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# July 17

## Large Recursive Ordinals and Large Cardinals

Let me preface this by saying: I'm just beginning, as in yesterday, to introduce myself to Ordinal analysis, and that this question comes more from a musing/stray thought than it does anything else. That said is there any sort of expected symmetry between large cardinal properties and large recursive ordinal properties - in other words, LCA some large cardinal axiom, does the structure of LCA tell us something about the nature of the proof theoretic ordinal of ZFC + LCA? Moreover, can we, from the relations of the various types of large cardinals, infer anything about the relations between their proof theory ordinals when added to ZFC? I apologize, in advance, if this question is not the most clear, or if it is stupidly glossing over something obvious. Thanks for any help/insight:-)Phoenixia1177 (talk) 06:58, 17 July 2014 (UTC)

It's not my field of mathematical logic, (except, or course, I knew ε0 for PA), but the article ordinal analysis states: "Most theories capable of describing the power set of the natural numbers have proof theoretic ordinals that are so large that no explicit combinatorial description has yet been given." Most large cardinals require power set in their definition. — Arthur Rubin (talk) 07:23, 17 July 2014 (UTC)
Thank you for the reply - I realize that we have no description of these ordinals, but are we able to say anything about them otherwise? For partial analogy, as far as I know, we have no way of describing large cardinals directly, but we can say that if a Huge cardinal and a Supercompact cardinal exist, then the first huge is smaller than the first supercompact. In this case it need not be size comparisons, but is there anything that could potentially be said? --addendum: another analogy: we can't directly describe the Absolute Galois Group of the Rationals, but we can say things about the group itself; is there something that might be said/expected to be said about the overall collection of ordinals that would result? I realize that this is vague, my brain just keeps nagging me that there is something interesting here.Phoenixia1177 (talk) 07:38, 17 July 2014 (UTC)
This is a little off the track from your question, but the example that you mention is kind of an interesting one, because although the first huge cardinal comes before the first supercompact, the existence of a huge cardinal is a much stronger axiom, in terms of consistency strength, than the existence of a supercompact.
There are other anomalies like this, but as far as I'm aware, they're all based on LCAs that are not "local" — that is, you can't say that a cardinal is supercompact just because there's some sufficiently large α such that Vα thinks the cardinal is supercompact.
Non-local LCAs are a bit problematic because they rely on knowing stuff about the whole universe. In Woodin's work around 2000, he gave an abstract definition of large cardinal (not widely adopted, but interesting) that simply made locality part of the definition. I don't know of any counterexample to the claim that, when you restrict attention to local LCAs, the LCA with the larger first witness always has larger consistency strength. (However, a properly stronger LCA might have the same first witness as the weaker one.)
Also, the order of consistency strength on LCAs seems to be the same as how far up the Wadge hierarchy they prove regularity properties for sets of reals (again with possible collisions — for example, weak large cardinal axioms, those consistent with V=L, don't prove any regularity properties beyond ZFC itself). --Trovatore (talk) 11:32, 17 July 2014 (UTC)
Your "slight tangent" answers a question at least as interesting to me as my original, doubly so since I had no idea about the Wadge Hierarchy - thank you:-) Do you know of any good introductions to Woodin's work, or where to start with it? For some reason, I feel like there is some sort of connection between all the various hierarchies in computation, logic, and the transfinite, like they're all just instances of some underlying thing - like the difference between classical and modern algebraic geometry. Anyways, I digress. Thank you again for the interesting ideas:-)Phoenixia1177 (talk) 16:55, 17 July 2014 (UTC)
Don't know if this is exactly what you are looking for, but it is Woodin and it is around 2000:
YohanN7 (talk) 20:37, 17 July 2014 (UTC)
Yes, that's it exactly. Anyone reading it should be aware that not all of it has held up perfectly in the intervening years. A result that Woodin thought he knew (that there's a limit to the large cardinals you can have in HOD) fell through. I don't know how much of that paper depends on that. --Trovatore (talk) 21:29, 17 July 2014 (UTC)
Thank you:-)Phoenixia1177 (talk) 03:33, 21 July 2014 (UTC)

# July 18

## Awk

Is there any valid reason why Awk (number) redirects to Large numbers? There's no explanation in the article. Rojomoke (talk) 09:58, 18 July 2014 (UTC)

