the end of the video was WAY cooler. still, i really enjoyed the video, beginning included:)

@@DcubedJ I'm not even sure it was really necessary. Like I said, what I liked the most about your video is how it deified all my expectations. The fact that you didn't oversell your problem at the start made me presumptuously assume that the topic was not really interesting. But, when around 5:00, I realize how wrong I am, you've hooked me indefinitely (since your video, I've been talking about this sequence to everyone I know 😉 ). Anyway, that was a month ago, and still, you video remains one of my favorites from #SOME2! Can't wait for the next one.

Yes. I had to go to Wikipedia too to be able to fully grasp what was happening

thanks!

Except base one. (does that count?)

Is that a pun?

Actually, it really isn't possible to convince youeself if you're not familiar with the concept of "infinities can be larger than other infinities". The only sequence that you can write in its entirety by hand (or even print out with a computer) is starting with the numbers 0, 1, 2 and 3. They're all pretty boring. 3 takes 6 steps to get to zero. A starting value of 4 will require a stunning 3*(2^27-27)-2 steps if I'm not mistaken.

Hey, please check the references in the description of this video. There is a link to the github repo titled “Goodstein Calculator”. Hope that helps!

Thanks!

Hey yeah it has been a while since I researched ordinals so I had to go back to my notes regarding your question. When we want to find the ordinal exponentiation of a successor ordinal i.e a^b where there exists a y where y+1=b, then a^b=(a^y).a where y=b-1. For ordinal exponentiation of a limit ordinal a^b this is equal to the supremum (or least upper bound) of {a^y where y

@@DcubedJ thanks, that helped! but I still don't quite understand how to construct w×2, is that just sup({w+n | n is a natural number})?

It's good to clarify that a sequence will terminate in a _finite_ number of elements. But, I'd like to see the mechanism by which the value shrinks and specifically the context in which a tower of powers can produce zero.

@@creativenametxt2960 For 2w, I think its easier to think of it as w+w. Remember, w+1={0, 1, 2, 3, ..., w} w+2={0, 1, 2, 3, ..., w, w+1} so if we follow this pattern, w+w={0, 1, 2, 3, ..., w, w+1, w+2, ...} and w+w+w={0, 1, 2, 3, ..., w, w+1, w+2, ..., 2w, 2w+1, 2w+2, ...} etc. Important note here is that as long as there is a lower bound (which is 0 for w, w+w, w+w+w), then these sets are well ordered therefore any subset of these sets cannot be infinitely descending.

That's... not true? An infinite sequence of positive values does not necessarily reach a given limit. Take 1+½+¼+... and note that it never becomes 3. You can set up your snail problem equivalently.

@@DavidGuild the rubber band as it stretches also advances the snail. Here is a short video that explains it intuitively with an ant (rather than a snail) and a stretching rope. ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-lbjTGqZspjE.html There are other videos that use the harmonic series to explain it more rigorously, which you can easily find by searching "ant on a rubber band paradox" (and the video to which I linked has other more detailed videos in the description).

Hey thanks for your comment. So regarding (1) all goodstein sequences (if we accept goodsteins theorem) will have an upper bound. Now the question is - what is the upper bound of a goodstein sequence starting at x. Well the easiest answer is when x=3 and this kind of answers your second question as well. hb-2 of 3: 2+1 2->3, -1: 3+1-1=3 3->4, -1: 4-1=3 4->5, -1: 3-1=2 5->6, -1: 2-1=1 6-7>, -1: 1-1=0 So the upper bound is 3 and the length of the sequence starting at x=3 takes 6 steps to terminate. The simplicity of x=3 can be quite misleading because the length of the goodstein sequence starting at 4 is known to take 3(2^402653211)-2 steps.

@@DcubedJ Thanks. It's always nice to see even a trivial example, so that the final result doesn't sound so crazy :)

To answer your second question, swapping the numbers for the hereditary notation is always going to make the number bigger if there are terms besides the units term, and if there are any terms besides the first degree term, it will grow by more than one. The sequence will flatten out once we have just the degree 0 and 1 terms. Consider 17 starting with base 10. We would have (Base 10) 17=10^1+7 (Base 11) 11^1+7-1=11^1+6=17 (Base 12) 12^1+6-1=12^1-5=17 One the base gets large enough (base 18) we start having only unit terms, which decrease by 1 (Base 18) 17 (Base 19) 17-1=16 (Base 20) 16-1=15 So what happens with this sequence is that we slowly eat away at these degree 0 and 1 terms until there are only higher degree terms, and when we subtract 1 we finally decrease the exponent. An example of this is (Base 3) 29=3^3+2 (Base 4) 4^4+2-1=257=4^4+1 (Base 5) 5^5+1-1=3125=5^5 (Base 6) 6^6-1=5*6^5+5*6^4+5*6^3+5*6^2+5*6+5 These numbers aren’t getting smaller but when we cross to base 6 we no longer have the both the base and exponent growing. Very slowly, we are able to chip away at the terms that grow quickly and replace them with terms that grow more slowly until we only have linear terms left.

