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December 28

Lebesgue measure versus Borel measure

Hi. I was searching the web recently for an example of a set of real numbers that is Lebesgue measurable but not Borel measurable. Such sets must exist, because I know that the cardinality of the family of Lebesgue measurable sets is greater than the cardinality of Borel measurable sets ( $2^{\mathfrak {c}}>{\mathfrak {c}}$ ). However, this argument does not furnish an example.

So, I found a website claiming that the following set is Lebesgue measurable but not Borel measurable: The set of real numbers with continued fraction expansions of the form $\left[a_{0};a_{1},a_{2},\ldots \right]$ such that there is some sequence of integers $i_{0}<i_{1}<\cdots$ with $a_{i_{n-1}}$ dividing $a_{i_{n}}$ for all $n\geq 1$ .

Evidently, this set can be seen to be Lebesgue measurable because it is an analytic set, which I don't really understand. I'm not sure how to see that it's not a Borel set, although there is some talk about it here, which I again don't really understand. On to my question: Can someone either help me better understand the example I've presented, or else demonstrate another example where we can see why it's Lebesgue-but-not-Borel measurable? Thanks in advance for any illumination anyone can provide. -GTBacchus^(talk) 00:50, 28 December 2009 (UTC)[reply]

So if all you want to know is that there is some Lebesgue measurable set that's not Borel, here's a cheapo way of doing it: Look at the almost-bijection between the Cantor set (thought of, let's say, as the set of infinite sequences of zeroes and ones) and the interval [0,1] given by putting a binary point before the infinite sequence and reading it off in binary notation. I say almost bijection because there are countably many points in [0,1] that have two distinct binary representations and therefore are mapped to by two elements of the Cantor set.

Now do the Vitali set construction inside [0,1], and pull it back to the Cantor set. The resulting set cannot be Borel, because, if it were, the Vitali set would also be Borel, and therefore Lebesgue measurable (there are a couple of details here that I'm confident you can check for yourself).

But if you now think of the Cantor set as the usual middle-thirds construction inside the reals, well, that's a set of measure zero, so the pullback of the Vitali set is a subset of a set of measure zero, and is therefore Lebesgue measurable (specifically, measure zero).

This is almost cheating. I'd encourage you to pursue the more interesting stuff about an analytic set that's not Borel. Unfortunately this requires much more theory. It's theory well worth learning, though. --Trovatore (talk) 02:13, 28 December 2009 (UTC)[reply]

That's very good; thank you. I can understand that construction, and relate it to my study partner, who was asking me about this. As for the analytic set business... yeah. It looks like a lot of point-set topology. I know that stuff's good for the soul, but I'm a number theorist! One of these days. :) -GTBacchus^(talk) 02:29, 28 December 2009 (UTC)[reply]

(e/c) Since the Cantor space 2^ω is homeomorphic to a closed measure-zero subset of the real line (namely, the middle-thirds Cantor set K), you can use any nonmeasurable set in Cantor space to achieve the same goal. This is because a homeomorphism between two spaces gives an isomorphism of the Borel algebras of the spaces. So any standard example of a nonmeasurable set inside Cantor space – an uncountable set with no perfect subset, a non-determined set, etc. – will correspond to a non-Borel, Lebesgue measurable subset of K. This avoids the "almost-injective" argument.

The point of the "continued fraction" thing in the original post is to use this same trick: the map that sends a real number to its continued fraction expansion is a homeomorphism from the irrationals to ω^ω.

The original question seems to be how to show that the set in question is analytic, and how to show that it is not Borel. To show that a subset of Cantor space or Baire space is analytic, you just check that the set is definable by a logical formula of the correct complexity, namely $\Sigma _{1}^{1}$ in the analytical hierarchy. This is more complicated for the real line, which is one reason descriptive set theorists avoid working with the real line directly. But the desired result here can be obtained pretty easily using the isomorphism fact I just mentioned and the standard fact that the continuous image of an analytic set is still analytic.