I'd guess because awk is able to handle fairly long numbers, it holds them in double precision float. But that's not a good reason for having that link so it should just be deleted. Dmcq (talk) 10:07, 18 July 2014 (UTC)
Double precision float cannot compute N - (N - 1) correctly if N is Avogadro's number, so I'd say that awk doesn't deserve the redirect. (Neither is temporary real aka extended precision aka long double.) It would take a 79-bit significand to handle the "mol" version of Avogadro. Double and long double are still inadequate with 52 and 64 bits. - ¡Ouch! (hurt me / more pain) 12:00, 18 July 2014 (UTC)
I found the redirect while looking for the AWK article, but I was wondering if there was a separate mathematical meaning to the term. Rojomoke (talk) 12:28, 18 July 2014 (UTC)
Just seconding Dmcq, after creating the redirect the same editor made a bunch of test edits, so it appears to me that the redirect was one as well, which would mean a speedy delete is in order. I can't find any mathematical meaning for awk.--RDBury (talk) 18:16, 18 July 2014 (UTC)
In the future, this sort of notice would be better placed at Wikipedia talk:WikiProject Mathematics. --Trovatore (talk) 18:21, 18 July 2014 (UTC)
It wasn't a notice, it was a question. I wonder if the poster was thinking of the Ackermann numbers, a sequence which rapidly gets into very large numbers. I think I've seen the word abbreviated as "ack", which might be confused with "awk". --50.100.189.160 (talk) 19:32, 18 July 2014 (UTC)
It was a question, but not about math. It was more of a "what do we do about this page" sort of question, or at least that's my take on it. Those are more wikiproject-like. I don't want to belabor this; there was no real harm done. I just want people to be aware that the wikiproject exists, and that this is the sort of thing it's there for. --Trovatore (talk) 20:10, 18 July 2014 (UTC)

# July 19

## Random walk on polygon

What is the easiest way to show that, for a symmetrical random walk starting on a vertex of a polygon, the expected value of the number of steps to visit all n vertices is n(n-1)/2? And how would one derive the expected value of the number of steps to cover all n edges? →86.146.61.61 (talk) 12:30, 19 July 2014 (UTC)