Yes - ordinals have a fixed ordering to them, and you can define some operations on them (although this wasn't covered in the video). omega + 1 is different than omega, as shown. Then omega*2(=omega+omega) is the set of all numbers you get by adding 1 from that (all ordinals of the form n or omega + n). omega*3 is by doing that again, and then omega*4, etc. Then you have omega^2(=omega*omega) which comes from doing that multiplication increasing repeatedly (so it's the set of all ordinals omega*a+b), then you can continue on in similar fashion from there to get omega^2 * 2, and then keep going to omega^3, ..., to omega^omega, then omega^(omega+1), etc.

"w-1" isn't a number. Neither is "w-2". Say for example you are at w in the chain. You know the next ordinal must be something less than w. Well all the numbers less than w are Natural Numbers, so whatever that number is it is a finite Natural Number. And then there can only be a finite number of decreasing steps after that before you get to the minimum fixed point 0.

@@Bodyknock So if I've understood correctly the reasoning is that the natural numbers in the 'Goodstein sequence' will always be less than w and thus bounded by it, at the same time the 'new sequence' is clearly decreasing meaning for any finite value other than w the sequence would finally terminate at 0. It still seems so paradoxical that an increasing base and exponent could ever be overcome by the subtracting of one. Really mind boggling if that is the case. Very interesting though

@@oskarwallberg4566 Not quite. Here's a quick proof that there are no infinite strictly decreasing sequences of ordinals. - By definition, the ordinals are well-ordered, meaning that any set or sequence of ordinals has a minimum. - Assume you have an infinite strictly decreasing sequence of ordinals. As above, it has a minimum, let's call it aₘ. - Because aₘ is a minimum, and the sequence is strictly decreasing, there can be no elements in the sequence after it because if there were then the next element would be smaller than the minimum. Thus there can be no elements in the sequence after aₘ. - But that's a contradiction because infinite sequences don't have a "last element", meaning if the sequence is infinite there must be some next element after that element aₘ which, since the sequence is strictly decreasing, would be smaller than it. Therefore by contradiction there are no infinite strictly decreasing sequences of ordinals, and so any strictly decreasing sequence of ordinals must be finite and terminate at some number. But in this Goodstein example, there's only one possible such number they could terminate at, namely 0, because 0 is the only possible fixed point in the sequence (i.e. the Goodstein number 0 generates the ordinal 0 and then the next Goodstein number remains at 0 so the ordinal also doesn't change.) Therefore these ordinal sequences generated from Goodstein numbers are finite and always end in 0.

Hey, thanks for your question.. I don't think I did a good job of explaining the proof in detail so understandably there seems to be some misunderstandings here. The first thing I want to highlight is that we never perform "minus 1" or any operation on the "new sequence". The parallel sequence is simply constructed from its corresponding Goodstein sequence value by replacing all n of the current HB-n with omega. Eventually, there won't be any "n"s left for the base changing operation to have any effect so it will continue to just decrease by 1 from that point onwards. This can be shown with the Goodstein sequence starting at 3. hb-2: 2+1 2->3, -1: 3+1-1=3 3->4, -1: 4-1=3 4->5, -1: 3-1=2 (no more n's left for the base change to have any effect) 5->6, -1: 2-1=1 6->7, -1: 1-1=0 Another point (and again this is probably my fault) is that the Goodstein sequence being bounded by the parallel sequence actually does not matter. The important point is that the parallel sequence is simply constructed from the Goodstein sequence so if a value in the parallel sequence is 0 then the only way that it can be 0 is if the corresponding Goodstein value is also 0. Hope that helps!

@@Bodyknock This clarified, thank you.

Thanks for your feedback :)

yeah, I believe the proof takes similar approach

Hey thanks for your kind words. So one point that may help you understand is that the new sequence is simply constructed from its corresponding goodstein sequence value. The change of base from n -> n+1 and the minus 1 is never performed on the parallel sequence. It is only done on the goodstein sequence.

@@DcubedJ thanks for the reply! I understand "how" the new sequence is construct, but not "why" it was constructed this way. In other words, why would the comparison/conclusion be valid if the new sequence is not following the sequencing of the original one?

