Talk:Epsilon number

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Cantor's Transfinite numbers are also called aleph (aleph is the first letter of the hebrew alphabet, and slightly assembles a gothic N). Aleph zero is the first of these transfinite numbers, and the ordinal of the infinite countable sets, such as N (natural numbers), Z (whole numbers), and Q (rational numbers). Proved by Cantor with the diagonal proof, it's considered one of the most visual demonstrations..it is in fact included in what Paul Erdös called "The Book", a book in which were included the perfect proofs for Mathemathical theorems, alongside with it there is the subsequent Continuum Hypothesis, which is also the first of the 23 Hilbert Problems. This Continuum Hypothesis states that there is no set whose size is strictly between that of the integers and that of the real numbers, that is aleph zero and aleph one, respectively.
 * What does this have to do with &epsilon;0? It is not an aleph; it's a countable ordinal. &mdash; Carl (CBM · talk) 00:30, 2 December 2007 (UTC)

$$\gamma$$ or $$\gamma-1$$

 * "The $$\gamma$$-th ordinal $$\alpha$$ such that $$\alpha=\omega^\alpha$$ is written $$\varepsilon_\gamma$$. These are called the epsilon numbers. The smallest of these numbers is &epsilon;0."

If the smallest is called &epsilon;0, isn't the $$\gamma$$-th called $$\varepsilon_{\gamma-1}$$? Otherwise $$\varepsilon_0=\varepsilon_1$$ --SuneJ (talk) 07:34, 6 December 2007 (UTC)


 * I added the clarification "counting from zero". Using an ordinal to specify the position of an element in a well-ordered set always counts from zero, but I agree that the formulation should not be ambiguous.--Patrick (talk) 09:32, 6 December 2007 (UTC)

Definitions?
I'm just starting to look at transfinite numbers, and I'm wondering if there is a fully expandable definition of epsilon numbers other than $$\epsilon_0$$. If $$\epsilon_0=\omega^{\omega^{\omega^{.^{.^{.}}}}}$$, is there a similar way to define $$\epsilon_1$$ based on lower ordinals like $$\epsilon_0$$ or $$\omega$$?Eebster the Great (talk) 19:04, 20 April 2008 (UTC)


 * Just as $$0, 1, \omega, \omega^{\omega}, \omega^{\omega^{\omega}}, \ldots$$ approaches $$\epsilon_0 \!$$.
 * So also $$\epsilon_1 \!$$ is the limit of $$\epsilon_0 + 1, \omega^{\epsilon_0 + 1}, \omega^{\omega^{\epsilon_0 + 1}}, \ldots$$ and also the limit of $$0, 1, \epsilon_0, \epsilon_0^{\epsilon_0}, \epsilon_0^{\epsilon_0^{\epsilon_0}}, \ldots$$.
 * For any ordinal $$\alpha \!$$, $$\epsilon_{\alpha + 1} \!$$ is the limit of $$\epsilon_{\alpha} + 1, \omega^{\epsilon_{\alpha} + 1}, \omega^{\omega^{\epsilon_{\alpha} + 1}}, \ldots$$ and also the limit of $$0, 1, \epsilon_{\alpha}, \epsilon_{\alpha}^{\epsilon_{\alpha}}, \epsilon_{\alpha}^{\epsilon_{\alpha}^{\epsilon_{\alpha}}}, \ldots$$.
 * If $$\lambda \!$$ is any limit ordinal, then $$\epsilon_{\lambda} \!$$ is the supremum of $$\epsilon_{\beta} \!$$ for $$\beta < \lambda \!$$. JRSpriggs (talk) 02:11, 21 April 2008 (UTC)

Thanks, also, is this an actual mathematical limit we're talking about (a behavior) or is it actually a power tower omega (or some other transfinite ordinal) stories high? Another way of looking at it is, if the operation were defined, would this simply be omega tetrated to the omegath power? Eebster the Great (talk) 00:30, 10 May 2008 (UTC)


 * Both. Using the order topology, this is the topological limit (is there any other kind?). If the hyper operator notation were extended to ordinals, we would get $$\operatorname{hyper4} (\epsilon_{\alpha}, \omega) = \epsilon_{\alpha + 1}$$. In other words, it is tetration as you surmised. JRSpriggs (talk) 21:14, 10 May 2008 (UTC)


