Uselessness of set theory?
Tuesday 2007 December 25
Mathematics is supposedly based on set theory. But if one looks closely, it is amazing (to me) just how little set theory is really used. Almost all “ordinary” mathematics seems to use no sets larger than the continuum. In other words most of mathematics can be carried out in second order arithmetic: roughly speaking this means that one can use the integers and subsets of the integers but nothing more complicated. Although third order and higher order constructions occasionally occur, it usually seems easy to replace them by second order constructions. (By ordinary math I mean things like the Atiyah-Singer index theorem, Fermat’s last theory, the Weil conjectures, and so on; more or less everything that is not some sort of set theory or logic.)
In fact one seems to need a lot less than this if one is willing to work harder. I suspect that a lot of stuff can be encoded in Peano arithmetic if one is willing to work hard (for example, encoding constructible reals in terms of integers is tiresome but possible). The few examples known of theorems that cannot be proved using Peano arithmetic (as in the Paris-Harrington theorem) tend to have extremely rapidly growing functions appearing, and I would guess that proofs without such large functions somewhere can usually be encoded in Peano arithmetic. One can probably go a lot lower: Peano arithmetic has much weaker fragments, such a primitive recursive arithmetic. In practice it seems rare to have functions that require more than a finite tower of exponentials to describe, which presumably corresponds to something even weaker than primitive recursive arithmetic (does anyone know what this is called?).
Harvey Friedman has a program called “reverse mathematics” to identify what axioms are really needed to prove various results, but this seems to concentrate on various fragments of second order arithmetic. He has found many examples of results about the integers, often with a Ramsey-theoretic flavor, that need reasonably strong subsystems of second order arithmetic to prove.
It is easy to find mathematical results whose proofs need much stronger systems using Godel’s theorem: for example, the consistency of second order arithmetic is a perfectly good statement about integers that is (presumably…) true, but cannot be proved in second order arithmetic, though it can easily be proved even in weak set theories. I do not know of any mathematical theorems about the integers that cannot be proved in second order arithmetic that are not related to consistency results.
So my question is the following: why do we use so little set theory in ordinary mathematics? Is it because almost all interesting mathematical results only need tiny fragments of Peano arithmetic to use, or is is because we are too stupid to make proper use of the vastly more powerful axioms in set theory?
Tuesday 2007 December 25 at 8:04 am
Personally, I don’t go out of my way to use a powerful axiom if a simple one will do. Why should we multiply axioms (like entities) unnecessarily? We can do so much with a small collection of relatively intuitive rules already.
Saturday 2009 December 5 at 5:26 am
All that I have seen of all set theory, whether from Frege, or Russel, is that it shows cleverness on the part of the originator, but it ultimately reveals flaws in its arguments. Set theory was likely invented for no other reason, but for some mathematician to gain fame for being the first for making a generalized theory of mathematics. It was not invented to be in any way useful. The fact that math existed long before set theory was formalised is proof of the uselessness of set theory.
Tuesday 2007 December 25 at 4:31 pm
To the extent that mathematics models the universe… Perhaps God didn’t know much set theory?
Thursday 2007 December 27 at 2:24 am
Dear Richard,
Good to hear from you again!
One place where a little bit of higher order set theory comes in useful is when various types of ultrafilters are deployed to make limits converge and do all sorts of other handy things. To set up Zorn’s lemma on the set of all filters on Z in order to exhibit even a single ultrafilter requires dealing with sets of sets of sets of integers, if I have counted correctly. Admittedly, a lot of things one can do with ultrafilters can eventually be rearranged so as not to use those ultrafilters, but the proofs often become significantly messier as a consequence.
At a more philosophical level, perhaps what is going on is that statements which involve really convoluted hierarchies of sets tend to be either (a) trivially true or false, (b) collapsible to lower hierarchies, (c) very hard for human mathematicians to internalise, or (d) undecidable, leaving only a small fraction of actually useful statements that would end up being used in a mathematical proof.
The other thing, of course, is that mathematical formalism should follow the ideas, not the other way around; if one sets up some overly intricate abstract structure to attack a problem, there had better be some good motivation and evidence for doing so; abstracting for its own sake tends to be a poor problem-solving strategy in general. And most of the mathematical ideas and tricks we have seem to not be recursive enough to merit using things really high up in the set hierarchy, except of course when one deals with parts of mathematics which are recursion-intensive, such as logic and certain types of combinatorics.