@@diediepet3 Ahhh yep, i think i misunderstood your question. I guess the reason it was constructed this way is to use it as a tool to prove that goodstein sequences reach 0. As long as the construction of the parallel sequence is consistent, then it is totally fine for us to use. Many sequences are derived by this "replace" operation. The goodstein sequence itself is derived by replacing the n (of the current HB-n notation) with n+1 then minus 1. The parallel sequence is simply derived by replacing n with omega of the current goodstein sequence value in HB-n notation. Edit: An analogy could be, imagine we had an unknown sequence A. We construct a parallel sequence B by doubling the corresponding value in A. If we somehow figure out that one of the values in B is 0. Then we can conclude that the corresponding value in A is also 0 because the only value doubled that equals 0 is 0 itself. As long as the construction of the parallel sequence is consistent, then we can make this conclusion.

Yeah, i believe Hydra game(sequence?) also had a similar solution to Goodstein theorem.

​ @John Długosz Hey John, thanks for your feedback. Yeah I see what you mean and I guess I had to kind of choose what I thought were the interesting elements to the definition and proof of this theorem which may differ with others. The actual mechanism that starts reducing the sequence is actually the -1 after increasing the base. When the sequence reaches a value where the HB-n notation looks like n+x for an x

Hey, thanks for your comment. If I am understanding your question correctly, you are asking how the new sequence can decrease to 0 if the value is w^w^w or 2w, since subtracting 1 from these values is not defined. And you are correct in that we cannot subtract 1 from these infinite ordinals. The minus 1 actually occurs on the Goodstein sequence value which we then use to construct a corresponding "New sequence" value. So we never actually subtract 1 in the new sequence at all, although it may look like it. Hope that helps, but let me know if I am misunderstanding your question.

@@DcubedJ See, the thing is, in order for the Goodstein sequence to be bounded by the new sequence, we must eventually subtract from the new sequence. The problem is that it seems like the new sequence could take an infinite amount of time to reach 0.

@@blockmath_2048 you are always subtracting from the Goodstein sequence, so when what looks like subtracting 1 from 2w at the new sequence, it's actually doing 2n+0-1 in base n,which turns it into 1n +(n-1) and 2w becomes w+(n-1)

@@DcubedJ The reason we can’t subtract 1 from ω an infinite number of times is because the ordinal ω-1 is undefined (that is a surreal number, but not an ordinal)

The 2nd last term for all Goodstein sequences is 1.

@@DcubedJ Oh. Right, duh! More generally, then, if one takes an input of a Goodstein sequence, is it possible to predict the ending terms?

Yep, so the sequence actually starts decreasing when it reaches a value where the HB-n notation looks like n+x. This value actually stays the same until x=0 because we increase n by 1 then decrease x by 1. This continues until x=0 then the value starts decreasing by 1 until 0. So to answer your question the last values of a Goodstein sequence will be n, ..., 5, 4, 3, 2, 1. However, I am not too sure if there is an easy way to determine when the decreasing actually begins though...

​@@DcubedJthe decreasing begins once a(n)=n, and goes until a(2n)=0. n is difficult to compute for any given starting seed number, though.

The first term in the sequence often ends up looking like n^n while the rest look like n^2 or just a constant, so n^n will dominate the value of the expressions. When n is a power of two, n^n will also be a power of two with the rest acting like a negligible error term.

Same

Hey, thanks for your comment and im glad you enjoyed the video. Your question about when/how the minus 1 breaks the omegas is an interesting one and this is how i think of it. Say for the current hb-2 notation of our goodstein value is 2. The parallel sequence value for this value is omega. Now for the next step we increase 2 to 3 then subtract 1 so we are left with 2 again. However since the current hb-n is n=3 our parellel sequence value is 2. So we have dropped from omega to 2 in the new sequence. For powers of omega, a similar process occurs. Say the current hb-2 goodstein value is 2^2. The parallel sequence value is w^w. Now we change 2 to 3 and -1 so we get 3^3-1. Without fully calculating we can already see that the highest power will drop because of the minus 1. So the highest power in the new sequence will have dropped from omega to some finite number.

@@DcubedJ Thank you for your reply! So if I have well understood, when we switch from n to n+1 in the normal world, we are not doing w to w+1 but rather w to n+1 in the omega world, subtracting 1 and then switching to w again? I am sure that I have confused something because if what I think is true, then everything happens as if there was no omega. Or maybe omega is here to tell us that even though everything appears to increase, we are in fact not adding true value to the sequence, and the -1 will eventually terminate the sequence? So it is a matter of the -1 catching up with the replacement of n to n+1 so that at one point, we cannot logically apply the successor in the tail of the k-th term which will eventually kill it at one point. Then we apply the same reasoning with the successive power chains. Is that right, or I got it wrong?