 * Thanks a lot. That's interesting, because although $$\omega < $$infinity, $$lim$$x->inf x$$\omega = $$$\omega$$$\omega$$.  And I was talking about a calculus limit, by the way, not really knowing enough about topology to discuss it one way or another.  My understanding of topology amounts to the most cursory of all cursory explanations in the context of quantum physics. Eebster the Great (talk) 02:45, 15 May 2008 (UTC)

History?
Who first introduced &epsilon;0? Cantor? What about&epsilon;1 ? linas (talk) 01:36, 5 June 2008 (UTC)


 * The &epsilon; notation has been around since at least 1908; Felix Hausdorff uses it in "Grundz&uuml;ge einer Theorie der geordneten Mengen" with the modern meaning. I don't know if Cantor used the same notation but there is no doubt that he was aware of the concept. Mr Death (talk) 19:17, 12 September 2008 (UTC)

countability
The article states that &epsilon;0 is "countable", can this be made more explicit? I can see how $$\omega^2$$ is countable, but don't understand the leap to counting $$\omega^\omega$$, then extending this to &epsilon;0. linas (talk) 01:43, 5 June 2008 (UTC)
 * The union of countably many countable sets is countable. --Trovatore (talk) 02:17, 5 June 2008 (UTC)
 * See Ordinal notation. Every ordinal less than &epsilon;0 can be described uniquely by a finite string consisting of just two symbols &mdash; "0" and "ξ". Clearly there are only countably many such strings. JRSpriggs (talk) 04:52, 5 June 2008 (UTC)
 * Thanks, right. I was thinking.. well, I wasn't actually thinking about ordinals when I asked this question. linas (talk) 16:12, 5 June 2008 (UTC)

Move to "epsilon numbers"?
Given the recent change of focus, should the article be moved/renamed to epsilon numbers/epsilon number? After all, the article currently starts with the words “In mathematics, the epsilon numbers are a collection of transfinite numbers whose defining property is…” — Tobias Bergemann (talk) 06:54, 12 September 2008 (UTC)


 * I could be persuaded either way. There's only so much to say about &epsilon;0 itself, and I think that what there is to say about it is better expressed in the context of "epsilon numbers" generally.  On the other hand, I suspect that there are more direct references to &epsilon;0 elsewhere in Wikipedia than to the general concept, and that the external world contains more references to &epsilon;0 without discussion of epsilon numbers generally than vice versa.  Michael K. Edwards (talk) 15:35, 12 September 2008 (UTC)


 * Either we should do this or redirect ordinal numbers to $$\omega_0$$. (For the sarcasm-impaired, I am in favour.) Mr Death (talk) 19:27, 12 September 2008 (UTC)

I don't really think this is a good idea. I think ε0 has a modest intrinsic interest (first place that Cantor Normal Form runs into a circularity, first discovered proof-theoretic ordinal) that doesn't really extend to the epsilon numbers in general. A few words here or at Cantor normal form should suffice for the latter. --Trovatore (talk) 19:43, 12 September 2008 (UTC)
 * Epsilon numbers after ε0 (i. e., the series ε1, ε2, ε3, ...) are not so interesting, but the idea of an "epsilon number" (a fixed point of an exponential map) is. It seems to me that material that connects ε0 to the next "interesting" ordinal &Gamma;0 belongs somewhere, and that no one who doesn't know the field is going to know to look for it under Veblen function.  And there's also the material about surreal epsilon numbers, which I think deserves a reasonably close linkage with the ordinal version given the common motivation (enumeration of fixed points of an exponential map) and the natural embedding of the ordinals in the surreals.  Michael K. Edwards (talk) 21:14, 12 September 2008 (UTC)
 * I'm certainly not going to try to talk you out of being interested in what you're interested in. I submit that a lot of people who might be interested in reading about ε0 are not interested in epsilon numbers in general. --Trovatore (talk) 08:32, 13 September 2008 (UTC)

Definition as an equation?
The article says that epsilon numbers can be defined by the equation
 * $$\varepsilon = \omega^\varepsilon$$