Thursday 2007 December 27 at 4:36 am
Do you ever really need ultrafilters to make things converge?
I’ve never come across any use of ultrafilters outside logic and set theory that couldn’t easily be avoided. In any case sets of sets of sets of integers is still only 4th order arithmetic, which is way below most set theories in power. Do you know any examples of any “ordinary” math that cannot be easily converted to 2nd order arithmetic? (I tried asking Friedman once and he didn’t offhand know any examples.) Or to put it another way, if Congress banned everything beyond 2nd order arithmetic, would it make any serious difference?
Godel showed that proofs can become vastly longer if converted from (say) 2nd order to first order arithmetic, but I dont know of any “natural” examples of this. (It is easy to get slightly artificial examples by unwinding the Paris-Harrington theorem, and so on.)
Thursday 2007 December 27 at 5:04 am
I think the reason that we don’t need more set theory is due do the way mathematics is teached at universities today. Mathematicians often proove their theorems with standard techniques they learned before and the standard Math one learns first is Analysis and Algebra.
Analysis the way it is commonly taught, is a theory which deals with items that are easyly visualizable plus the notion of infinitely large and infinitely small.
For this, not much of the abstract set theory is needed. And most mathematicians try to use these standard techniques in their more advanced prooves.
Maybe we need a new Bourbaki to have more set theory in a new era of mathematics. But It may very well be, that this would be a math, which is broader and even more abstract than ours, and the standard parts of our subject we have now, will show beiing just a very small peace of a math that is build of the more of the abstract parts of set theory.
Thursday 2007 December 27 at 5:19 am
Do you know of this guy? http://www.math.wustl.edu/~nweaver/conceptualism.html
He argues for a different foundational scheme for math, in a countable universe called J2. I don’t have enough understanding to tell whether it’s bogus. There is an interesting exchange of letters between him and S. Feferman linked at the bottom of the page.
Feferman also has an essay called “Is Cantor Necessary?” (answer: no) in his book In The Light of Logic. I have a copy from the library but haven’t read it yet.
Anyway the idea that set theory isn’t needed has some currency among foundationalists. I’m sure you know a lot more than I do about what they’re doing. I’m new at this stuff.
Thursday 2007 December 27 at 5:31 am
Also, I forgot to ask, I’m trying to read a rant by JY Girard and he presents Gentzen’s cut-elimination theorem as an induction on epsilon-0 if I understand it right. Does that mean cut elimination can’t be proved with first order PA? Forgive me if this question is basic and dumb.
Also, H. Jervell has a proof that finite trees are well ordered. It’s a straightforward lexicographic order whose order type is way above epsilon-0, so unless I’m confused (very possible) there is no PA proof. There are several cites here: http://folk.uio.no/herman/ and the most detailed presentation is in chapter 11 of his book-in-progress, starting at page 93: http://folk.uio.no/herman/incompleteness.pdf
His ordinal diagram notation very cool.
Thursday 2007 December 27 at 7:51 am
Re Richard’s remark on avoiding the use of ultrafilters: is it easy to avoid ultrafilters in proving the following theorem? “A space X is compact if for every space Y, the projection from X x Y to Y is closed.”
(The proof I have in mind runs as follows: given X, let Y be the space of ultrafilters on the underlying set of X. The ultrafilter convergence relation as a subset of X x Y is closed; its image under the projection is dense in Y [by considering principal ultrafilters], and closed under the hypothesis of the theorem. It follows that the image is all of Y. This means precisely that every ultrafilter converges, and hence [assuming AC] that X is compact.)
Thursday 2007 December 27 at 8:00 am
Sorry to keep posting here but there was a big interesting rant by Harvey Friedman a while back that I couldn’t find earlier. Here it is (url below). He is pro-set-theory but also observes the lack of examples of where it’s needed. Among other things he claims that almost all normal math is or should be Pi_0^1 sentences! I wonder what he thinks should be done with Waring’s problem or with P vs NP. He mentions Kruskal’s tree theorem and the graph minor theorem as needing something like set theory to prove. I had wondered about these because of their mention in the Jervell papers linked above.
http://www.cs.nyu.edu/pipermail/fom/2006-January/009526.html
Thursday 2007 December 27 at 10:33 am
[...] Something we won’t be ignoring is relevant discussion on other blogs, and here’s some of that going on at Richard Borcherds’ blog. They’re wondering why so little of [...]