To show that the set described in the original post is not Borel is harder, and I don't know what the original proof could have been. The way I would prove it is by showing that some standard $\Sigma _{1}^{1}$ -universal set reduces to the set in question. That would just be a standard textbook exercise, but as Trovatore says you would have to invest some time in the textbook to do it. The two places this sort of thing comes up are in descriptive set theory and in generalized recursion theory. Kechris' book Classical descriptive set theory has detailed information about universal $\Sigma _{1}^{1}$ sets. — Carl (CBM · talk) 02:57, 28 December 2009 (UTC)[reply]

I've seen this "divisibility" argument many times, and it always seemed like it should be trivial to check that it's just encoding either the problem of whether a linear order is a wellordering, or whether a tree is wellfounded. When I actually started thinking about it, I realized it's not quite as easy as maybe I expected.

But it's still not hard. I think this works: Assume given a transitive relation R on the naturals, that we want to check whether it's wellfounded. Code R as the following element of Baire space:

f(n)=p_{n}\Pi _{k<n,nRk}f(k)

, where p_n is the nth prime. Then for k<n, f(k) divides f(n) just in case nRk. So the continued fraction given by f is in the set if and only if R is illfounded. --Trovatore (talk) 08:23, 28 December 2009 (UTC)[reply]

Yes, that will work. The difficulty I was referring to is in showing

(*): the set of (codes of) non-well-founded trees is

\Sigma _{1}^{1}

complete

in the first place, which you would need to do to really believe that you had solved the original problem. In jargon, (*) is just the normal form theorem for

\Sigma _{1}^{1}

formulas plus the normal form theorem for

\Pi _{1}^{0}

formulas. None of this is truly hard, but it takes a little time to go through it all. I think it could be done very well to a general mathematical audience in two 50-minute seminars. — Carl (CBM · talk) 13:36, 28 December 2009 (UTC)[reply]

Oh, well, I'm used to seeing

\Sigma _{1}^{1}

defined as projections of trees, which makes your (*) pretty trivial. If you have some other characterization in mind, sure, there's a bit of work in showing all the characterizations are equivalent. --Trovatore (talk) 19:11, 28 December 2009 (UTC)[reply]

Universal enveloping algebra

Is every (associative) K-algebra a universal enveloping algebra of some Lie algebra (over K)? -- Taku (talk) 05:23, 28 December 2009 (UTC)[reply]

No. Universal enveloping algebras are infinite-dimensional (see Poincaré–Birkhoff–Witt theorem). 86.15.141.42 (talk) 16:09, 28 December 2009 (UTC)[reply]

Right. Assume associative algebras are infinite-dimensional too. Or how about a quotient of a universal enveloping algebra? (By the way, I tend to think the answer is yes (if the question is formulated correctly)) -- Taku (talk) 22:07, 28 December 2009 (UTC)[reply]

You are probably going to have to work hard to make the answer yes. Universal enveloping algebras have PBW-type bases, this is very special. The question about quotients is easy because the UEA of a free Lie algebra is a free algebra. 86.15.141.42 (talk) 01:59, 29 December 2009 (UTC)[reply]

Good point. So, the question really boils down to which associative algebra admits a PWB-type basis; that's certainly not the case in general. (I guess a quotient one is trivial.) Anyway, thanks. -- Taku (talk) 06:52, 29 December 2009 (UTC)[reply]

How to solve math poblems without calculating

Many thanks for all responses! My high school teacher said though that there isn't math after 12th grade; I read the Lockhart's lament suggested, but my math teacher has a math degree and he's always right. He also says that he knows all math like all math teachers. He won many math competitions and came first place, and he says math is about problem solving. He also says that all problems have been solved, and that's why it matters to show working in math. Who's the world best mathematician? Who's won the most math competitions? I understand that math is not about calculating but how else do you solve math problems like x + 2 = 5? Thanks but please don't insult me though. —Preceding unsigned comment added by 122.109.239.199 (talk) 09:06, 28 December 2009 (UTC)[reply]

If there's no math after 12th grade, how did your teacher get a math degree? -- Meni Rosenfeld (talk) 10:33, 28 December 2009 (UTC)[reply]

By doing high school math, and solving complex math problems? I don't know. That's sort of what I'm asking. Why's there math after 12th grade and what math's there? My high school teacher says there isn't and he's always right about math (he's a math nerd as my friends at school used to say). I'm trying to understand what sort of math is done if calculating isn't there. My math teacher is an expert on calculus, and isn't calculus the highest math? He said that calculus was taught at uni. So does he have a degree in calculus or something? Please explain? Many of youy say my high school teacher's wrong, but how's that? He has a math degree. Many thanks. —Preceding unsigned comment added by 122.109.239.199 (talk) 11:18, 28 December 2009 (UTC)[reply]

Look, we've already answered your questions. We told you that mathematics does involve calculations, but there's much more to it than only that.