However, this doesn't make sense to me. By analogy, we should be able to define $$\alpha = \omega+\omega+...+\omega+\omega$$ as the smallest $$\alpha$$ that satisfies the equation $$\alpha = \alpha+\omega$$, which clearly isn't true, since $$\omega+\omega+...+\omega+\omega = \omega*\omega = \omega^2$$, and $$\omega^2+\omega$$ is a distinct ordinal from $$\omega^2$$. Am I missing something? The definition in terms of tetration, $$\epsilon_0 = \omega\uparrow\uparrow\omega$$, makes much more sense to me. —Deadcode (talk) 20:16, 22 April 2010 (UTC)
 * It's not analogous. There simply does not exist an ordinal &alpha; such that &alpha;=&alpha;+&omega;, but there does exist an &epsilon; such that &epsilon;=&omega;&epsilon;.
 * On the other hand, there does exist &alpha; such that &alpha;=&omega;+&alpha; (the smallest such &alpha; is &omega;2). --Trovatore (talk) 23:22, 22 April 2010 (UTC)


 * To Deadcode: You seem to be assuming that $$\omega = 1 + 1 + \ldots + 1 + 1$$. However, that is false, and indicates that you do not understand what an ordinal is. It would be closer to the truth to say that $$\omega = 1 + 1 + \ldots$$ . Notice that it has no last element . JRSpriggs (talk) 21:56, 23 April 2010 (UTC)


 * I realize that it has no last element. I only typed it out that way because I was thinking of the definition of Graham's number, which is finite, hence my mistake. Incidentally, the Ordinal Number page on WolframMathWorld makes the same error (for example it has $$\omega+...+\omega$$) so I'm in good company!
 * Please accept this rephrased analogy:
 * By analogy, would should be able to define $$\alpha = \omega+\omega+\omega+...$$ as the smallest $$\alpha$$ that satisfies the equation $$\alpha = \alpha+\omega$$, which clearly isn't true, since $$\omega+\omega+\omega+... = \omega*\omega = \omega^2$$, and $$\omega^2+\omega$$ is a distinct ordinal from $$\omega^2$$.
 * I understand the difference between $$\omega+1$$ and $$1+\omega$$; the difference is that $$1+\omega=\omega$$. However, as I understand it, $$\epsilon_0=.^{.^{.^{\omega^{\omega^{\omega^\omega}}}}}$$, not $$({({({({\omega^\omega})^\omega)}^\omega)}^\omega)}^\omega...$$, so $${\omega}^{\epsilon_{_0}}$$ should be a distinct ordinal from $$\epsilon_0$$. The key is that exponentiation has right-to-left evaluation, and I believe this has been overlooked and has misled people into believing that $$\epsilon=\omega^\epsilon$$. —Deadcode (talk) 07:24, 24 April 2010 (UTC)