For the second part, it will take as long to reach all vertices of the polygon as to reach n (necessarily consecutive) integers in a random walk on the integers, and it will take as long to traverse all edges as to reach n+1 integers, so if n(n-1)/2 is the right answer to the first part then n(n+1)/2 is the right answer to the second. -- BenRG (talk) 03:01, 20 July 2014 (UTC)
I don't know about the easiest way, but here is a way. Suppose you know that it takes on average n(n-1)/2 steps to visit every vertex for n vertices. Now suppose you're on an n+1-gon. The number of steps needed to visit all n+1 vertices = (the number of steps needed to visit n vertices) + (the number of steps needed to reach the n+1st vertex having just reached the nth vertex). Look at a random walk, basically as sequence of lefts and rights, where you have just visited the nth vertex on the last step. Perform that same walk on an n-gon and you also get a walk where you have visited the nth vertex on the last step. Also, any walk on an n-gon where you where you visit the nth vertex on the last step will, if performed on an n+1-gon, will give a walk that visits exactly n vertices and reaches the nth vertex on the last step. This is using the fact that the walk can't make a complete circuit without having first reached all n vertices. So the number of walks on an n-gon where you reach n vertices after exactly S steps is the same as the number of walks on an n+1-gon where you reach the n vertices after exactly S steps. In other words, the (the number of steps needed to visit n vertices) part of the sum given above is the same for an n+1-gos as it is for an n-gon, and we already know that number is on average n(n-1)/2. So if the second number is, on average, n, then the sum is n(n-1)/2 + n = (n+1)n/2 then we have proved the formula for n+1 and the result follows by induction. Therefor it is sufficient to show that the second term in the sum is n, that is, from a position where a walk has visited n vertices and visited the nth vertex on the last step, it takes on average n steps to get to the n+1st vertex.
So lets start in a position where n of the vertices have been visited, and the nth vertex was visited on the last step. The n vertices visited must be contiguous and the last vertex must be at one end, since if the walk ended in the middle it must have previously reached both ends. Therefor the vertices that have been visited form a path, with our current position at one of the ends, and we want to know how long on average will it take to fall off that path if the walk is continued. Falling off the path is equivalent to reaching the n+1st vertex since going off the path at either end will result in reaching the n+1st vertex. Number the vertices visited so far from 1 to n, then going off the path would be the same as reaching 0 or n+1.
Now we have the following restatement of the problem, given a random walk starting at 1 or n, show that the expected number of steps required to reach either 0 or n+1 is n. To do this it's probably easier, to actually do more than is required. Namely, given a random walk starting at any number k between 1 and n, find the expected number E(k) of steps before the walk reaches either 0 or n+1. Take E(0)=E(n+1)=0. From k after 1 step you land on k-1 with probability 1/2 and k+1 with probability 1/2. So E(k) = 1 + (E(k-1)+E(k+1))/2. This gives you a system of n equations in n unknown, namely
2E(1)-E(2) = 2
-E(1)+2E(2)-E(3) = 2
-E(2)+2E(3)-E(4) = 2
...
2E(n-1)-E(n) = 2
The coefficient matrix for the system is nonsingular, in fact it's easy to show by induction that its determinant is n+1. So there is a unique solution to the system. Since we suspect that E(1)=n, start with that value and plug in the equations to the the remaining ones, which come out as E(2)=2(n-1), E(3)=3(n-2), ... and in general E(k)=k(n-k+1). It's not hard to verify that these values satisfy the equations and since a solution is unique these must be the values of the E(k). Plug in k=1 and k=n to get E(0)=E(n)=n as desired. I've glossed over a point here in that I'm assuming without proof that E(k) is finite. This isn't extremely hard to show, though it's a bit messy, and this response is already overly long. So I'll leave it as an exercise.
Random walks are well studied and so are Markov processes which are a generalization, so I assume most of the above can be simplified a great deal by using results from those theories, but I'm giving a completely naive proof that does not rely on such specialized knowledge. According to my calculations, if instead of just moving to adjacent vertices you can go to any vertex with equal probability, the result is the same. So there may be a larger class of graphs for which the expected time to visit all vertices is n(n-1)/2. Also, it might be interesting to what happens if the probabilities going left and right are not the same. It seems clear that as the probability of going left approaches 1 then the expected number of steps will approach n-1. --RDBury (talk) 06:00, 20 July 2014 (UTC)
Thanks, that's probably as easy as it gets. I came across a paper which was really beyond me, and the n(n-1)/2 result for an n-gon with equal left/right probability just appeared in a load of confusing detail, with extra notation being defined from time to time along the way. It does seem strange if the result isn't in some sense standard (I deduced it by considering a triangle and quadrilateral from first principles, having been unable to find it by direct searching), likewise n being the expected number of steps to escape from one end of an n-path.
Yes, covering edges is equivalent to covering one more vertex.→86.146.61.61 (talk) 16:18, 20 July 2014 (UTC)

## Is it a metric space?

I am not a mathematician but I need this for a project I am involved with. Given a final set of complex numbers; if each complex number defines one dimension of a topological space is this space a metric space? In short can a distance be defined on it? I feel it can be but want to make sure. I think this follows from "Metrics on vector spaces" in the article "Metric" here. Thanks, --AboutFace 22 (talk) 22:48, 19 July 2014 (UTC)

You'll have to be more specific about how exactly you construct your space. Any set can be given a metric- the discrete metric can always be used no matter what the set is. Probably you're wondering if some "nicer" metric exists, but it's not clear from your question exactly what you're talking about. Do you mean the vectorspace defined by taking your finite set of numbers as a basis? Staecker (talk) 23:02, 19 July 2014 (UTC)

Thank you for the answer. Honestly I am surprised my question is unclear. I will need a day or two to think about what you've said. --AboutFace 22 (talk) 01:27, 20 July 2014 (UTC)

L² norm states:
On an n-dimensional complex space Cn the most common norm is
$\|\boldsymbol{z}\| := \sqrt{|z_1|^2 + \cdots + |z_n|^2}= \sqrt{z_1 \bar z_1 + \cdots + z_n \bar z_n}.$
Are you asking if this yields a metric space? -- ToE 08:45, 22 July 2014 (UTC)

