Theorems that are 'obvious' but hard to prove - MathOverflow

Theorems that are 'obvious' but hard to prove

2011-01-09T11:19:49Z

There are several well-known mathematical statements that are 'obvious' but false (such as the negation of the Banach--Tarski theorem). There are plenty more that are 'obvious' and true. One would naturally expect a statement in the latter category to be easy to prove -- and they usually are. I'm interested in examples of theorems that are 'obvious', and known to be true, but that lack (or appear to lack) easy proofs.

Of course, 'obvious' and 'easy' are fuzzy terms, and context-dependent. The Jordan curve theorem illustrates what I mean (and motivates this question). It seems 'obvious', as soon as one understands the definition of continuity, that it should hold; it does in fact hold; but all the known proofs are surprisingly difficult.

Can anyone suggest other such theorems, in any areas of mathematics?

Answer by Georges Elencwajg for Theorems that are 'obvious' but hard to prove

2011-01-09T12:02:34Z

That $\mathbb R^n$ has topological dimension $n$. In a similar vein that affine space $\mathbb A^n_k$ over a field $k$ has Zariski dimension $n$.

Answer by Ricky for Theorems that are 'obvious' but hard to prove

2011-01-09T12:43:12Z

The Jordan curve theorem! Of course in this case the real problem is the meaning of "closed curve".

Answer by Wadim Zudilin for Theorems that are 'obvious' but hard to prove

2011-01-09T13:36:07Z

Fermat's Last Theorem. (Should I state it?)

Answer by Daniel Moskovich for Theorems that are 'obvious' but hard to prove

2011-01-09T13:40:40Z

The notorious Dehn's Lemma, ~~and it's generalizations, the Loop Theorem and the Sphere Theorem~~. It was a bone in the throat of 3-manifold topology for almost half a century, despite being `obvious', until proven by Papakyriakopoulos in 1957.
A comment on why it is obvious: the only singularities one can possibly imagine a disc having are things like "stretch a feeler out and around and around through the disc"- and it's obvious how to re-embed to get rid of those. Dehn's Lemma is a statement in the PL category, not in the topological category, so nothing pathological can occur. There's nothing which could possibly go wrong- no way you could possibly create a singularity in a DISC which you can't kill by re-embedding. But... prove it!

Answer by Steven Landsburg for Theorems that are 'obvious' but hard to prove

2011-01-09T13:45:41Z

That every continuous vector field on ${\bf S}^2$ has a zero is pretty "obvious" (when you think about the image of trying to comb the hair on a billiard ball) yet takes considerable machinery to prove.

Answer by Denis Serre for Theorems that are 'obvious' but hard to prove

2011-01-09T13:46:45Z

A Theorem is 'obvious' when one does not see an immediate obstruction (for instance a counter-example). Of course it may be true or false, depending on how you are lucky or not. An obvious true theorem whose proof is notoriously difficult is the existence of solutions to linear PDEs $P(i\nabla_x)u=f$ for constant coefficients operators (Malgrange-Ehrenpreis theorem). I don't mean elliptic, hyperbolic, parabolic PDEs, or PDEs of principal type. No, just PDEs. It is not only true but somehow accurate, because it becomes false when the coefficients are non constant, even with analytic coefficients (H. Lewy counter-example).

Dick: At first glance, the Fourier transform reduces the question to the resolution of an algebraic equation $P(\xi)\hat u(\xi)=\hat f(\xi)$. The difficulty is whether $\xi\mapsto\hat f(\xi)/P(\xi)$ is the Fourier transform of a distribution. Because $P$ may vanish, and $P^{-1}(0)$ can be quite singular, this is not a piece of cake. Malgrange had to prove his division theorem to solve it.

Answer by Zen Harper for Theorems that are 'obvious' but hard to prove

2011-01-09T14:10:42Z

Maybe on the boundary of what's allowed, but I would say most basic geometric things like Pythagoras' Theorem, trigonometry with sine/cosine, the area of a circle, etc. etc. etc.; here of course the difficulty is in defining what we mean by length, area, angle, etc. - in which case some of these become axiomatic, but then the difficulty is shifted onto proving that things do work correctly.

