Long time since I put out some music for my faithful readers out there! So here goes … . The other day I was listening to a beautiful kori sevimpa raare in kharaharapriya by M D Ramanathan. I was idly listening when the mridangist went to a different plane for about 15 seconds, just as the pallavi ended and the anupallavi was about to begin. And then I noticed that the rendition was full of small delights. The niraval and swarams were extremely playful! Giants like Shri Tiruchi Sankaran have said that it is a great challenge to play to MDR’s slow tempo, and I think both the violinist and the mridangist did an admirable job here.

As with everything else, this too reminded me of Madurai Mani Iyer’s playfulness, and I next found an inta saukhyamani ne in my collection that I hadn’t heard very often. Half an hour of complete joy! T N Krishnan’s alapana was sweetness personified. There are even ragamalika swarams at the end!

Listen and enjoy!

Update
Some of you would rather I provided download links. Here they are:

• Kori sevimpa raare
• Inta saukhyamani ne – Part I
• Inta saukhyamani ne – Part II

More music and other things!

How much can I torture my dear readers? Well, seeing as there are four and a half of them, I will not be increasing the overall pain in the world by much! Today’s post is intended to serve only one purpose – to record for myself some things that I want to remember! And so, dear reader, if you’re not down with it, we’ve got two words for you! – actually three – bear with me!

First up, probably the best concert I attended this season – my sixth Vijay Siva! Marvelous it was! With R K Sriramkumar on the violin, Tiruchi Sankaran Sir on the mridangam, and B S Purushothaman on the kanjira. He seems to be at the peak of his musical form! The concert was total sowkhyam throughout, everything kept simple and pure, with no gimmickry at all! And what choice of ragas! Hard to see so many weighty ragas packed into one concert!

1. vallabhanāyakasya – begaḍa.
2. saṅgītaśāstrajñānamu – mukhāri. Nice niraval, swaram.
3. ambikāyām abhayāmbikāyām – kedāram. Short alapana, and niraval in the kriti.
4. mīnalocana brova – dhanyāśi. I thought this was the best piece of the concert. Beautiful alapana, but the hightlight was the niraval, swaram in kāmapālinī.
5. jānakīpate – kharaharapriya. Imagine singing kharaharapriya for a filler between two main pieces!
6. śrisubrahmaṇyāya namaste – kāmbhoji. Majestic! The main piece, with an ālāpanā that is still fresh in memory, elaorate niraval and swarams.
7. tani āvartanam – tisra ekam. I don’t know enough to say anything knowledgeable about Tiruchi Sankaran Sir’s playing except that it was scintillating, with B S Purushothaman ably matching him.
8. rāgam tānam pallavi – ṣaṇmukhapriya – tisra tripuṭa (2). Not much time could be spent on the ālāpanā, but the pallavi had all the standard elements, singing in four tempos, etc.
9. ini enna pechu – sahānā
10. viruttam – yamunākalyāṇi, maṇiraṅgu, jañjūṭi
11. tiruppugazh – jañjūṭi
12. harivāsarada – sindhubhairavi
13. maṅgalam

The other thing I want to record is something I fought hard to figure out more than once, and promptly forgot every time! After the latest episode (three or four days back) I decided to record it somewhere. (And a nice place I have chosen! Let me see if I find this the next time I am stuck on the same problem!) It has to do with a Lemma in Kenneth Kunen’s Set Theory that is dismissed with a one line proof! The claim has to do with relativizations. (The relativization of a formula with respect to another formula is got by replacing all quantifiers in by and by replacing all quantifiers in by . )

The claim is that if is a formula with one free variable, and are sentences such that (this notation stands for: there is a derivation (using the axioms and rules for first-order logic) of from the as assumptions), then (here etc. means the relativization of the formulas with respect to ).

I was trying to transform a derivation of the former to a derivation of the latter. It didn’t work! Then I realised that one has to work with derivations that consist only of sentences. Whenever the assumptions and conclusion are sentences, one can manage things so that only sentences are used in the derivation, no matter what sound and complete axiom system for first order logic one uses. Then it’s a routine matter to check that the relativization of each axiom holds and that the relativized versions of the rules are admissible. But … even after many attempts I couldn’t pin down exactly where the extra assumption was being used in the second derivation. Finally I did find it, and that is what I want to record here!