# July 20

## When mathematicians speak of "Zermelo–Fraenkel set theory", do they mean ZFC or ZF?

I had always assumed that mathematicians mean ZFC when they use the phrase "Zermelo–Fraenkel set theory". Historically, the axiom of choice (the "C" in "ZFC") was part of Ernst Zermelo's earliest forms of an axiomatization of set theory (see Zermelo set theory), and it was the one axiom that caused the most debate, making it perhaps the most prominent part of Zermelo's axiomatization. This is why I myself had assumed that the phrase "Zermelo–Fraenkel set theory" always meant ZFC. However, some time ago, when editing Dedekind-infinite set, I came to doubt this. Now I wonder: when mathematicians speak of "Zermelo–Fraenkel set theory", do they conventionally mean ZFC or ZF ("ZFC minus C") or even some other variant? Or do mathematicians avoid to talk about "Zermelo–Fraenkel set theory" at all, directly using the unambiguous terms ZFC and ZF and so on instead?

BTW: Just for fun, I have looked at the way the phrase "Zermelo–Fraenkel set theory" is used on wikipedia, checking some of the pages that link to the article with the title "Zermelo–Fraenkel set theory".

Tobias Bergemann (talk) 10:26, 20 July 2014 (UTC)

(Partially) answering my own question: Apparently I am somewhat mistaken in my assumption that ZF universally refers to "ZFC minus C". Indeed, most works on set theory appear to use this convention (e.g. the canonical works by Jech or Kunen or Levy or Smullyan/Fitting or Fraenkel/Bar-Hillel/Levy/van Dalen, and of course those books that specifically talk about the axiom of choice like Gregory H. Moore's "Zermelo's Axiom of Choice"). So I think it is safe to say that this use is conventional. But: I also found a few books on set theory that explicitly use "ZF" to refer to the axioms of Zermelo–Fraenkel set theory with the axiom of choice included (e.g. Cohens "Set Theory and the Continuum Hypothesis", and Mary Tiles's "The Philosophy of Set Theory"), at least if I'm not mistaken. However, they are certainly a tiny minority.
And the term "Zermelo–Fraenkel set theory" indeed appears to be ambiguous, with some authors including the axiom of choice and others not. On the one hand, for example, Keith Devlin's "The Joy of Sets" says (on p. 45): "The theory whose Axioms are 1–8 above is usually denoted by ZF. If we add Axiom 9, we denote the resulting theory by ZFC. This is at slight variance with the fact that 'Zermelo–Fraenkel set theory' has all nine axioms as its basic assumptions, but the nomenclature is now standard." On the other hand, Azriel Levy's "Basic Set Theory" says (chapter 5.24, p. 23): "The system consisting of the axioms of extensionality, union, power-set, replacement, infinity, and foundation is called the Zermelo–Fraenkel set theory and is denoted by ZF."
I am somewhat confused as several authors explicity or implicitly claim that the axiom of choice was added later to Zermelo–Fraenkel set theory. E.g. Smullyan/Fitting (Part I, Chapter 1 §9 "Zermelos set theory") talk about a "Zermelo set theory" without the axiom of choice (basically ZFC without choice, substition and replacement), then present substition and replacement (the additions of Fraenkel and, independently, Skolem) and then refer to the resulting system as "Zermelo–Fraenkel set theory" (ZF), without even mentioning choice, and (at least to me) implicitly suggesting that the axiom of choice was a much later addition to the axiomatization of set theory when it was in fact Axiom VI in Zermelo's 1908 article "Untersuchungen über die Grundlagen der Mengenlehre. I". (And, of course, Zermelo explicitly used choice in his 1904 proof of the well-ordering theorem, and he explicitly formulated and used the choice principle again in his modified proof of the well-ordering theorem presented in his 1908 article "Neuer Beweis für die Möglichkeit einer Wohlordnung".)
At least the meaning of "ZFC" is unambiguous.
Tobias Bergemann (talk) 13:22, 20 July 2014 (UTC)
I have heard some vague claims that Zermelo's original theory, as he conceived it, was sort of implicitly in second-order logic, which could be part of the confusion. If you take Z to have full second-order separation and consider it as a theory in second-order logic, then it's categorical up to the first inaccessible cardinal. (More precisely, the only thing that can differ between two models is their height.) So if you think AC is true, then you think that all models of Z in this sense satisfy AC, so it maybe doesn't seem so important whether it is explicitly included or not.
All these ZF-like set theories have the same "picture" of the universe of sets (namely the von Neumann universe). Workers who study theories with a significantly different picture of the universe (like New Foundations) are sometimes heard to use the term "ZF" to refer, not to a precise formal theory, but to all set theory that views sets in that standard way. --Trovatore (talk) 05:23, 21 July 2014 (UTC)