 * To Trovatore: Could you please explain why that is true? I understand that $$\omega+{\omega^2}=\omega^2$$, because evaluation of addition is left-to-right, but that doesn't extend by analogy to $${\omega}^{\epsilon_{_0}}$$ because exponentiation has right-to-left evaluation. —Deadcode (talk) 06:57, 24 April 2010 (UTC)
 * Right-to-left or left-to-right doesn't matter for addition &mdash; addition is associative. Somehow you've gotten things backwards here.  The fact is that &alpha;&omega; is always bigger than &alpha;, but &omega;&alpha; is not.  I could go into a formalistic explanation, I suppose, but I think it's beside the point; somehow you've just gotten something backwards in your head, and you just need to write it down carefully and examine it, and I think you'll see it.
 * (Just by the way, and not meaning any offense, being in company with MathWorld is in my experience not really something to brag about.) --Trovatore (talk) 08:41, 24 April 2010 (UTC)
 * Regarding MathWorld, I was joking, not bragging. But this is probably not the best space in which to continue this discussion, since it would probably become longer than anything else in this discussion page. —Deadcode (talk) 17:22, 24 April 2010 (UTC)
 * To Deadcode: &epsilon;0 is neither of the two things you mentioned. It is $$\epsilon_0 = \omega^{(\omega^{(\cdots)})} \,.$$ More precisely, it is the supremum specified in the lead of the article with the rightmost exponentiations executed first. JRSpriggs (talk) 22:43, 24 April 2010 (UTC)
 * Yes, thank you. I have learned how to properly think about $$\epsilon_0$$ after discussing it on a math forum — I no longer have any issue with its definition. BTW, I knew that it was evaluated right-to-left, and that is what caused the problem for me in the first place — I thought it broke the analogy of the ladder of operations. Now I see that all of the limit ordinal operations take place from right-to-left. For example, $$\omega=1+(1+(1+(1+...)))$$ and not $$(((1+1)+1)+1)+...$$ or $$(((...+1)+1)+1)+1$$, and that is why $$1+\omega=\omega$$ (and so on, up the ladder of operations). This is what was the key in helping me understand $$\epsilon_0$$. —Deadcode (talk) 02:18, 28 April 2010 (UTC)
 * I really can't figure out what you mean by this. Ordinal addition is associative; the order of evaluation is irrelevant.  If you have a wellordered sequence of ordinals and you want to add them up, you can do it all in one fell swoop &mdash; you just concatenate them all, and take the order type. --Trovatore (talk) 03:08, 28 April 2010 (UTC)
 * To Trovatore: That is only true for a finite number of additions. When you add $$1$$ to $$\omega$$, or $$\omega$$ to $$\omega^2$$, addition is no longer associative. Notice that if $$\omega=1+(1+(1+(1+...)))$$, then $$1+\omega=1+(1+(1+(1+(1+...))))$$; all you see is an extra level of nesting out of an infinite nesting, and it's obvious that $$1+\omega=\omega$$. However, if it were true that $$\omega=(((1+1)+1)+1)+...$$, then $$1+\omega=1+((((1+1)+1)+1)+...)$$, and it would no longer be obvious that $$1+\omega=\omega$$. Notice that I wanted a deeper justification for the definitions of ordinal arithmetic; when I had a problem with $$\epsilon_0$$, it was because I was thinking in terms of $$\omega+1=1+1+1+1+...+1$$, and $$\omega^2+\omega=\omega+\omega+\omega+\omega+...+\omega$$, and $$\omega^\omega*\omega=\omega*\omega*\omega*\omega*...*\omega$$, which is an analogy that breaks at exponentiation (when constructing $$\epsilon_0$$) — but now I realize that this is a nonsensical way of thinking about it; it's like saying that $$1-0.99999...=0.00000...1$$. —Deadcode (talk) 22:59, 29 April 2010 (UTC)
 * No, you're wrong. Ordinal addition is associative, period.  The expression 1+&omega; is precisely the order type of one single point followed by &omega; single points. --Trovatore (talk) 23:37, 29 April 2010 (UTC)
 * I'm saying that addition is associative for a finite number of additions, but that it is not associative for an infinite number of additions. &omega; represents an infinite number of additions of 1 to itself. How you associate these additions with each other determines whether $$1+\omega=\omega$$ and $$\omega+1\neq\omega$$ or $$\omega+1=\omega$$ and $$1+\omega\neq\omega$$; as it so happens, &omega; is defined such that $$1+\omega=\omega$$ and $$\omega+1\neq\omega$$. I'm basically saying that the non-commutativity of ordinal addition depends on the non-associativity of the addition used to construct &omega;. If you prefer to think of those statements $$1+\omega=\omega$$ and $$\omega+1\neq\omega$$ as axioms of ordinal arithmetic, then do so; I'm sure it's the more standard way of thinking; but I prefer my way, as it gives a deeper reason for the behavior of ordinal addition. —Deadcode (talk) 00:50, 30 April 2010 (UTC)
 * You're not being nonstandard; you're being wrong. The difference between &omega;+1 and 1+&omega; is a question of the order in which the terms appear, not the order in which the operations are evaluated. --Trovatore (talk) 00:53, 30 April 2010 (UTC)
 * Yes, my hypothesis is indeed wrong, even as a way of thinking about ordinal arithmetic. Given that way of thinking, $$\gamma=1+(1+(1+(...)+1)+1)+1$$ should result in a set that stretches off to infinity in both directions — it is of course no longer a well-ordered set, and subsequent addition in either direction should result in a new order type (if this concept is allowed for non-well-ordered sets). However, using the justification I formerly used to show why $$1+\omega=\omega$$ now shows that in this case, $$1+\gamma=\gamma$$ and $$\gamma+1=\gamma$$. So it would appear for the moment that it's better to express the concept of &omega; as $$\omega=1+1+1+1+...$$, with unspecified associativity. But this re-opens the whole can of worms regarding infinitely recursive non-commutative operations that bothered me in the first place. Luckily I now understand how to justify $$\omega^{\epsilon_{_0}} = \epsilon_0$$ in terms of sets rather than just notation and definitions. —Deadcode (talk) 01:38, 30 April 2010 (UTC)
 * To Deadcode: I think you are focusing too much on the operations and not enough on the ordering of the elements of the set. JRSpriggs (talk) 05:29, 28 April 2010 (UTC)
 * To JRSpriggs: On the contrary. It's just that I wanted to visualize arithmetic operations on ordinals in a way that is isomorphic to the corresponding operations on well-ordered sets of those order types. It is impossible to visualize a set with the order type of $$\epsilon_0$$ from the top down. That is why I wasn't able to justify $$\epsilon_0 = \omega^{\epsilon_{_0}}$$ in terms of thinking about nested sets of natural numbers. However, now I realize that it's easy to visualize $$\epsilon_0$$ in a different way. Like any other ordinal, it is the order type of the ordered set of ordinals smaller than itself; for $$\epsilon_0$$, this is the ordered set of ordinals in Cantor normal form. This lends itself to a bijection showing that $$\omega^{\epsilon_{_0}} = \epsilon_0$$. —Deadcode (talk) 23:05, 29 April 2010 (UTC)
 * I want to demonstrate some more of what I've learned since starting this discussion section, by describing a bijection that shows that $$\omega^{\epsilon_{_0}} = \epsilon_0$$. I'm not studying set theory formally, so my terminology is going to leave something to be desired, but here goes. The bijection is this: Take set $$C_0$$ to be the set of all Cantor normal form ordinals (all ordinals less than $$\epsilon_0$$) in ordinal order. Define a set schema $$P(A)$$ as this: for every element in $$A$$, create a natural number element in $$P(A)$$, preserving order. Every natural number value of an element of $$P(A)$$ corresponds to $$\omega$$ raised to the power of the corresponding ordinal in $$A$$, multiplied by the natural number value as a coefficient. Define the set $$C_1$$ to be the set of all sets following the schema $$P(C_0)$$, in reverse lexicographical order, where each set can also be thought of as an ordinal in Cantor normal form. $$C_1$$ is still the set of all ordinals in Cantor normal form; this shows that the order type of $$C_1$$ is the same as the order type of $$C_0$$, which shows that $$\omega^{\epsilon_{_0}} = \epsilon_0$$. —Deadcode (talk) 00:26, 30 April 2010 (UTC)