The troublesome axiom is this: But since our original proof consists only of sentences, the axiom that actually gets used is a generalization, Now the relativized version of this is One can check easily that this is a validity, and hence provable (or construct a derivation directly!).

So where is the extra non-emptiness assumption used? Precisely in proving the relativization of the above axiom, once we realize our error! The point is that might not occur free in , and so might just be . In this case, our original derivation might have just used the axiom The relativized version is And this is not necessarily provable, unless we also assume “nonemptiness” of ! The antecedent says that for every , either holds or holds. Now if it is the case that for every , holds, then it is not necessary for to hold. Precisely this is ruled out by the assumption .

Now I can go back to doing some actual work, after having disposed of this irritating gap in my understanding!

Update In my focus on the Kunen lemma, I forgot to mention another highlight of the concert. Padma and her father were there for this, too! And it was great company!

The music season!

After five Vijay Siva concerts, I went to listen to Prof. T N Krishnan and Viji Krishnan at Sivagami Pethachi Auditorium. To my surprise I was able to recognise all the songs they played! (I almost always attend vocal concerts, so had forgotten that this is an additional challenge!) I was trying to impress Padma (who had also come for the concert, along with her father) by (mis)identifying the ragas almost instantly. But she didn’t look too impressed! :(

But what to do with all the things I noted down about the concert? No way I shall let it go waste! Here it is for your edification, my dear readers!

Gajananayutam – Vegavahini – Adi
Nannu palimpa – Mohanam – Adi
Azhimazhaik kanna – Varali – Adi
Arivar yaar unnai – Mukhari – Mishracapu
Tolijanma – Bilahari – Khandacapu
RTP – Shanmukhapriya – Adi (the pallavi was “Sharavanabhava guruguha shanmukha”, rendered perhaps a thousand times by the evergreen Madurai Mani Iyer, perhaps five hundred of those accompanied by TNK himself! I am listening to one such rendition right now, as I am writing this post!)
Ramajogi – Khamas – Adi
Vangak kadal kadaindha – Surati – Rupakam (two Tiruppavais in one concert! The first one followed a Varali alapana, and featured niraval swarams!)
Something or the other – Sindhubhairavi

I liked the concert very much, though I felt let down by the mridangam of the veteran, Shri Guruvayur Dorai! Somehow it felt like the valanthalai (the right hand or treble side) didn’t “speak” at all! It must be me who has to learn more to appreciate him, I guess!

Now what was the use of writing all this, I am wondering… . Oh well, I can’t hope to reach the heights of my previous post with every piece of my writing, can I? So there you are!

Update: Ah yes! I forgot to say that I was completely stumped by the Varali piece till Padma said “Tiruppavai maadhiri irukke!” Of course it was Azhi mazhaik kanna!

Another significant event worth reporting! After reading about it so many times, this time I wanted to find it! So I took a detour into Karpagambal Nagar on the way back home from Pethachi Auditorium, and looked carefully at each house. And sure enough, there it was, near the door of house number 9 (but not near the gate, so you can’t see it unless you’re looking for it!), the magical words on the nameplate: “Ganakaladhara Madurai Mani Iyer”! Yes, there lived one of the greatest musicians of all time, and I have seen the place! (And what a beautiful cascade of swaras in his rendition of Koluvaiyunnade in Bhairavi, which is playing as I am writing this update!)

Poem ahoy!

Oh pretty girl so potty,
marry early and party.
Lest you turn forty
and people call you paatti!

I am glad my blog is finally on its way to gaining a modicum of respectability with the appearance of original poems, but one has to admit that it involves considerable effort. Given below is the genesis of this post. Read it and decide for yourself!

S: Sir, I am looking at this website of a friend who is getting married. She also has a blog, which you’ll absolutely love!

Me: Wow, poetry and all! What k!

S: I don’t understand why everybody who has a blog has to write poetry! Why haven’t you, by the way?

Me (displaying a streak of brilliant आशुकवित्वम्): Here’s one! Think I can post it?

S: Moshtu definitely!

Two days later …

Me: Sir, the moment I came up with a poem, I am starting to display finer sensibilities like a poet. I am wondering how well it will be received, and all that!

S: But this is precisely thinking unlike a poet, illaiya? Post it, I say!

A toga party!

Heard this on the September 4 edition of BBC Radio Four’s I guess that’s why they call it the news.