Yes, I think you are on the right track. This may help your understanding as well. We never perform any operation such as n->n+1 or -1 on the parallel sequence (all of these happen to the goodstein sequence). The parallel sequence values are simply constructed from its corresponding goodstein sequence value. And i think your comment about the "-1 catching up with the replacement of n to n+1" in the goodstein sequence is spot on. This is the mechanism that will slowly decrease the sequence to terminate at 0!

@@DcubedJ Thank you very much, now I understand the matter clearly :)

@@miguetyann8266 for a bit of additional explaining, each natural number is less than w but none come before it. Because of this w-1 is not an operation you can do. In the ordinals, you can always ask “what comes next” but you can’t always ask “what comes before.”

That is a very interesting question. I’m not too sure if there is an easy way to calculate the length of ALL Goodstein sequences, however it is known that the sequence starting at 4 takes (3 x 2^402653210) - 1 steps to terminate (no idea how someone calculated that).

@@DcubedJ wow, super interesting, ty!

​@@DcubedJ so we know what is the "number" before 0? What is that number? Kinda bother me that what kind of number lead to zero in the sequence, i mean before zero there must be some non zero number

@@puturavindrawiguna3025 its a 1

@@gwenoleroger6262 thank you sir, it makes sense.

The last but one term will be one by definition. As to where the biggest term will appear that is a very good question. Given the slow decline, it must occur "relatively" near the start of the sequence(keep in mind these sequences can be extremely long, so it still might take a huge number of steps to compute)

One reason 2 is defined as {{}, {{}}} is because that way the number of elements in the set is 2. In other words the set associated with a Natural Number n has n elements. If you did it the way you're asking then there would always only be one element in the set. Also by using the method in the video you get the property that subsets are like the < relation (i.e. 2

Hey, so the notion of an omega-th term isn’t really defined as that implies that you have reached step n where n is the last finite number. However assuming that we are able to somehow reach the n-th step, Goodstein’s theorem states that all sequences terminate in a finite number of steps so we wouldve reached 0 long before step n.

I don't understand how this makes sense... couldn't you say that each step you add 9 balls? then it must be increasing...

​@@santiagovidal4497 You can search Ross-Littlewood paradox for more info. Basically need to number the balls with natural numbers. First step put 1 to 10 into the box, then remove ball 1, second step put 11 to 20 into the box, then remove ball 2. After infinitely many step, the infinitieth ball is removed, so no ball left in the box. This is also a supertask

Hey thanks for your question.. the length of goodstein sequence starting at 3 takes 6 steps. But any starting number >3 gets quite large. i.e sequence starting at 4 takes 3x2^402653211-2 steps! No clue how someone calculated that..

They hit a maximum, then they start the very slow decrease to 0.

😵

Hey, thanks for your question. The new sequence is constructed by replacing the current n of the HB-n notation to omega. So if the current n is 2 then we replace all 2s with omegas and if the current n is 3 then we replace all 3s with omegas. Hope that helps!

What would be the next ordinal after w? The question is not an infinitely descending set, because sets are unordered - it's an infinitely descending sequence.

Hey, thanks for your question. We never perform -1 directly on the parallel sequence and this is probably my fault for not explaining clearly. The -1 is always done on the Goodstein sequence which only conatins finite numbers. The parallel sequence is simply constructed from its corresponding Goodstein sequence value by replacing all n’s from its HB-n notation to omega. Hope that helps!

@@DcubedJ makes a lot more sense now, thank you so much!

Hey, so the number that becomes 0 is always 1. The mechanism that eventually decreases the sequence is the -1. For example say for the current n=10 of the hb-n notation is 2. Then we increase 10->11 and -1 which will give 1. Then 11->12 and -1 gives 0. Hope that helps!

Hey, thanks for your question. Just a couple of points that may help. Omega is the first transfinite ordinal number and is defined as the set containing all the natural numbers i.e. w={0,1,2,3,...}. What this means is that w-1 is not defined as that implies the "last finite ordinal number". The "minus 1" is never performed on the new sequence containing omega and is only done on the goodstein sequence which only contains finite ordinals. The new sequence is simply constructed from the corresponding goodstein sequence value. From this, if we believe that the new sequence is decreasing then it cannot be infinite and must reach 0 (well-ordering). Since the new sequence is directly constructed from the goodstein sequence, it follows that the goodstein sequence also eventually reaches 0. Hope that helps!