 * Trovatore is correct that even the addition of an infinite number of ordinals is associative.
 * Actually, &epsilon;0 can be understood "top down" as the ordering of the &xi;-notations described at Ordinal notations. JRSpriggs (talk) 01:26, 30 April 2010 (UTC)
 * But isn't this still "bottom up", like Cantor normal form, in that what you're really visualizing the set that results from the bottom-up construction of all ordinals? Given smaller transfinite sets, I really can think of them top down; for example, $$\omega^{\omega^{\omega^2}}$$ is the order type of the set, sorted in reverse lexicographic order, of all ω-dimensional matrices of ω-dimensional matrices of ω-dimensional matrices of ... [ω times] of natural numbers where elements are eventually zero in all directions and dimensions. That's nearly as high as I think I can go, in top-down visualization. —Deadcode (talk) 01:54, 30 April 2010 (UTC)

(unindent) Suppose I is a totally ordered set which is used to index other totally ordered sets Si for i&isin;I. Then the sum $$\sum_{i \in I} S_i \,$$ can be understood as the obvious ordering of $$\bigcup_{i \in I} (\{i\} \times S_i) \,$$ which is clearly associative.
 * I don't understand this, unless it's meant to be an intentional abuse of $$\sum$$ notation — a mathematical joke, a play on symbols. In that case, I see your point. The ellipsis in 1+1+1+1+... is not formal notation and should not be taken literally. —Deadcode (talk) 19:19, 1 May 2010 (UTC)

The ordering of the ξ-notations can be gotten all at once from: ξ(α,β)<ξ(γ,δ) if and only if either (α=γ and β<δ) or (α<γ and β<ξ(γ,δ)) or (α>γ and ξ(α,β)≤δ). The algorithm is highly recursive, i.e. it calls itself. You also have to assume that 0 is less than any other element. JRSpriggs (talk) 03:48, 30 April 2010 (UTC)
 * Yes, ξ-notation is described nicely in the Ordinal notation article, and it is pleasingly austere. However, I was talking about ways to visualize a set with order-type $$\epsilon_0$$, so I don't see what relevance this has. If I understand correctly, the less-than operator only recurses down through the lower ordinals used to define ξ(α,β) and ξ(γ,δ) within the ξ-notation, and does not construct or recurse through the infinite set of all ordinals less than ξ(α,β) and ξ(γ,δ). I don't see what relevance this has to visualizing $$\epsilon_0$$, which can't be expressed in basic ξ-notation anyway.
 * Constructing all ordinals under $$\epsilon_0$$ with ξ-notation is very simple (using infinite recursion starting with a set containing just 0) — but it's not great as a visualization tool, because you end up with long strings of ξ and 0 which aren't instantly meaningful to the human eye (including the mind's eye). To think about the ordinals thus constructed, you have to fall back to Cantor normal form, for example. —Deadcode (talk) 01:20, 1 May 2010 (UTC)