Thursday 2007 December 27 at 4:26 pm
In reply to various comments above:
1. Cut elimination for Peano arithmetic requires transfinite induction up to
which cannot be proved in Peano arithmetic.
2. Finite rooted trees can be well ordered in a natural way that has order type exactly
3. Kruskal’s tree theorem does not need set theory to prove: it is easy to prove in 2nd order arithmetic, but cannot be proved in 1st order arithmetic. Friedman has some examples of special cases of it that can be proved in Peano arithmetic but whose shortest proof is ridiculously long. The graph minor theorem is harder than Kruskal’s theorem; I would guess it can still be proved in 2nd order arithmetic but I don’t know for sure.
Friday 2007 December 28 at 4:20 am
My favourite non-trivial application of ultrafilters is to provide an extremely short proof of Hindman’s theorem: if the natural numbers
are finitely coloured, then one of the colour classes 
contains all the finite sums 
of some infinite set of integers A. The proof proceeds by first constructing an ultrafilter 
which is idempotent in the sense that 
, where we identify the ultrafilter p with an element of the Stone-Cech compactification 
in the usual manner. Then one chooses the colour class C that lies in the ultrafilter and uses the idempotent property repeatedly to build up the infinite cube. (The proof that idempotent ultrafilters exist, incidentally, is short but requires three(!) applications of Zorn’s lemma.) This is arguably the shortest and most conceptual proof of Hindman’s theorem currently known.
A nice discussion of these topics is at
http://www.math.ohio-state.edu/~vitaly/vbkatsiveli20march03.pdf
Returning to the general discussion, I also wanted to comment that while “useful” results in mathematics may require only low-order set theory, there are often “meta-results” surrounding that useful result that are also worth knowing (if only for philosophical or aesthetic reasons), but require higher-order set theory to prove. For instance, there may be an implication
that yields a desirable property Y from a checkable condition X which is useful and uses only low-order sets, while the converse implication 
, which is proven using higher-order sets, is not actually used in practice but is worth knowing because it tells you that one is not “losing” anything by always reaching for the easy 
implication. A lot of results that involve universal spaces (such as the Stone-Cech compactification) seem to fall into this category; it is rare that one genuinely needs the full power of a universal space, but it is still comforting on a metamathematical level to know that such spaces exist (for instance, Stone-Cech tells us, morally, that different notions of limits of divergent sequences can always be “reconciled” if necessary).
Friday 2007 December 28 at 7:03 am
Hindman’s theorem is still provable in a fairly weak fragment of 2nd order arithmetic, so the Stone-Cech compactification of the integers (which seems to be a 3rd order object) is “unnecessarily large” for the proof. (Trying to think about or visualize the S-C compactification of N always gives me a bad headache.) I don’t know if Hindman’s theorem is one of those that cannot be proved in Peano arithmetic; it seems vaguely like the Paris-Harrington theorem and the Kruskal theorem that cannot be.
(Added slightly later) Come to think of it, I cant even see how to STATE Hindman’s theorem in the language of Peano arithmetic,
never mind prove it.
Friday 2007 December 28 at 6:56 pm
I think you’re right that Hindman’s theorem cannot be stated in Peano arithmetic; it seems roughly analogous to the infinite pigeonhole principle in this regard. But various corollaries of Hindman’s theorem can be stated in PA. Just as the infinite pigeonhole principle easily implies the Paris-Harrington Ramsey theorem, Hindman’s theorem has a similar analogue: for any k there exists N such that whenever {1,…,N} is k-coloured, one of the colour classes contains the finite sums of some set A where |A| > min(A). Given that this statement implies the Paris-Harrington Ramsey theorem, it can’t be proven in PA.