@@DcubedJ Right. You're not actually subtracting 1 from omega at 14:25. You're defining an operation F(x) which is x-1 for finite ordinal, and defined as some arbitrary finite value for F(omega).

Hey, thanks for your comment. The only full goodstein sequence i can give you is the one starting at 3: hb-2 of 3: 2+1 2->3, -1: 3+1-1=3 3->4, -1: 4-1=3 4->5, -1: 3-1=2 5->6, -1: 2-1=1 6-7>, -1: 1-1=0 The reason this is the only one i can provide is because the goodstein sequence starting at 4 takes 3(2^402653211)-2 steps, and the goodstein sequence starting at 5 takes 10^10^10^20000 steps!

Hey, yep good pick up! I need to be careful of these things..

Hey, yep so it really comes down to how we define ordinals and "infinite" ordinals. The first transfinite ordinal (omega) is defined as the first ordinal number after all the natural numbers i.e. w = {0, 1, 2, 3, ...} and because order matters, it is conceivable to define w+1 = {0, 1, 2, 3, ..., w} and so on. Even with this definition, their are limitations such as w-1 is not defined because it implies the "last finite ordinal" which doesn't make sense.

Thanks for your feedback.. Yeah.. its a good point however the only examples that can be shown of a sequence going up and down to 0 are trivial examples of starting numbers 1, 2 and 3. The sequence starting at 4 takes 3x2^402653211-2 steps and the sequence starting at 5 takes approximately 10^10^10^20000 steps!

Hey, yep good catch. I guess the point still holds in your clarification in that the highest power in the parallel sequence will decrease (which shows that the parallel sequence is decreasing). Regarding your statement about the "class of ordinals is not a set", I am not too sure what you mean here. We have defined ordinals as a set containing all the elements of the previous set along with the previous set added as an element. i.e. n+1 = n union {n}. Let me know if I am misunderstanding you here.

@@DcubedJ the collection of all ordinals, if it were a set, would be an ordinal (because any set where [all elements of it are ordinals, and where for every ordinal it contains, it also contains all smaller ordinal], is an ordinal), but, then, it would contain itself. But no set contains itself. So it cannot be a set. The class of all ordinals is, in a sense, “too big to be a set”, and is therefore called a “proper class” (Just like the class of all sets is a proper class, rather than being a set.) (Of course, my statement that “no set contains itself (as an element)”, well, that depends on what foundations you are using, but I assume that we are working in ZFC (or, I guess, that one conservative extension of ZFC, the name of which I forget. Unlike ZFC it has a finite axiomitization, but as for the statements that can be expressed in both it and in ZFC, they prove exactly the same ones.))

I agree with you that the set of all ordinals is not an ordinal itself because that would imply that this set is an element of itself (similar to what you mention). I never mentioned that the set of all ordinals is an ordinal and if I did then this is a mistake. The main point of that section is to show that if a set contains all the ordinals then there is a minimum element hence showing that any set of ordinals will have a minimum as well.

@@DcubedJ No, I’m saying that, if the collection of all ordinals were a set, then it would satisfy the definition of an ordinal, and therefore be an element of itself, and therefore the collection of all ordinals is not a set. The contradiction that one obtains by assuming that the class of all ordinals is a set, is known as the “Burali-Forti paradox” .

Yep good point and I agree with you that the set of all ordinals implies a set containing itself (which leads to the paradox). But the point of that section is to show that IF we were somehow to find a set containing all the ordinals then there is a minimum element. I am not claiming that such set exists but was using it to help demonstrate that IF we construct a set with all ordinals then it will always have a minimum element. Similar analogy is a set containing all the natural numbers is not defined until we use "omega" notation for ordinals.

Hey thanks for your comment. So the function is replacing all n of the current hb-n notation with omega. So if the current hb-2 notation is 2^2^2, then the parallel sequence value is w^w^w.

Hey, thanks for your kind words.. The Goodstein sequence starting at 1, 2 and 3 takes 2, 4, and 6 steps respectively. But the sequence starting at 4 takes 3(2^402653211)-2 steps! I have no idea how this was calculated..

One reason 2 is defined as {{}, {{}} } is because the number of elements in that set is 2. This way the set that's related to a Natural Number n always has n elements. Also it has the nice property that if m < n then m's set is a subset of n's set.

@@Bodyknock Ooh ok I see, in fact itseem a good reason ^^ Thanks for this learning :)