Higher epsilon numbers by tetration
There's a new thing that's bothering me, now that I've accepted that $$\omega^{\epsilon_0}=\epsilon_0$$. The article says that
 * $$\varepsilon_1 = \omega \uparrow \uparrow \omega^2 = \sup \{ (\varepsilon_0 + 1, \varepsilon_0 \cdot 2, {\varepsilon_0}^2,) \, {\varepsilon_0}^{\varepsilon_0}, \ldots, \varepsilon_0 \uparrow \uparrow k \, (k \ge 2), \ldots \}$$

However,
 * $$\omega\uparrow\uparrow(\omega\cdot 2) = \sup\{\omega\uparrow\uparrow\omega,\omega\uparrow\uparrow(\omega+1),\omega\uparrow\uparrow(\omega+2),\omega\uparrow\uparrow(\omega+3),...\} = \sup\{\epsilon_0,\omega^{\epsilon_0},\omega^{\omega^{\epsilon_0}},\omega^{\omega^{\omega^{\epsilon_0}}},\ldots\} = \sup\{\epsilon_0,\epsilon_0,\epsilon_0,\epsilon_0,\ldots\} = \epsilon_0$$

Similarly,
 * $$\omega\uparrow\uparrow(\omega\cdot 3) = \sup\{\omega\uparrow\uparrow(\omega\cdot 2),\omega\uparrow\uparrow(\omega\cdot 2+1),\omega\uparrow\uparrow(\omega\cdot 2+2),\omega\uparrow\uparrow(\omega\cdot 2+3),...\} = \sup\{\epsilon_0,\omega^{\epsilon_0},\omega^{\omega^{\epsilon_0}},\omega^{\omega^{\omega^{\epsilon_0}}},\ldots\} = \sup\{\epsilon_0,\epsilon_0,\epsilon_0,\epsilon_0,\ldots\} = \epsilon_0$$

By induction,
 * $$\omega\uparrow\uparrow(\omega\cdot n) = \epsilon_0$$

As a result,
 * $$\omega\uparrow\uparrow\omega^2 = \sup\{\omega\uparrow\uparrow\omega,\omega\uparrow\uparrow(\omega\cdot 2),\omega\uparrow\uparrow(\omega\cdot 3),\omega\uparrow\uparrow(\omega\cdot 4),\ldots\} = \sup\{\epsilon_0,\epsilon_0,\epsilon_0,\epsilon_0,\ldots\} = \epsilon_0$$

I could continue this on and on, and it would prove that $$\omega\uparrow\uparrow\alpha=\epsilon_0$$ if and only if $$\alpha\geq\omega$$. What have I done wrong? —Deadcode (talk) 09:49, 3 May 2010 (UTC)