I think you’re also right that in most cases where really high order arithmetic is used, one could usually get by with much weaker arithmetics also. (For instance, there are some arguments in non-standard analysis which are extremely extravagant with regards to repeatedly taking non-standard extensions of various things to create huge hierarchies of infinitesimals and whatnot, which is almost certainly much more than absolutely necessary to prove any given task at hand.) But I think there can be a real tradeoff between how “elementary” the proof is (where I define elementary in terms of what order of arithmetic one uses) and how short and conceptual the proof is, and how easy it is to generalise or otherwise perturb the proof. (One could argue though that all the conceptual difficulty in slick higher-order proofs have been moved into setting up the set theory in the first place, but at least this is a conceptual “cost” that only needs to be paid once, no matter how many things one is trying to prove.)
Friday 2007 December 28 at 7:30 pm
[...] Set theory useless? In other words most of mathematics can be carried out in second order arithmetic: roughly speaking this means that one can use the integers and subsets of the integers but nothing more complicated. [...]
Saturday 2007 December 29 at 3:05 am
There is indeed a tradeoff between how “elementary” the proof is and how short and conceptual the proof is; in fact this is a theorem due to Godel, and is a minor variation of his incompleteness theorem. As an example, the statement
“This statement cannot be proved in less than
symbols in Peano arithmetic”
can easily be encoded in first order arithmetic using “Godelization”. It also has a very short proof if you assume Peano arithmetic is consistent, for example in 2nd order arithmetic. Since it is true, it can obviously be proved in Peano arithmetic just by checking every case. But again since it is true, any proof in Peano arithmetic must be monstrously large. So the proof of this in 2nd order arithmetic is vastly shorter (and more conceptual) than any proof in Peano arithmetic.
The proof of Hindman’s theorem using 3rd order ideas is not really a good example of Godel’s theorem, because it does not reduce the length of the proof all that much. Using ultrafilters in this case is rather like using your car to visit the neighbors because it is 1 second faster than walking. (Though I guess in LA this is normal behavior…)
Saturday 2008 January 5 at 1:40 pm
In response to Todd Trimble’s question, I don’t rememeber how, but in my Topology class last semester we proved that very result, an ultrafilters weren’t mentioned at all in class. If I remember correctly, the proof used what Munkres calls the tube lemma and not much more.
Saturday 2008 January 5 at 11:25 pm
Are you thinking of the converse statement, Marlowe: PI? I believe that’s the easy direction of the equivalence, and is found in Munkres’s textbook.
Wednesday 2008 January 16 at 5:02 am
It seems that the claim that not much of set theory is used in classical mathematics is a rather subjective view. Why, for instance, the result that all sets obtained from closed sets via the basic operations, i.e., projections, and complementation, are Lebesgue measurable and have the Perfect Set Property, is not viewed part of the classical mathematics? The result follows from PD (Projective Determinacy) and one can actually prove that infinitely many Woodin cardinals are needed (this is due to Martin, Steel and Woodin).
True, set theory hasn’t produced many deep results about integers in the sense you want (i.e., other than consistency results), but it has never been the goal of set theory to investigate integers. It has, though, produced many interesting results about reals and sets of reals.
Wednesday 2008 January 16 at 6:18 am
Forgot to mention. It seems that the only natural example of a left distributive algebra with one generator occurs in set theory. One gets such an algebra from j: V->V where j is a nontrivial elementary embedding and V is the universe. Then j generates algebra, i.e., j(j), j(j)(j), and so on are all embeddings from V into V.
Thursday 2008 January 17 at 10:33 pm
Grigor, did you mean the free left distributive algebra on one generator?
As mentioned here, there is also a braid group representation of this structure, so I’m not clear on what you meant by “only natural example”.
Thursday 2008 January 17 at 10:46 pm
Yes free, sorry. This is a new development that I didn’t know about. In general, I am no export on this topic, I was just mentioning it as an interesting example of applications of higher set theory in ordinary math. Sorry for misleading. Thanks for finding this abstract by Jech, it actually conveys the point I was trying to make better, so I decided to copy and paste it here.
>>>
A left-distributive algebra is a set with one binary operation satisfying the left distributive law
a(bc)=(ab)(ac)
(the operation of conjugation in a group is an example).
The subject of my lectures is the free left-distributive algebra on one generator. This algebra arises in the study of elementary embeddings in set theory, and the word problem for it was first solved by R. Laver using the theory of large cardinals. Subsequently, P. Dehornoy found an elementary solution of the problem by giving a representation in the braid group.