## Great circles through antipodal points

Is there any discontinuity-free procedure for choosing a great circle through any given pair of antipodal points on Earth -- i.e. so that the chosen great circle never suddenly jumps with a very small movement of the antipodal points? I feel that this may be impossible, but I'm not certain. Assume that the Earth is a perfect sphere. 86.167.19.47 (talk) 19:49, 20 July 2014 (UTC)

If your conditions are met - a pair of antipodal points on a perfect sphere - there are an infinity of identical great circles through them. Roger (Dodger67) (talk) 20:13, 20 July 2014 (UTC)
I am aware of that (though they are not "identical"). I want to know whether there is a systematic procedure for choosing ONE great circle through each pair of antipodal points so that no discontinuities occur. For example, a simple procedure would be to choose the great circle that also passes through the North Pole. However, this creates a discontinuity when the antipodal points are the poles. 86.167.19.47 (talk) 20:34, 20 July 2014 (UTC)
1) For the initial pair of poles, choose any third random point on the equator between the poles. Use that point and the two poles to construct the great circle.
2) Find the midpoint of one of the two semi-great circles between the poles (choose the semi-great circle which is closest to the last midpoint), and use that as the third point for the next great circle. Repeat step 2 indefinitely. StuRat (talk) 20:32, 20 July 2014 (UTC)
Given any two antipodal points, the procedure, or algorithm, needs to independently return a unique great circle without any reference to any previous history. Also, I am interested in solving this in an exact mathematical sense, not anything related to precision of variables. 86.167.19.47 (talk) 20:37, 20 July 2014 (UTC)
Then no, I don't think that's possible. Note that if you aren't worried about precision of variables, then the chance of either point being exactly coincident with your random 3rd point is zero, however. StuRat (talk) 20:44, 20 July 2014 (UTC)
However, that is no consolation to me. Except in pretty exotic cases, mathematical discontinuities do occur only at single points. 86.167.19.47 (talk) 20:49, 20 July 2014 (UTC)
Also, you probably won't be interested in this "trick", but if you can limit the selection of the antipodal points to a certain set, say if they can only specify each coord up to a trillionth of a degree, then just choosing a random 3rd point not on that grid of points would work. StuRat (talk) 20:44, 20 July 2014 (UTC)
Right, I'm only interested in solutions that are completely theoretically valid. I'm not looking for practical fudges or kludges. 86.167.19.47 (talk) 20:54, 20 July 2014 (UTC)
Then I think you're out of luck. May I ask what this is for ? StuRat (talk) 21:00, 20 July 2014 (UTC)
It is just for interest. 86.167.19.47 (talk) 22:43, 20 July 2014 (UTC)
This isn't wholly rigorous, but:
• Assume that the method you're looking for exists.
• This means that for any point A on the sphere, a function f produces a circle f(A). (We have f(A) = f(B) iff B=A or B=-A.)
• Let's define a vector v along f(A) from A, such that v also varies continuously. (This appears to follow from the definition of f, that it is continuous and unique.)
• v for all A is a vector field.
• The hairy ball theorem applies. Therefore the initial assumption is false - v cannot be continuous, so either v's continuity doesn't follow from f's, or f isn't continuous. If anyone can show that f can be continuous without v being so as well, I'll be surprised. AlexTiefling (talk) 23:15, 20 July 2014 (UTC)
I like it! I must be careful not to confuse with the hairy balls theorem... 86.167.19.47 (talk) 00:29, 21 July 2014 (UTC)