Your teacher may have a math degree, but the people who have responded to your query have about a dozen math degrees between them. If you trust him so much, and since he appears eager to discuss these matters with you, and you also have an algebraist friend, then what are you asking us for?

Personally I don't believe you are asking these questions seriously. But if I'm wrong and you are, please read our replies carefully and don't ask us to repeat ourselves. -- Meni Rosenfeld (talk) 12:25, 28 December 2009 (UTC)[reply]

Yeah, so what math's there higher than 12th grade if it isn't calculating. Please give an example of such math. Give a mind-boggling math problem to me please. I'm dead serious about these questions. I just wnat to understand the purpose of math, and what really math is after 12th grade math. My high school teacher says that teachers know most about math since that's the main profesional math job. Applied math's there also. That's why I'm confused why my math teacher's wrong. —Preceding unsigned comment added by 122.109.239.199 (talk) 13:05, 28 December 2009 (UTC)[reply]

Ask your teacher to define "math." I've heard it said the exact opposite way at the beginning of calculus. Some calc teachers say that everything before calc is not math at all. Without having a discrete definition of what math means, you are just having a stupid semantic argument that gets nowhere. Further, nobody here appears to care what your teacher thinks. So, attempting to form an argument between your teacher and everyone else is a trollish waste of time. As for math beyond 12th grade - there is a lot of it. For example, given two vectors of unequal size, how to you find the optimal alignment of the shorter one against the longer one? That is nothing but a bunch of calculations. Have you done vectors or vector alignment? It is very important if you get into health informatics (which is a blend of math and computer science and health). -- kainaw ™ 13:18, 28 December 2009 (UTC)[reply]

[ec] I think it's safe to say Mathematical logic is higher than 12th grade and isn't calculating. For mind-boggling math problems, you needn't look further than the Millennium Prize Problems, which involve varying amounts of calculations.

The amount of mathematical knowledge currently in existence is too great for even a brilliant individual to learn in a lifetime. So no, neither your teacher nor any other math teacher knows most of what there is to know in math. -- Meni Rosenfeld (talk) 13:23, 28 December 2009 (UTC)[reply]

Whoever claims there is no math beyond highschool level is like a kid that believes that the universe ends just out his little garden. There are thousands of streets instead, towns, mountains, seas and oceans, woods and deserts as large as a state, and rivers and lakes.... and this is just what we know; then there are planets stars and galaxies... The only thing that gives me a comparable feeling of immensity is the stupidity of your teacher's claim! :-) --pm a (talk) 15:16, 28 December 2009 (UTC)[reply]

Have you seen Pleasantville? --Trovatore (talk) 21:52, 28 December 2009 (UTC)[reply]

I haven't; after reading the plot summary it sounds a nice movie --pm a (talk) 15:26, 29 December 2009 (UTC)[reply]

Is it not obvious that our esteemed original poster is amusing himself by seeing how much we try to enlighten a fool? Michael Hardy (talk) 01:55, 29 December 2009 (UTC)[reply]

Yes, an obvious timewaster. As the masochism poster, I feel that those who've replied since are exhibiting that very thing.→→86.152.78.244 (talk) 11:27, 29 December 2009 (UTC)[reply]

You people shouldn't insult the OP so bad. When I was in high school I thought just like him "I know all the math, I'm the best in the world". Things started to change after I finished year 12 and now I feel like my knowledge is a needle in the ocean (and in reality it's less than a bacterium). It takes time for people to realize and appreciate the real stuff. And for a mind boggling problem, how about theory building? It's all so easy when you read it but trying to develop the tools for the first time is, to me, only a genius can do. Money is tight (talk) 12:39, 29 December 2009 (UTC)[reply]