Answer by algori for Theorems that are 'obvious' but hard to prove

2011-01-09T15:01:25Z

I would mention the triangulation and smooth stratification theorems for algebraic varieties and variants thereof (analytic, real analytic etc.) These results are quite useful and I would say they seem obvious, at least from my experience. However, it is tricky to find complete proofs in the literature (especially in the real analytic case, which implies all the rest). I would say this is not because the proof is that difficult; it is not, but it's a bit tedious to spell out all the details.

While I'm at it, let me also mention that proofs of these theorems can be found in the references given in the answers to this MO question: http://mathoverflow.net/questions/36050/embeddings-and-triangulations-of-real-analytic-varieties (many thanks again to Mohan and Benoit).

Answer by aorq for Theorems that are 'obvious' but hard to prove

2011-01-09T15:46:43Z

The trefoil knot is knotted.

Answer by Patrick I-Z for Theorems that are 'obvious' but hard to prove

2011-01-09T16:32:18Z

In the same vein, chating: My advisor used to say "An interesting theorem is a theorem true which looks false". Well, tastes and colors... ;-)

Answer by gowers for Theorems that are 'obvious' but hard to prove

2011-01-09T16:37:52Z

If $I_1,I_2,\dots$ are intervals of real numbers with lengths that sum to less than 1, then their union cannot be all of $[0,1]$. It is quite common for people to think this statement is more obvious than it actually is. (The "proof" is this: just translate the intervals so that the end point of $I_1$ is the beginning point of $I_2$, and so on, and that will clearly maximize the length of interval you can cover. The problem is that this argument works just as well in the rationals, where the conclusion is false.)

Answer by Paul Siegel for Theorems that are 'obvious' but hard to prove

2011-01-09T16:48:02Z

I think that the ergodic theorem is a good example of this. In down-to-earth terms it says that if you have a box full of gas then the average velocity of all of the gas particles at a given time (the space average) equals the average velocity of a single given particle over time (the time average). This can be regarded as at least a partial theoretical justification for the fact that gas in a container reaches an equilibrium state over time. And what could be more obvious than that?

Yet the ergodic theorem revealed itself as frustratingly difficult to prove. You might think that the challenge would be to just come up with the right precise formulation of the problem; indeed, I don't think it was until people started to identify the measure theoretic underpinnings of probability theory that this was really possible. But while any student with a semester of measure theory under his/her belt can understand the modern formulation of the pointwise ergodic theorem, I highly doubt that very many could supply a correct proof without a hint. For some reason, the proof simply demands an ingenious combinatorial trick.

Answer by aginensky for Theorems that are 'obvious' but hard to prove

2011-01-09T16:56:12Z

Speaking of thesis advisers, mine said, "I think something should be called obvious only if it is obvious in the logical sense of if A implies B and if B implies C then A implies C". All else is subjective and hence capable of misuse. I have tried, but not necessarily succeeded, to follow this. I am constantly amazed/amused at how people coming at a problem from different points of views will find certain facts obscure or well known.

Answer by Richard Borcherds for Theorems that are 'obvious' but hard to prove

2011-01-09T17:29:34Z

Chess is not a forced win for black.

Answer by Mark Bennet for Theorems that are 'obvious' but hard to prove

2011-01-09T17:32:38Z

That the surface of a sphere is not homeomorphic to the real plane.

This may be unfair, in that it requires a good understanding of continuous functions. But it is intuitively obvious at a significantly lower level of mathematical sophistication than is required for the proof.

Then again, it is almost equally "obvious" at the same level of sophistication that you can't turn a sphere inside out.

So the notion of "obvious" in this sense is too crude to distinguish between true statements and false ones, and the question shows hides a bias. It would be more balanced also to ask whether there are "obvious" statements which it is hard to prove false.

Answer by Bruno Martelli for Theorems that are 'obvious' but hard to prove

2011-01-09T20:12:07Z

As an undergradued, I remember having precisely this feeling when encountering (a version of) the weak Nullstellensatz, which says that the maximal ideals in $\mathbb C[x_1,\ldots, x_n]$ are the sets of all polynomials vanishing on a fixed point $(x_1,\ldots, x_n)$. This must be pretty obvious, what else a maximal ideal could be?