I wouldn’t go to a toga party … unless it was a low-key event involving giants wearing fake animal fur and dancing wildly in a caravan inside a small retreat in a Caribbean winery … because that would make it a meagre bogus tiger toga ogapogo(?) Winnebago party in a Tobago yoga(?) bodega!

As you can see, I couldn’t figure out one or two words properly, which is why I ask my dear readers to please help me out! It is vital that I get this completely right!

Boiled or fried!

Brilliant opening paragraph of a ChessBase report on the NH Chess Tournament. The reporter is the ever entertaining Steve Giddins.

It always used to be said that one knows one is getting old, when the policemen start looking younger. But now it would appear that the chess world has developed a new test. Chess is bad enough as it is, with the annual emergence of ever-younger chess talents. In my day, it was a sensation when Nigel Short qualified for the British Championships at the age of 12, but at the time, he was not even an IM, let alone a GM. Nowadays, 12-year old GMs are becoming almost commonplace. I myself never became a GM, but I still have good chances of becoming a GOM – a Grumpy Old Man. Like every such chessplayer, I hate playing against juniors. The grisly sight of a small head, clad in a baseball cap, peering between its own pieces, guzzling Coke, stuffing crisps and breathing through its mouth all at the same time, whilst simultaneously rattling out 25 moves of Najdorf theory – such a sight has always made me sympathize with W. C. Fields, who when asked how he liked children, replied “Boiled or fried!”

Well, what can we say? Srikanth would wholeheartedly agree with W C Fields, I am sure! And what about me, you ask? No way! I am a strict vegetarian, you should know.

If only I had a sharper wit!

New student (Ananth Shankar) comes and tells me “Excuse me, Hrishikesh tells me to call you sir.”

I came up with the following brilliance thirty seconds too late: “Oh, please don’t! Just call me Suresh.” And have to be content with only posting it on my blog.

One of those near misses …

Logic on NPR

Was pleasantly surprised to encounter the following logic puzzle in the latest edition of NPR’s Sunday Puzzle Podcast, which is mostly about word puzzles.

A waitress approaches a breakfast table with five logicians, and asks “Do all of you want a coffee?” The first logician says, “I don’t know.” The second, third and fourth also say (in that order), “I don’t know.” The fifth says, “No!” Who does the waitress give coffee to, and why?

Axiom of Choice and Zorn’s Lemma

The Axiom of Choice (AC) asserts that for all sets of nonempty sets, there is a choice function such that for all , .

Zorn’s lemma says that for every partially ordered set such that every chain in has an upper bound in , there is a maximal element of .

The primary aim is to prove that the Axiom of Choice implies Zorn’s Lemma. For this we use the notion of wellordered sets, which are just linearly ordered sets such that every nonempty subset of has a least element.

To gain some intuition about the proof, let us instead try to define a wellordering of a given set, assuming the Axiom of Choice. For a given set , consider a choice function for the set of all nonempty subsets of . Then one could try to define a wellordering of as follows: pick as the first element, as the second, as the third, and so on. But all the ’s might not exhaust all of , so one might have to extend this process of building a wellorder by choosing the element following all the ’s to be . Then we continue with , and so on. But even this might not exhaust all of and we have to “jump to another limit point” and so on, to form the desired wellorder by a transfinite procedure. But we do not wish to build the machinery (ordinals) necessary to describe such a procedure. And indeed we do not need to, since the problem we are after is slightly different—that of proving Zorn’s Lemma assuming the Axiom of Choice!

Let us keep the above procedure in mind in trying to prove Zorn’s lemma. We are given a partially ordered set each of whose chains has an upper bound. And we suppose, towards a contradiction, that there is no maximal element, and use this to show that there is indeed a chain of with no upper bound in . The chain will be built essentially following the intuition in the above paragraph.

To begin with, notice that every chain in has some upper bound (, say). Now is not a maximal element of , and hence there is some such that . It follows that for every . The Axiom of Choice allows us to pick one such — , say — for each chain .

Now we can try to build our desired chain as follows: let (this is the same as !), , , and so on. The ’s form a chain , but there is an element that is strictly greater than all the ’s. So we add to the end of to get a larger chain. We can repeat this procedure “forever” (even taking into account “jumps to the next limit point”) to get a chain without an upper bound. This is the idea, but let’s do it a little more formally.