 * The article on hyperoperations does not specify what happens when the right argument is a limit ordinal. So we can draw no conclusion, except perhaps that that section of this article should either be deleted or clarified by providing such a specification. JRSpriggs (talk) 06:30, 4 May 2010 (UTC)
 * Yeah, I hadn't really bothered to look very closely at the arrow-notation stuff in this article before. I still haven't tried very hard to figure it out, but from what effort I have put in, I'm not convinced it makes a lot of sense as written.  It appears to have been added by User:Michael K. Edwards; maybe one of us should contact him to see if he has any references, or whether this is some formulation of his own. --Trovatore (talk) 17:29, 4 May 2010 (UTC)
 * I am working on a way to define ordinal operations higher than exponentiation such that something similar to what Michael K. Edwards added might be consistent; however, it's not working out very well at all so far. I've attempted to contact Michael K. Edwards through his talk page. In the meantime, should his section be removed, or peppered with citation-neededs, or prefaced with a notice indicating that it is questionably original research? —Deadcode (talk) 23:28, 4 May 2010 (UTC), edited 03:13, 5 May 2010 (UTC) and 10:15, 5 May 2010 (UTC)
 * Update: My attempts to work out a rigorous meaning for tetration $$\omega\uparrow\uparrow(\omega+1+\alpha)>\omega\uparrow\uparrow\omega$$ have failed, so it appears quite clear now that $$\omega\uparrow\uparrow(\omega+\alpha)=\omega\uparrow\uparrow\omega$$, and that $$\varepsilon_\alpha>\varepsilon_0$$ must be defined either as a supremum, or as $$\varepsilon_{\alpha+1}=\varepsilon_\alpha\uparrow\uparrow\omega$$. This is in direct conflict with what Michael K. Edwards said about tetration in the article. —Deadcode (talk) 19:59, 5 May 2010 (UTC)
 * Really I think even the $$\omega\uparrow\uparrow\omega$$ thing is problematic, and should be sourced or removed. --Trovatore (talk) 20:51, 5 May 2010 (UTC)
 * Sorry, I haven't looked at Wikipedia in quite some time. I'm not a professional in this area, and will happily defer to anyone who is.  But as far as I'm concerned the Conway approach to transfinite ordinals (placing them in the context of the surreal number system) is far more straightforward than the traditional Cantor exposition, and makes the extension of the hyperoperations to general surreal operands fairly routine.  It's certainly not original research; if I recall correctly (I lent my copy to someone), it's all there in On Numbers and Games, although Conway's exposition is elliptical and his notation positively unhelpful.  Epsilon numbers are also discussed towards the end of Gonshor, although I can't make heads or tails of his proof tactics myself.
 * You are quite right that I did a sloppy job of translating results from the surreal number exposition to Knuth notation. $$\omega\uparrow\uparrow\omega$$ is to be read as the supremum of $$\omega\uparrow\uparrow k,\, k < \omega$$, i. e., $$\varepsilon_0$$.  $$\omega\uparrow\uparrow(\omega+1)$$ should then be read as $$(\omega\uparrow\uparrow\omega)\uparrow\omega = {\varepsilon_0}^\omega$$, not $$\omega\uparrow(\omega\uparrow\uparrow\omega) = \omega^{\varepsilon_0} = \varepsilon_0$$.  (This isn't just "magic notation"; it follows from the surreal-number definition of tetration, which is a recursive definition much like that for multiplication.  Essentially, $$\omega\uparrow\uparrow(\omega+1) = (\omega\uparrow\uparrow\omega)\uparrow\omega$$ is the simplest surreal number larger than $$(\omega\uparrow\uparrow\omega)\uparrow k$$ for all natural numbers $$k$$, just like $$\omega\cdot(\omega+1) = (\omega\cdot\omega) + \omega$$ is the simplest surreal number larger than $$\omega\cdot\omega + k$$ for all natural numbers $$k$$.)  Then $$\omega\uparrow\uparrow(\omega\cdot 2) = (\omega\uparrow\uparrow\omega)\uparrow(\omega\uparrow\uparrow\omega) = {\varepsilon_0}^{\varepsilon_0}$$, and so forth, until $$\omega\uparrow\uparrow(\omega\cdot\omega) = {\varepsilon_0}^{{\varepsilon_0}^{{\varepsilon_0}^\cdots}} = \varepsilon_1$$.
 * If this explanation helps justify the material that has been deleted since [], feel free to restore it. (The jump from epsilon numbers to the Veblen hierarchy, which I had attempted to explain by connecting the Veblen normal functions to the surreal hyperoperations, seems completely unnatural in the current article.)  I think this material is important, since it helps explain where the Feferman–Schütte ordinal comes from, and the sense in which it is still an "epsilon number" even though it is not reachable from 0 using the Veblen construction.  Michael K. Edwards (talk) 22:04, 6 June 2011 (UTC)

Surreal epsilon numbers and the exponential map
I'm going to delete the section of "surreal epsilon numbers" dealing with the map x&mapsto;2x. I deleted the same thing from the surreal numbers article. Reason being, I can't find any reference for it. There were no references listed (and there are none on this page either), and when I tried looking up the surreal exponential map myself, I found definitions that disagreed with the ones listed here. I then asked about it on MathOverflow and nobody there had heard of this definition -- only the ones I had found in the books, which as I said earlier, is inequivalent. I've tried to find User:Michael K. Edwards, who added it to both of them, but he appears to be long inactive and has left no way to contact him. Thus, I deleted it from that page, and I'm going to delete it from this page now. Sniffnoy (talk) 04:37, 7 February 2013 (UTC)