However, the statement now does not look so "obvious" to me anymore... and I don't know if this is a good or a bad thing :-)

Answer by Dick Palais for Theorems that are 'obvious' but hard to prove

2011-01-09T20:20:37Z

How about the fact that a sphere is the surface of minimal area that bounds a given volume? (BTW, if it isn't geometrically obvious to you, and you understand a little physics and about surface tension, then the roundness of bubbles is a "proof".)

Answer by Qiaochu Yuan for Theorems that are 'obvious' but hard to prove

2011-01-09T20:52:01Z

Subgroups of free groups are free. The plausible argument is that any relation satisfied in a subgroup must somehow translate to a relation satisfied in the larger group. Nowadays I guess most people see the proof which proceeds through the fact that the fundamental group of a graph is free, but it's not trivial to set up this machinery (even if one uses a purely combinatorial definition of fundamental group). I don't know how hard the algebraic proof is; perhaps it's easier.

Answer by Igor Rivin for Theorems that are 'obvious' but hard to prove

2011-01-09T21:26:20Z

The independence of the Parallel Postulate (especially since the proof, which consists of demonstrating that elliptic geometry satisfy the other axioms, is not hard, yet too 2500 years to find).

Answer by Timothy Chow for Theorems that are 'obvious' but hard to prove

2011-01-09T22:24:12Z

There is a whole class of examples of the following general form: There is an obvious candidate for the solution to an optimization problem, and the obvious candidate is in fact best, but it's very hard to prove that it's best. Two of the examples mentioned in the comments—isoperimetric inequalities and sphere packing—fall into this class. Lower bounds in computational complexity furnish other examples, although our knowledge in this area is so pitiful that the best examples are still conjectural.

I like these examples better than the topological ones like the Jordan curve theorem and the invariance of domain, because there is room to argue that (for example) what makes the Jordan curve theorem hard is that modern mathematics has an exceedingly general definition of a Jordan curve that includes monsters that are non-rectifiable, nowhere differentiable, etc. The "man in the street" doesn't have these monsters in mind when judging that the Jordan curve theorem is obvious. In contrast, if we take something like "the kissing number of the sphere is 12," the man-in-the-street's conception of a counterexample is really no different from the mathematician's. It's just that the man in the street will be convinced after a few minutes of playing with velcro balls and the mathematician won't.

Answer by John Stillwell for Theorems that are 'obvious' but hard to prove

2011-01-09T22:36:38Z

The consistency of Peano Arithmetic. This is provably hard to prove, and I think that most would agree that it is obviously true (if not, why are we still doing mathematics?)

Answer by Jim Conant for Theorems that are 'obvious' but hard to prove

2011-01-09T22:57:07Z

Inspired by ``the trefoil knot is knotted" answer, how about the fact that Reidemeister moves generate isotopy of PL knots? This is pretty obvious but a full proof requires a lot of machinery. ~~Indeed, the proof was not known to Reidemeister, who took the fact that his eponymous moves generate isotopy as an unproven axiom.~~ (See Daniel Moskovich's comment.)

Answer by David Speyer for Theorems that are 'obvious' but hard to prove

2011-01-09T23:38:16Z

Inspired by the trefoil knot example: "If two knots are smoothly* isotopic, then their complements are homeomorphic." I'm not sure exactly how hard the proof is, but it certainly seems obvious, and I don't think there is a one line proof.

*Thanks to Richard Kent for pointing out that I need this adverb.

Answer by roy smith for Theorems that are 'obvious' but hard to prove

2011-01-10T03:39:19Z

I agree this question is interesting, but only in a psychological rather than mathematical sense, i.e. the only reason the jordan curve theorem seems obvious is that we do not appreciate the generality of the definition of "continuous", rather taking our simplest intuitive examples as typical. Indeed the proof for smooth functions is pretty easy (cf. Guillemin and Pollack), and how many of us distinguish intuitively between (piecewise) smooth and continuous functions? For instance young students assume the intermediate value theorem is obvious because they do not appreciate the local nature of the definition of continuity, i.e. they intuitively assume that the intermediate value theorem is the definition of continuity, as indeed it was in a less rigorous time. Of course the proof of the IVT is a justification of the reasonableness of the definition of continuity. As Moishezon remarked to us as students: " even if it is obvious, you still have to prove it". Or as Tate said after giving an irresistible pictorial argument in first year honors calc. for the continuity of a composition of continuous functions; :"Of course this is NOT a proof! I have merely rendered it intuitively plausible!" (a statement i did not believe at the time.)