Actually I do not think that the OP's antics surprise anyone, since it is well known that a majority hold the belief that mathematics is "arithmetical computations" (educated people and high school teachers inclusive). What is somewhat awkward is that the OP expresses his ignorance openly. Most people are so confident that they are "right about mathematics", that they believe it is not necessary to explicitly note its purpose. Rather thay feel they may implicitly assume its "arithmetical nature" in their assertions. That said, I agree with you, and especially with your last statement (Galois being an excellent example of this). --PS T 13:21, 29 December 2009 (UTC)[reply]

You are missing the point. Of course most people are confused to some extent about mathematics. It's the specifics of what the OP said and how that triggered our troll-detectors. Things like

Asking us what his algebraist friend does, instead of asking him.
Telling us he wants to be a mathematician although he knows little about it and hates what little he knows.
Explaining that his teacher is always right. Virtually nobody believes that some other individual is always right.
Reporting that his math teacher said there is no math after 12th grade and that math teachers know all math, when of course anyone with a math degree would not say that.
Not updating his questions based on our answers - e.g. "[Without calculating, how] do you solve math problems like x + 2 = 5" after we explained that there are calculations in math, and asking for examples of higher math after we've given them.
And of course other subtleties in his style.

-- Meni Rosenfeld (talk) 15:27, 29 December 2009 (UTC)[reply]

Why feed the trolls (per Don't feed the trolls)? If a majority feels that the OP's posts were not made in full seriousness, the posts should simply be ignored. In fact, a better alternative would be to remove the posts altogether. Although I certainly agree with your judgement of the OP's questions, insulting him is counterproductive; whatever the circumstance (anyway, I think he is enjoying our insults, as Michael points out). --PS T 02:34, 30 December 2009 (UTC)[reply]

Well I guess he is somewhat trolling. I was just trying to "assume good faith" but Meni made his point. Money is tight (talk) 03:30, 30 December 2009 (UTC)[reply]

Assuming good faith is something I completely agree with - so much so that I have the userbox on my page. But the point of WP:AGF is to assume good faith when lacking evidence to the contrary. In this case there is certainly no lack of evidence.

In my view, the other point of AGF is to always consider, if possible, the distant possibility that the posts are genuine, give the OP a chance to explain himself, and not do anything that would be incredibly offensive (which is why I disagree with PST's suggestion of removal). Of course, in particularly disruptive cases this is not an option.

Anyway, posts like those of 86 and Michael Hardy are not meant to insult the OP, only to signal to others that they should not waste their time. -- Meni Rosenfeld (talk) 08:55, 30 December 2009 (UTC)[reply]

And avoid the fate that nearly befell Matilda's aunt in Matilda Who told Lies - and was Burned to Death. :) Dmcq (talk) 09:19, 30 December 2009 (UTC)[reply]

After all I think is OK being polite with maths trolls, especially at christmas times... Go and look at a troll's pub... all the other trolls treat them like scum you know, I mean real trolls working in categories such as history politics sexuality social sciences women studies...--pm a (talk) 12:07, 30 December 2009 (UTC)[reply]

Considering that two interesting queries (Wikipedia:Reference desk/Mathematics#transcendental_functions and Wikipedia:Reference desk/Mathematics#No_math_after_12th_grade.3F) have resulted in the troll's silly questions, I would say that he has done more good than he intended... (of course, it would have been more productive if his posts were ignored sooner, but I think that he has left permanently by now). --PS T 12:32, 30 December 2009 (UTC)[reply]

Sum and Difference of Two Tangents

Hello. Can a sum or difference of two tangents be expressed as a product or quotient? For example,

\sin x-\sin y=2\cos \left({\frac {x+y}{2}}\right)\sin \left({\frac {x-y}{2}}\right).

Thanks in advance. --Mayfare (talk) 21:03, 28 December 2009 (UTC)[reply]

Just from the definition of tan, we have

\tan x\pm \tan y=\left({\frac {\sin x}{\cos x}}\right)\pm \left({\frac {\sin y}{\cos y}}\right)={\frac {\sin x\cos y\pm \sin y\cos x}{\cos x\cdot \cos y}}={\frac {\sin(x\pm y)}{\cos x\cdot \cos y}}.

Is this the kind of thing you had in mind? ~~ Dr Dec (Talk) ~~ 23:37, 28 December 2009 (UTC)[reply]

Yes. Thank you very much Dr Dec. Have a happy new year. --Mayfare (talk) 18:34, 29 December 2009 (UTC)[reply]