The free algebra can also be expressed as a limit of finite algebras. These cyclic algebras have been studied by R. Dougherty and myself. The statement that the limit algebra is free is equivalent to a simple statement of elementary number theory, and follows from Laver’s work (using large cardinals). It is however unprovable by elementary methods (viz. in primitive recursive arithmetic), and it is an open problem whether the large cardinal assumptions are necessary.
The study of cyclic left-distributive algebras leads to some interesting questions on computability and complexity.
T. Jech
The fact that the statement that the limit algebra is free cannot be done in primitive recursive arithmetic and that it is open whether the large cardinals are needed is an intriguing situation.
Wednesday 2008 January 23 at 1:01 am
One can sharpen your obervation that: “any proof in Peano arithmetic must be monstrously large…”
Great to see you back! I was away too, undergoing emergency manifold surgery. When done, and necrotic tissue removed before it killed me, I had a small intestine 12 centimeters shorter, battled postoperative complications, and shall be coauthoring a paper (applied mathematics and Complex Systems Theory) of the more-art-than-science medical situation, an in the Giome, with the surgeon.
Classical example of a system weakened “just enough” [lazily excerpted from wikipedia, below, as it know it so from more citable sources] is Presburger arithmetic, the first-order theory of the natural numbers with addition, named in honor of Mojżesz Presburger, who published it in 1929. It is not as powerful as Peano arithmetic because it omits multiplication.
Mojżesz Presburger proved Presburger arithmetic to be:
* Consistent. There is no statement in Presburger arithmetic which can be deduced from the axioms such that its negation can also be deduced.
* Complete. For each statement in Presburger arithmetic, either it is possible to deduce it from the axioms or it is possible to deduce its negation.
* Decidable. There exists an algorithm which decides whether any given statement in Presburger arithmetic is true or false.
The decidability of Presburger arithmetic can be shown using quantifier elimination, supplemented by reasoning about arithmetical congruence (Enderton, A Mathematical Introduction to Logic, p. 188).
Peano arithmetic, which is Presburger arithmetic augmented with multiplication, cannot be decidable as a consequence of the negative answer to the Entscheidungsproblem. By Gödel’s incompleteness theorem, Peano arithmetic is incomplete and its consistency is not internally provable, unless Peano arithemetic is inconsistent, in which case all sentences are provable.
The decision problem for Presburger arithmetic is an interesting example in computational complexity theory and computation. Let n be the length of a statement in Presburger arithmetic. Then Fischer and Rabin (1974) proved that any decision algorithm for Presburger arithmetic has a worst-case runtime of at least 2^{2^{cn}}, for some constant c>0. Hence, the decision problem for Presburger arithmetic is a rare example of a decision problem that has been proved to require more than exponential run time. Fischer and Rabin also proved that for any reasonable axiomatization (defined precisely in their paper), there exist theorems of length n which have doubly exponential length proofs. Intuitively, this means there are computational limits on what can be proven by computer programs. Fischer and Rabin’s work also implies that Presburger arithmetic can be used to define formulas which correctly calculate any algorithm as long as the inputs are less than relatively large bounds. The bounds can be increased, but only by using new formulas.
Going 7 days without food was no big deal; some folks do that deliberately. But going 9 days with no Math, and 9 days with no Internet — that was agony that narcotics could not weaken. Fortunately, I had mathematical physicists to talk to on the phone, and one astrophysicist (Dr. Thomas McDonough) visit me in hospital.
Thursday 2008 January 24 at 1:18 pm
A very recent example of a something interesting being preovable inside a very weakened arithmetic is:
arXiv:0801.3639
Title: The sum of irreducible fractions with consecutive denominators is never an integer in a very weak arithmetic
Authors: Victor Pambuccian
Subjects: Logic (math.LO); Number Theory (math.NT)
Two theorems of elmentary arithmetic, one stating that the sum of the reciprocals of any number of consecutive positive integers is never an integer, and a generalization thereof by Trygve Nagell, are shown to be provable inside a very weak arithmetic, Richard Kaye’s PA^-, in which there is no induction axiom whatsoever.
Thursday 2008 January 24 at 5:52 pm
Jaakko Hintikka in his book “Principles of Mathematics Revisited” http://www.fetchbook.info/fwd_description/search_9780521624985.html
offers a new perspective on the nature of mathematics and mathematical abstraction, along with an explanation why set theory is mostly used as a language, while its more serious results remain somewhat marginal in the mainstream mathematics. I found it a fascinating reading.