One clarification: If f is continuous, one could obviously choose a discontinuous v (but there's no reason to); but if you can't make v continuous, that's either because there's something peculiar about the choice of v (that I've missed) or else f isn't continuous either. Hence why I say this isn't rigorous. I can easily see an argument that if f(A) isn't unique for each A, v can't be guaranteed continuous, but not if f(A) is unique. (For example - distinguish the hemispheres into which f(A) divides the sphere; require v to be a unit along f(A) from A with a specified hemisphere on its left. If f(A) is continuous, the identity of the chosen hemisphere should be too.) AlexTiefling (talk) 11:38, 21 July 2014 (UTC)
The fundamental polygon of the projective plane
This problem is clearly what one might call a variant of the hairy ball theorem on a different space: the elliptic plane, topologically the real projective plane, since antipodal points are identified, and we are trying to find a field of tangent lines over this manifold that does not have topological discontinuities. A tangent line is equivalent to the pair of oppositely pointing units vectors, which is to say we can tolerate a 180° flip when we get back to the point via any path in the projective plane. Or: can one draw a set of non-crossing lines that covers the real projective plane? Looking at the fundamental polygon of the real projective projective plane, if we rule it with vertical lines, it would suggest that it might be possible. There is an inherent 180° shift in the vector field, but that is accommodated by the lines being non-directed. If one examines this more closely, this corresponds to the fields making two U-turns around the tips of the arrow B, creating two singularities in the field. This corresponds to two pairs of antipodes where small circular path around them will result in the vector direction spinning 180°, or if one shrinks them to one point, this becomes one pair of antipodes where there will be a spin of 360° (corresponding to diagonal lines in the diagram). While this is not rigorous, it does lend support to AlexTiefling's argument from a topological perspective. —Quondum 17:54, 21 July 2014 (UTC)
I think I agree. But is there necessarily a mapping from any ruling (even partial) of the fundamental polygon of the real projective projective plane onto a set of great circles on the corresponding (hemi)sphere? Projective geometry's not my strong point, I'm afraid. AlexTiefling (talk) 08:57, 22 July 2014 (UTC)
No, there isn't. The projective plane is a topological construct, so what is a straight line in the fundamental polygon isn't necessarily anything very nice in the projective plane (or in the sphere when taking preimages). For this, you need more structure than a topology. YohanN7 (talk) 14:14, 22 July 2014 (UTC)
That is to say, the concepts "straight line" and "great circle" both lack meaning in this topological setting. YohanN7 (talk) 14:37, 22 July 2014 (UTC)
Agreed, regarded as a topology, though this doesn't detract from the argument. (Actually, the real projective plane does have straight lines, and they do map to great circles in our case, but we are not interested in these, only in the topology. These lines are not straight in the diagram here, though.) If the real projective plane cannot be ruled with wavy lines that are locally parallel everywhere, this is a proof that there is no procedure as outlined in the original question. —Quondum 15:58, 22 July 2014 (UTC)
How do you define "straight" in the real projective plane? Its elements, in one construction, are straight lines but these are points in the space, not lines, so you mean something else obviously. I guess you could probably use the projection from the sphere (using its standard embedding in R3) to define straight in one incarnation of the real projective plane. YohanN7 (talk) 16:21, 22 July 2014 (UTC)
The geometry of the projective plane is off-topic here. —Quondum 18:42, 22 July 2014 (UTC)
Don't bring it up then. YohanN7 (talk) 19:27, 22 July 2014 (UTC)