Problems in freshman calculus: 1) Give a characterization of a function g such that g is a primitive of a given Riemann integrable function f. Is it enough to assume that g is continuous and differentiable wherever f is continuous, and that g has derivative equal to f at such points? E.g. is a continuous function which is differentiable with derivative zero a.e. a constant function? If not, what assumptions do you have to add?

2) Give an intrinsic characterization of a function g that is a primitive of some unknown Riemann integrable function on [a,b]. Is it enough to assume that g is Lipschitz continuous?

I guess i would give more credence to this if it concerned say theorems that have physically compelling arguments that are hard to make mathematically rigorous, such as Riemann's arguments for the existence of meromorphic functions of second kind with arbitrary poles.

When someone says it is "obvious" that Euclidean space R^n has dimension n, they are really saying that any definition for which this is false is a bad definition, not that it is easy to give an appropriate definition, nor that it is easy to prove the theorem even for a good definition. So this is just an imprecise use of language.

Let me pose a little fun question: Since everyone knows that if n < m, there can be a continuous surjection, but no homeomorphism from R^n to R^m, what about a continuous injection from R^m to R^n? What is the obvious answer? Is it also the correct answer? How much does your response draw on some non obvious mathematical reasoning?

My best idea in the direction of the original question is: "why is a straight line the shortest smooth curve joining two points?"

Answer by Jason for Theorems that are 'obvious' but hard to prove

2011-01-10T03:50:08Z

I think this answers your question in a perverse way: All statements in the theory of Natural Numbers provable from the ZFC axioms of set theory. They are obviously true by definition.

EDIT: Looking at this objectively, it probably sounds like I'm saying if a statement is true, then it's obviously true. However, that was not my intent, and I apologize for what may have sounded like a thoughtless response. This is how I see it:

All statements expressible in the language of arithmetic can be represented by formulas in the language of set theory that are only $\Delta_1$ in the Levy hierarchy. In particular, all transitive models will agree on whether they are true. If we further restrict ourselves to only consider the true statements in $\mathbb{N}$ that are ZFC theorems, then all ZFC models will agree that these statements must be true so they are about as obvious to ZFC models as possible. Now if you are an oracle having knowledge of all such true statements, then you will probably develop an intuition that makes them all seem "intuitively obvious." This reflects the answers suggesting that a theorem is obvious after you prove it.

To add one more related point here, when addressing G$\ddot{\textrm{o}}$del's Incompleteness Theorem, one can naively ask about completing PA in the "obvious" way, i.e., by extending it to be the theory consisting of all true statements in $\mathbb{N}$. But of course such a completion is not computable.

Answer by Frank Thorne for Theorems that are 'obvious' but hard to prove

2011-01-10T05:23:44Z

Here's a "trick" answer:

$[0, 1]$ is connected.

But I consider this a trick answer because the real difficulty is turning this into a completely rigorous statement. Understanding the rigorous definition of connectedness (and understanding the point of making definitions like this) can be a substantial hurdle, but once this hurdle is crossed, the proof is not difficult.

Answer by David Sprehn for Theorems that are 'obvious' but hard to prove

2011-01-10T07:39:40Z

All isometries of the plane are affine linear.

Answer by ABayer for Theorems that are 'obvious' but hard to prove

2011-01-10T08:56:02Z

$\mathbb R^n$ is not homeomorphic to $\mathbb R^m$ unless $m = n$.

Answer by Daniel Moskovich for Theorems that are 'obvious' but hard to prove

2011-01-10T14:20:42Z

The most deadly example I know is the Hauptvermutung in dimensions 2 and 3 (in dimension $>3$, it is the ultimate "obvious but false" theorem). The Hauptvermutung, or "Main Conjecture" states that any two triangulations of a polyhedron are combinatorially equivalent, i.e. they become isomorphic after subdivision.
The Hauptvermutung is so obvious that it gets taken for granted everywhere, and most of us learn algebraic topology without ever noticing this huge gap in its foundations (of the text-book standard simplicial approach). It is implicit every time one states that a homotopy invariant of a simplicial complex, such as simplicial homology, is in fact a homotopy invariant of a polyhedron, unless one also proves independence relative to triangulation.
The Hauptvermutung for 2-manifolds was proven by Radó, and for 3-manifolds by Moïse in 1953. It is a genuinely deep, difficult theorem.
Edit: This answer is essentially taken from Page 4 of The Hauptvermutung Book.