Monday 2008 January 28 at 6:03 am
Two perspectives in response to the original post and its question:
1. “Cultural”: The western mathematical mind made by Greek philosophy and world view lead mathematicians to develope a certain kind of mathematics. Math itself as a (much?) vaster object may be universal, but the one that we have can be very specific in form and contents. One can see a similar phenonmenon in medicine. Chinese medicine and western medicine are hardly anything in common while they are both valid in their own ways. One can see this in logic (defined in a looser sense) too. Buhddist logic (very complex and most exhaustive)gives me a two-dimensional feel (Greek is linear), and that of China’s I-Ching (Book of Transform) gives me a three-dimensional feel.
2. Structure of math itself: The ordinary math that we pratice is of a difference hierachy from set theory. As an analogy: Theory of color (as of part of optics) is the fundamental of oil painting, but the artist practices art at the hierachy of composition and harmony of color, creation of sense of space, rendering of light, creation of testure etc. underneath which theory of color is implicit and automatically included. Logicians and set theorists are more similar to color theorists and ordinary mathematicians the practicing artists. A sort of like the hierachy of linguistics and literature. It is natural that important results in linguistics are hardly represented in literature.
Friday 2008 February 1 at 8:22 pm
I see what you are saying, but not too many logicians will agree that they are more like color theorists. It is not the main goal of the modern day set theory (or other areas of logic) to develop a background theory for the rest of the mathematics. Set theory, at its best, investigates the consequences of higher infinities on reals and sets of reals. Recursion theorists have been occupied with things like randomness, and model theorists have been occupied with applications of their methods in other areas. All three areas have made considerable contributions to many other fields of what you might call an ordinary math. For instance, Scanlon recently found a proof of Pop’s conjecture and he is a model theorist. No, we are not color theorists, we are just mathematicians.
Friday 2008 February 8 at 10:48 am
The main usefulness of set theory relies in its existence.I think Richard Borcherds is missing the main point by lack of historical perspective :try to imagine what mathematics was like before Cantor !
His first few lines in December 25th message could not have been written ,because mathematical objects had only an individual existence ,no collective one (the set of points on the line,the set of integers .. ).For example as one of the earliest applications,the genious Jacques Hadamard immediately saw the future development of functional analysis (quite used by physicist ,isn’t it ?).
In the same way,one can discuss the marxist view of history,but the concept of class is quite useful.
My remark is sort of underlying Terence Tao’s “philosophical level” in his answer of December 27
Best wishes
Sunday 2008 February 17 at 9:24 pm
The fragment of arithemetic that contains finitely iterated exponentials is known as “exponential function arithmetic” or “EFA”.
Thursday 2008 April 3 at 4:13 am
Thanks to “reverse mathematics” and related work, it is well known that:
*Nearly all of mathematics employed in science and engineering requires nothing beyond the continuum;
*A great deal of mathematics can be done with second order arithmetic, sometimes augmented with some ingenious constructions;
*Nearly all of the vast power and generous ontology of set theory is of interest only to set theorists.
Conjecture: the “general set theory” of George Boolos, augmented with axioms of Infinity and Choice, should suffice as foundations for nearly all algebra and analysis.
It is also conceivable that some sort of second order arithmetic will be become the canonical foundation of mathematics within 30-40 years.
I have learned much from John Burgess’s “Fixing Frege.”
Tuesday 2008 May 13 at 12:58 am
Sorry to have taken so long. It was definitely proved both ways… it was an iff. The proof was rather long, and we used a rather strange condition on nets over spaces and so on. It was actually left as an excercise, and I remember it was pretty damn hard, but we did it without ultrafilters…
Saturday 2008 May 24 at 1:15 am
Marlowe:PI, I can easily believe that you don’t absolutely need ultrafilters (for the proof of the statement I mentioned above on December 27, 2007), but I also strongly suspect that your proof used rather more than the tube lemma (as you seem to intimate now). In particular, I would be a little surprised if you managed to avoid some form of the axiom of choice [or something slightly weaker like the Boolean Prime Ideal Theorem, to be technical]. If I’m wrong about that, I’d be interested! I can be reached at topological[dot]musings[circled a]gmail[dot]com if you wish to discuss this further.