──────────────────────────────────────────────────────────────────────────────────────────────────── Alex, can you please further explain your uniqueness condition. As I understand it, nothing in the definition of the original problem explicitly prevents f(A) = f(B) where A ≠ B and A ≠ -B, but that the uniqueness is allowing you some method of creating a continuous v from a continuous f. I don't understand that method -- your "distinguish the hemispheres" example -- and I don't see how uniqueness comes into play.

Trying to map the tangent lines of a continuous f() to a vector field, it seems to me that given an A, and choosing an arbitrary hemisphere of f(A) to be on the left of the vector derived from f(A), and calling the point in the center of that hemisphere X, the continuous nature of f means that there is a neighborhood of A in which for all elements C of that neighborhood, f(C) will be sufficiently far from X that we can assign a vector where the hemisphere containing X is to its left, and that the resulting vector field in that neighborhood will be continuous. Does that work, and if f is uniformly continuous, should it be possible to paste together a continuous vector field for the entire sphere? But what if f is continuous, but not uniformly so? -- ToE 14:29, 22 July 2014 (UTC)

I was worried that I was out of my depth, and it looks like I might be. My attempt to clear up as much as I can:
• You're right about more than two points being capable of producing the same circle; however, the uniqueness is not essential to the argument. I was over-specifying. The original problem only requires that f(A) is uniquely determined by choice of A, not that it is actually unique.
• 'Distinguishing hemispheres' was simply my way to try and visualise consistently assigning a vector to each A. But yes, I imagined that if f were continuous, small changes in A would result in small changes to v, and my original draft included identifying a point equivalent to your X - such that small changes to A would result in equivalent-sized small changes to X.
• Your argument about neighbourhoods of A makes sense to me. But this is a proof by contradiction...
• I'm weak on uniform continuity, but I'm not sure it's essential to the argument. Let's say I think my argument is true for uniform continuity, and might well be otherwise.
• The argument about neighbourhoods of A implies that the function can be locally continuous, but the hairy ball theorem implies that there must be point discontinuities somewhere in the global function.
Does that make any sort of sense? AlexTiefling (talk) 14:58, 22 July 2014 (UTC)
I haven't followed the reasoning closely ITT, but it seems like the Hairy ball theorem is incorrectly interpreted. It says there is no nonvanishing continuous vector field on the two-sphere. Continuous vector fields do exist. YohanN7 (talk) 14:44, 22 July 2014 (UTC)
The definition of v(A) that I was using assumed it has unit magnitude everywhere. If it's capable of having zero magnitude, it can't correspond to the orientation of a uniquely specified circle through A. AlexTiefling (talk) 15:00, 22 July 2014 (UTC)
I think that we are on the same page, Alex. Clearly, any combing of a hairy ball would yield the f() asked for. But since no such combing exists, if we can show that a continuous f() can be used to construct a continuous unit vector field, then such an f() does not exist.
S2 is compact, so I suppose that any continuous f() is uniformly continuous, so we should be able to piece together a continuous unit vector field by covering the sphere in finitely many neighborhoods. So yes, I think that the hairy ball theorem does preclude a continuous f(). -- ToE 15:39, 22 July 2014 (UTC) (I *know* that I'm out of my depth here, but this sounds reasonable to me.)

Fantastic question! In my mind, the question is precisely asking for a global section of the projectivised tangent bundle of $\mathbb{RP}^2$ (for each pair of antipodal points, choose a direction (tangent line at either point), through which the great circle then goes). Nonexistence of such a section does not, to my eyes, immediately follow from the hairy ball theorem. Crucially, there is a difference between the following two circle bundles on the sphere $S^2$:

• the unit circle bundle associated to the tangent bundle, which has no global sections precisely by the hairy ball theorem, and
• the projective bundle associated to the tangent bundle, which is the quotient of the previous circle bundle by the antipodal map (on the fibers).

The first bundle has Euler class $2 \in H^2(S^2,\mathbb{Z})$, as the Euler characteristic of $S^2$ is 2. This precludes the existence of a global section (essentially, this is the proof of the hairy ball theorem via Poincaré—Hopf). The second bundle will then have attached Euler class 1 and the same argument applies to rule out a global section. (Note that I don't have to worry about twisted Euler classes as $S^2$ is simply connected so every vector bundle is orientable). This was for the projectivised tangent bundle of $S^2$, but as I noted you really care about the projectivised tangent bundle of $\mathbb{RP}^2$, in which case we do need to worry about the orientation sheaf to get a twisted Euler class, but the same argument should apply. --137.73.15.6 (talk) 16:12, 22 July 2014 (UTC)

# July 22

## Is every infinite field with cardinality aleph-0 isomorphic to the rationals?

Just wondering. --2404:2000:2000:5:0:0:0:C2 (talk) 00:31, 23 July 2014 (UTC)

No. Consider the field Q(sqrt(2)) (I'm no longer sure of the notation; what I mean is the smallest field containing the rationals and sqrt(2)). Any isomorphism between them would have to fix the rationals, so there's nowhere for sqrt(2) to go on the Q side. --Trovatore (talk) 00:47, 23 July 2014 (UTC)
Even worse: the algebraic closure of a finite field is countable but doesn't even contain an isomorphic image of the rationals. —Kusma (t·c) 07:29, 23 July 2014 (UTC)