Redirect?
I'm a physics student, and upon typing epsilon nought, was redirected to this page, which was very annoying. Because the vacuum permittivity constant is probably in much higher demand than this pure mathematics, I have created a disambiguation page; I believe that ε₀ should also point to this disambiguation page. Ε₀ (disambiguation)
 * Hmm, honestly, I can see your point. But foo should never point to foo (disambiguation); that's just silly.  If we adopt your approach, then it's ε₀ itself that should be a disambiguation page.  This needs admin assistance &mdash; we would first move ε₀ to ε₀ (ordinal) (or whatever disambiguated title is chosen), delete the redirect left behind, move ε₀ (disambiguation) to ε₀, and then do some cleanup. --Trovatore (talk) 03:40, 18 November 2013 (UTC)
 * How do we get an admin? Vacuum permeability works exactly like this; type in a name one thinks should direct them to vacuum permeability, and one gets vacuum permeability. — Preceding unsigned comment added by Feynmaniac (talk • contribs) 05:49, 18 November 2013 (UTC)


 * You could look at Category:Wikipedia administrators and try to find an administrator with whom you are acquainted. Or just pick one at random. JRSpriggs (talk) 07:21, 18 November 2013 (UTC)
 * The way issues like this are usually solved is through WP:RM. Keφr 07:47, 18 November 2013 (UTC)

Discussion at Wikipedia talk:WikiProject Mathematics
There is some discussion at Wikipedia talk:WikiProject Mathematics. I started it as a discussion about the redirects Epsilon zero, Epsilon 0, Epsilon naught, Epsilon nought, ε₀, and ε0, but it later took a turn towards the name and scope of this article. — Tobias Bergemann (talk) 08:47, 8 January 2014 (UTC)

Requested move 13 July 2022



 * The following discussion is an archived discussion of a requested move. Please do not modify it. Subsequent comments should be made in a new section on the talk page. No further edits should be made to this section.

Moved to Epsilon number, per the discussion. BD2412 T 02:31, 22 July 2022 (UTC)

Epsilon numbers (mathematics) → Epsilon numbers – Only topic by that name. Per Talk:Epsilon zero and Articles for deletion/Epsilon number, other topics at the Epsilon zero dab (formerly at Epsilon number) are not epsilon numbers. Rotideypoc41352 (talk · contribs) 14:18, 13 July 2022 (UTC)
 * Move to Epsilon number; the title should be singular. –LaundryPizza03 ( d c̄ ) 07:33, 14 July 2022 (UTC)
 * Note to closer: I am also okay the singular (in the style of Aleph number, for example). Rotideypoc41352 (talk · contribs) 18:52, 14 July 2022 (UTC)
 * I opined in Talk:Epsilon zero for the plural but after looking again I think the singular makes sense - even though the plural is common, there is some amount of usage of the singular such as the entry epsilon_number. Furthermore there are lots of article titles of the form "X number" but only a few pages like Large numbers. So agree with Laundry. Mathnerd314159 (talk) 22:16, 14 July 2022 (UTC)
 * The above discussion is preserved as an archive of a requested move. Please do not modify it. Subsequent comments should be made in a new section on this talk page. No further edits should be made to this section.

last part of intro section
says stuff about gamma numbers in terms of ω^β and ω^ω^β, but what is β in this context? a successor ordinal? 24.137.229.197 (talk) 05:04, 19 March 2024 (UTC)

Spelling "epsilon nought" vs. "epsilon naught"
According to the second paragraph of, there is a tendency in British English for "nought" to mean "zero" and for "naught" to mean "nothing". In light of that $ε0$ is the first in the sequence with next entry $ε1$, the former definition appears appropriate. According to the same article, Americans spell both definitions as "naught". I see no choice of American vs. British English for this article. So, what do we do? I am thinking to use the British usage because it is more precise, despite that I usually gravitate to American English. — Q uantling (talk &#124; contribs) 15:58, 5 July 2024 (UTC)


 * If all terms are correct we could have that written as 'pronounced epsilon naught, epsilon nought or epsilon zero', since my change was reverted despite the word nought not appearing in my dictionary and being underlined red. Vytron (talk) 16:07, 5 July 2024 (UTC)
 * I like that; I'll make an edit. If it could be better ... please edit there or reply here.  Thank you — Q uantling (talk &#124; contribs) 17:00, 5 July 2024 (UTC)