My original point was that (I suspect) some proofs are easier with ultrafilters than without them, and this may be a case in point. Despite their abstract nature, they can be technically useful, as Terence Tao was also suggesting.
Thursday 2008 June 5 at 7:36 pm
I have come to this discussion rather late, but wonder, in a rather imprecise way, whether there is something to do here with the fact that mathematics is written down using finite sequences containing a finite number of symbols.
All sorts of exceptional objects can be shown to exist, and we can prove that varous universes of discourse are inherently incomplete – but as soon as a particular example is demonstrated, this is done within an inherently countable part of mathematics, and cannot therefore be ‘too pathalogical’. So our discussions of ‘pathalogical objects’ cannot get out of hand. [My Greek teacher once said "no language has a lot of irregular verbs" - don't know enough languages to know whether this is true, but it has an analogue which I think is true of mathematics]
All this is worlds away from ultrafilters.
Another part of the answer to the original question is perhaps that once it is shown [or we believe - possibly wrongly] that our conceptualisations are compatible with set theory, then most of the time it seems easiest to use the concepts developed for a particular part of mathematics than to relate them all the time to sets. This is often convenient, but some of the subtlety and power of set theory may get lost or mislaid on the way.
Wednesday 2008 July 16 at 3:42 am
A mathematical why of the Big Bang
Outline
Let Ui be a set of locations of particles of the universe.
U1xU2x …… xUix ….. a set of infinite paths
(Cartesian product).
this set is equal to the void set by the
negation of the axiom of choice.
So there is no more space containing the particles.
The particles collapse on themselves: Big Crunch.
Then Big Bang.
The Big Bang has taken place thus the negation of the axiom
the choice is likely to considered as a good axiom.
Adib Ben Jebara.
Saturday 2009 January 24 at 5:28 am
I see that I’m late to this discussion, but I only just came across it.
In my view, there is no great mystery as to why so little set theory is “needed” for mathematics. The point is that mathematics historically did not start with the study of sets. Set theory, especially set theory as a tool for foundational studies, is a johnny-come-lately that happens to give a very simple and elegant foundation that *suffices* for (essentially) all mathematics. The goal was to find something that would be general enough to do the job, not to find the “lowest bidder” (i.e., the weakest set of axioms that would suffice). Thus we formulate axioms like “every set can be well-ordered” rather than “this set and that set and, oh, by the way, that other set too can be well-ordered” since the latter is pointlessly cumbersome given our aims. Naturally, the result is an extremely powerful axiomatic system.
Only if one has been mis-educated into thinking that mathematics is “really” based on set theory, as opposed to mathematics just being mathematics and just happening to have the property that it *can* be based on set theory, will one be surprised at how strong ZFC is.
On a different note, solrize wondered why Friedman says that math is essentially Pi^0_1 when obviously we have things like P != NP. Friedman’s point is that once we prove some Pi^0_2 or Sigma^0_2 statement, it is almost a conditioned reflex for us to ask immediately for a stronger version of the statement that gives explicit bounds. This drops us down to the Pi^0_1 level. So for example, if someone proves P != NP, we are not likely to be satisfied until we prove something stronger, like “for all n, there is no circuit for SAT having fewer than (1.1)^n gates” or something even stronger than that.
Monday 2009 January 26 at 3:40 pm
[...] Set theory useless?<br/> In other words most of mathematics can be carried out in second order arithmetic: roughly speaking this means that one can use the integers and subsets of the integers but nothing more complicated. [...]
Friday 2009 January 30 at 8:51 pm
i think we have a big problem here, since i used newton rhapson methode for calculate:
Y = a^(1/n)….,where “a” is arbitary number
including complex number, while “n” is every positive numbers.
i can’t use the first guest number by 0 (zero).
why does matlab can’t do this?
i think there is some mistake in matlab work. if it is truth that matlab can’t do this, so the conclusion is all the people in the world that used mathlab for numerical solution had make a big mistake…
is anyone can help me if iam wrong?????
Wednesday 2009 April 1 at 9:18 am
Uselessness of Numbers? I don’t see your point Richard. You may as well say that mathematics uses very few large numbers.
Monday 2009 May 18 at 8:55 pm
Great post and great discussion!