Answer by Eric Hsu for Theorems that are 'obvious' but hard to prove

2011-01-10T22:45:46Z

In the same genre, if not the same type: The Fundamental Theorem of Algebra. Easily understood by high schoolers, plausible, beautifully simple to state. As far as I know, there are no nice proofs understandable by a good (not brilliant) high school student.

Answer by gowers for Theorems that are 'obvious' but hard to prove

2011-01-11T14:34:06Z

There are a number of facts in multivariable calculus that are obvious but hard to prove. For instance, the change-of-variables formula in a multiple integral is very easy to justify heuristically by talking about little parallelepipeds but troublesome (as I discovered to my cost in a course I once gave) to justify rigorously. And the same goes for the inverse function theorem: although the proof can be made quite transparent and the need for continuous differentiability makes good intuitive sense, there seems to be an irreducible core of actual work needed (in particular, the use of a fixed-point theorem to replace the use of the intermediate-value theorem in the 1D case).

I'd be quite glad to be told that this answer was wrong. If anyone knows of a link to an exposition of these results, particularly the first, that does proper justice to their intuitive obviousness, I'd be very pleased to hear about it.

Answer by Timothy Chow for Theorems that are 'obvious' but hard to prove

2011-01-11T14:50:32Z

The Kneser-Poulsen conjecture says that if a finite set of (labeled) unit balls in $\mathbb{R}^n$ is rearranged so that in the new configuration, no pairwise distance is increased, then the volume of the union of the balls does not increase. This was finally proved by Bezdek and Connelly in dimension 2 but remains open in higher dimensions.

There are several other notorious elementary problems in geometry that might qualify, e.g., the equichordal point problem, though this one is not quite as "obvious" as the Kneser-Poulsen conjecture.

Answer by Richard Stanley for Theorems that are 'obvious' but hard to prove

2011-01-12T03:40:47Z

It takes Russell and Whitehead several hundred pages to prove that $1+1=2$ in Principia Mathematica. They then say that "the above proposition is occasionally useful."

Answer by Terry Tao for Theorems that are 'obvious' but hard to prove

2011-01-13T16:13:50Z

"Global regularity of the Navier-Stokes equation" is not yet in this category, but once a proof is found, I am sure it will be.

More generally, there are many PDE which are "obviously" solvable for physical reasons, but for which actually proving existence (particularly in "global", "non-perturbative" situations, and requiring strong (regular) solutions rather than weak ones), is extremely difficult. A typical example is the Boltzmann equation, for which good global regularity results have only become available recently, with the work of Villani and others.

EDIT: Admittedly, many of the global regularity problems become a lot easier if one applies a physically reasonable truncation. For instance, global regularity for Boltzmann is much easier if one can somehow restrict the particle velocities to never exceed some upper bound $c$. But then the non-obvious fact moves elsewhere; rather than global regularity, the issue is whether one has sufficiently quantitative bounds that these thresholds rarely get triggered. Physically, it is intuitively obvious that a Boltzmann gas is not routinely churning out particles travelling at close to the speed of light; but it is remarkably difficult to quantify and then establish this rigorously.

Answer by Daniel Moskovich for Theorems that are 'obvious' but hard to prove

2011-01-19T14:28:40Z

The first of the Tait Conjectures seems intuitively obvious:

Any reduced diagram of an alternating link has the fewest possible crossings.

This 19th century conjecture is difficult to prove, with the proof coming only in 1987 by Kauffman, Murasugi, and Thistlethwaite, using the Jones Polynomial. The discovery of this proof was a huge coup for quantum topology; a quantum invariant was used to prove a difficult classical open problem.
While this is certainly hard to prove, I don't think it's unexpectedly hard to prove. Knot diagrams modulo Reidemeister moves form a rather complicated algebraic structure; and there's no reason to expect that any statement about knot diagrams should be easy to prove.