Notice that any chain is an initial segment of this chain that we are hoping to build if and only if is well-ordered by and it has the following property:

Consider the collection of all chains that are well-ordered by and that satisfy condition (1). (Intuitively, is just the collection of initial segments of “our big chain”, but we will have to get at it in a slightly roundabout manner.)

Proposition 1 For distinct , in , either is an initial segment of or is an initial segment of .

Proof: Let us first prove that either or . Suppose neither holds. Then consider the least in , and the least . Thus and . Notice that and . Now, since is a total order, either is greater than every element of , or it is less than or equal to some element of and thus less than .

Suppose the former. We claim that is an initial segment of , for if not, there exist in such that and . But this is a contradiction, since and was supposed to be the least element of not in , so after all! But now we have , and has an element greater than everything in , and thus the least such, , belongs to , a contradiction!

Suppose . Then by arguing just as above, we can show that belongs to , a contradiction!

Thus either or . Suppose, without loss of generality, that . Since and are distinct, there exists a least that is not in . We claim that is an initial segment of . Suppose not. Then there is a least such that and . But this means that for all , if and only if . But then , a contradiction. Thus is an initial segment of .

Proposition 2 If is an element of , is also an element of .

Proof: Of course will be the largest element in , and it is easy to see that is a well-ordering of this set, and that (1) is satisfied by .

Proposition 3 is an element of .

Proof: Let such that .

We first show that well-orders . Suppose there is an infinite descending chain in . Let such that , for each . Now it is easy to prove that for all , . It is trivially true for . Suppose and consider . Now if is an initial segment of , it follows that (from , and ). If is an initial segment of then and hence . But and hence is a well-ordering of it, so there cannot be an infinite descending sequence in , and the same holds for .

We now show that every is an initial segment of . Clearly is a subset of . We need to show that for every such that and , . Let be some element of such that . Now either is an initial segment of , in which case (since ), or is an initial segment of , in which case also (since and such that ).

Now suppose that . Let such that . As we have seen many times now, and since satisfies (1), it follows that .

Thus .

Now we are almost done. We claim that above has to lack an upper bound in , contrary to the assumption that every chain has an upper bound. Because if did have an upper bound, then would be an element greater than everything in . But then would be a chain in . But this leads to an immediate contradiction, because this means that , but was supposed to be greater than every element of !

The upshot of all this is that for every partially ordered set , if every chain has an upper bound in , then has a maximal element. Thus the Axiom of Choice implies Zorn’s Lemma.

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Update: Many thanks to Shreevatsa, as usual, for pointing out a bug in Proposition 1, which has now hopefully been fixed!

A math problem …

Yesterday, I had a chance to sit through a problem solving session where Abhishek Dang and Ashwin Deopurkar (final year BScM students at CMI) solved some 2009 IMO problems for the benefit of some school students. According to them, the problems were “actually very easy”. In fact, I was surprised to find that I was able to understand the solutions quite well, but that’s more a testament to their impressive skill in exposition than anything else.

Anyway, here is a nice problem. I am writing this (without permission from either of them!) so that I can refer back to it when I want to, and since otherwise I will forget it completely! It will stay here as long as there is a take-down notice from Abhishek! :-) At which point, Shreevatsa will teach me how to make this post private, I am sure!

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Determine all functions from to such that for all and , the numbers and form the three sides of a non-degenerate triangle.

Here is the solution as described by Abhishek.

Take and let . Then we know that for all , and form the sides of a non-degenerate triangle. In particular, this means that , i.e. and , i.e. . Thus . But notice that this says that either or is a periodic function. But if is a periodic function, it is bounded. But then we can choose large enough (say more than twice the largest value that can take) so that there is no for which form a triangle. Thus, it has to be the case that .

Now we have determined that . Let’s play around a little more to gain more information about We could try to set and see what we can learn. Well, in that case and thus we conclude that for all form the sides of a nondegenerate triangle. Arguing as before, we see that for all . But this easily implies that is a bijection from the positive integers to itself.

What more? Try . Let . We know that since and is a bijection. Let . Now we see that for all form the sides of a nondegenerate triangle. But this means that for all . In particular, since is a bijection (and ), we know that for all , either or .

Now consider the value of on . It is an easy proof by induction (using the fact that is injective) that . But this would mean that is a bijection between and , and this can happen only if . And it follows that is the identity!

Nice, wasn’t it?