Answer by Lennart Meier for Theorems that are 'obvious' but hard to prove

2011-01-19T21:18:40Z

That the identity map of the circle is not nullhomotopic. [When one thinks about it, it is pretty much equivalent to the Brouwer fixed point theorem, which is not as obvious.]

Answer by ann o'nimous for Theorems that are 'obvious' but hard to prove

2011-01-20T11:12:30Z

On an elementary level, the intermediate value theorem is surprisingly deep.

On a less elementary level, the prime number theorem is "obvious" from $\sum_{p\leq x}1/p\sim \log\log x$ (that was noticed by Euler) and Dirichlet's theorem on primes in arithmetic progressions is "obvious" if you use the sieve of Erathostenes.

Answer by Daniel Moskovich for Theorems that are 'obvious' but hard to prove

2011-02-21T19:43:22Z

On page 33 of Stable Mappings and Their Singularities by Golubitsky and Guillemin (GTM 014; 1974), the following proposition is characterized as an "obvious, but surprisingly complicated result":

Proposition 1.10: Let S be a nonempty open subset of $\mathbb{R}^n$. Then S is not of measure zero.

Edit: As pointed out in the comments, Gowers already mentioned an equivalent result whose difficulty is more clear. Please vote this answer down (I can't vote down my own answers).

Answer by José Figueroa-O'Farrill for Theorems that are 'obvious' but hard to prove

2011-02-28T01:42:32Z

The Hodge decomposition theorem

It is obvious that there is a unique point in any given affine plane in a finite-dimensional euclidean vector space which is closest to the origin.

Therefore it would seem similarly obvious that every de Rham cohomology class on a compact oriented riemannian manifold should have a unique representative with minimal $L^2$ norm: namely, its harmonic representative.

Yet it takes some effort (elliptic regularity,...) to prove that the harmonic representative does in fact exist, i.e., that it is a smooth differential form.

Answer by kenneth for Theorems that are 'obvious' but hard to prove

2011-02-28T02:46:04Z

Dynamic programing principle (DPP) is one of the 'obvious' and also intuitive one, in the control problem. Many papers proves its validity in various setup, and all proofs are very complicated. But, there is rarely a counter example of DPP. I wonder, if there is general framework on it. See, http://mathoverflow.net/questions/22689/dynamic-programming-principle-dpp

Answer by Mark Biggar for Theorems that are 'obvious' but hard to prove

2011-02-28T18:31:52Z

No one has mentioned the 4-color map theorem yet. it would seem to be a classic example.

Answer by Ken Knox for Theorems that are 'obvious' but hard to prove

2011-03-05T07:53:17Z

The stability of Minkowski Spacetime

An asymptotically flat initial data set for the Einstein equations that is sufficiently "close" to the initial data for Minkowski spacetime generates a solution to the Einstein equations that approaches Minkowski spacetime asymptotically. (try saying that fast 3 times)

It is "obvious" because of our physical experience and intuition with gravity, and it is hard to prove because Einstein's equations are quite subtle and complicated.

There are other theorems in mathematical relativity that fall into this category, but this one is especially striking since it is particularly difficult to prove, while it "feels" blatantly true.

Answer by Buschi Sergio for Theorems that are 'obvious' but hard to prove

2011-03-05T10:47:03Z

No exatly what asked..

But historically: The fifth postulate of Euclid (to a point outside a straight line passes exactly one line parallel to this line). At first it seems an obvious fact, and tried to prove it....

Answer by Paul for Theorems that are 'obvious' but hard to prove

2011-08-26T07:48:44Z

I'm not sure if it is considered obvious, but I think the Collatz conjecture is also a good example (not just hard to prove, but still unproven!). I've always found it very frustrating that such a simple conjecture does not lend itself to a simple proof.

'Obvious' here sort of means: Try a lot of values, and you'll get a feeling THAT it indeed holds, but also WHY it holds.

http://en.wikipedia.org/wiki/Collatz_conjecture

Answer by Victor Protsak for Theorems that are 'obvious' but hard to prove

2011-08-26T20:01:58Z

The Carpenter's rule: a planar linkage can be straightened without the links running into each other. Although the statement had initially seemed obvious, its truth or falsity was a matter of debate among the experts for several years until Bob Connelly, Eric Demaine, and Günter Rote finally proved it. (The analogous statement in 3 dimensions is actually false.)

Answer by jjcale for Theorems that are 'obvious' but hard to prove

2011-08-26T20:46:00Z

A differentiable manifold M that is homeomorphic to the n-sphere is also diffeomorphic to the n-sphere . Obvious, but wrong ! (But right for 1-, 2-, 3-, 5-, 6- and 12-spheres).

Answer by Ron Maimon for Theorems that are 'obvious' but hard to prove

2011-08-27T10:35:03Z

Since this was resurrected, here is the statement that at this time seems to me to have the greatest gap between obviousness of truth and obviousness of proof:

There exists a natural set theoretic universe in which every subset of [0,1] is Lebesgue measurable, so that the reals admit no well order and do not inject into aleph-1.

Here are a different class of obvious theorems, these are only obvious in the sense of physical intuition. They took a long time to prove:

The existence of solid matter occupying space (in the lowest energy state, the electron-nucleus system occupies a volume proportional to the number of nuclei)
The positive energy theorem--- every asymptotically flat solution of GR obeying the appropriate energy condition has a positive mass at infinity, with zero mass only for Minkowski space.
Hard sphere collisions on a negatively curved space are ergodic.

Here is a physically obviously true statement, which can be seen from physical intuition, but which is not proven (as far as I know):

The asymptotic fate of GR with a positive cosmological constant is within any causal patch, and except for a set of initial conditions of measure zero, a deSitter space.

The reason this is obvious is because the deSitter space maximizes the horizon area, which is a measure of the entropy.

Answer by Woett for Theorems that are 'obvious' but hard to prove

2011-09-29T22:47:27Z

Very good candidates for this question are theorems that amount to saying that some sequence behaves randomly in a way. Both the fact that the statements are obvious and the fact that they are usually hard to prove, are explained by the fact that there is just no reason for the sequence to nót behave randomly. The primes are of course notorious for this. Easy example that springs to mind: there are about as many primes whose last (decimal) digit is $1$ as there are primes whose last digit is $3$.

Answer by James D. Taylor for Theorems that are 'obvious' but hard to prove

2011-09-30T02:02:59Z

I think Godel's completeness theorem is very intuitive. For example, can you imagine a first order theorem that would be true for all groups, that you wouldn't be able to prove (by Godel's definition of `prove'). Of course not! But the proof of the completeness theorem is hard.

Answer by none for Theorems that are 'obvious' but hard to prove

2013-05-04T04:00:56Z

P is not equal to NP. This is "obvious", is a straightforward arithmetic proposition doesn't involve any fancy set theory or spacefilling curves, and yet it's so hard that there have whole workshops ("Barriers") and important papers (BGS, natural proofs etc.) devoted to the question of what makes it so hard. Scott Aaronson describes "a would-be P≠NP prover who hasn’t yet grasped the sheer number of mangled skeletons and severed heads that line his path." P≠PSPACE is even more obvious and yet there is a comparable lack of progress.

Answer by Henr.L for Theorems that are 'obvious' but hard to prove

2013-05-04T04:15:00Z

I think a very good example is Kepler Conjecture:

http://en.wikipedia.org/wiki/Kepler%27s_conjecture

This conjecture stated that the most "tight" stack of same balls have only two kinds of arrangement with a fixed density.

Every physicist knows that it's true, no mathmaticans ever proved it.

Fortunately, Hales used computers to step forward a little little bit.

Answer by Burak for Theorems that are 'obvious' but hard to prove

2013-05-04T05:37:52Z

Here is an example proof of which is not conceptually difficult, but tedious to write down in full detail: There exists a universal Turing machine.

The reason I find this obvious is that, after learning the definition of what a TM is, a moment's thought will suggest that we should be able to encode information of any TM to a natural number since it is a finitary object. Then, it is conceivable that there is an algorithm which decodes any natural number and carry out the instructions.

Of course, constructing the universal machine and proving that it can actually simulate all the other machines involve some unpleasant details.