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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?

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    $\begingroup$ Perhaps the isoperimetric inequality. $\endgroup$
    – Péter Komjáth
    Commented Jan 9, 2011 at 11:28
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    $\begingroup$ Perhaps the Kepler Conjecture: en.wikipedia.org/wiki/Kepler_conjecture $\endgroup$
    – Aaron Meyerowitz
    Commented Jan 9, 2011 at 11:57
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    $\begingroup$ A former colleague of mine used to say (to students), "A theorem is obvious if a proof instantly springs to mind," a maxim I like a lot. I think what you are talking about is theorems where a plausible argument instantly springs to mind but falls short of being a proof. $\endgroup$
    – gowers
    Commented Jan 9, 2011 at 15:12
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    $\begingroup$ I am tempted to vote to close as subjective and argumentative given the comments on the existing answers. Can we narrow the definition of 'obvious' being used? Something like gowers' definition is good, but depends a lot on one's training. Perhaps something like "if you asked an undergraduate if it were true, they'd bet yes." $\endgroup$
    – Qiaochu Yuan
    Commented Jan 9, 2011 at 16:46
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    $\begingroup$ I disagree that the Jordan curve theorem is "obvious" but admits a surprisingly difficult proof. The proof for curves with reasonable regularity is not difficult, while the truth of the theorem for wild curves is not so obvious, I think. (At least, I think it is reasonable to argue that most people's sense of this being intuitively clear comes from imagining a rather regular curve in the plane, not a wild one.) $\endgroup$
    – Emerton
    Commented Jan 10, 2011 at 16:46

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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.)

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    $\begingroup$ Can you expand on this? $\endgroup$
    – Lennart Meier
    Commented Jan 9, 2011 at 20:40
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    $\begingroup$ @Lennart: enumerate the rationals, and take an interval of length $\epsilon / 2^n$ around the $n$th rational. You get a countable collection of intervals with lengths summing to $\epsilon$ whose union contains all the rationals (never mind just those in $[0,1]$). $\endgroup$
    – Chris Eagle
    Commented Jan 9, 2011 at 21:07
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    $\begingroup$ @Joe and @Harry, it is of course trivial if you know that there is a countably additive measure that extends lengths of intervals. But that result is not trivial, and many people are tempted to think that the simple statement about intervals is just plain obvious. $\endgroup$
    – gowers
    Commented Jan 13, 2011 at 12:16
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    $\begingroup$ Using Lebesgue measure, you can do it. However, you probably need to know this fact (or something equivalent) in order to construct Lebesgue measure. The proof in my book uses compactness and reduces to the case of finitely many intervals. (Measure, Topology, and Fractal Geometry, Lemma 5.1.1) $\endgroup$
    – Gerald Edgar
    Commented Mar 4, 2011 at 17:50
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    $\begingroup$ The proof (any proof) of Heine Borel can be adapted to simultaneously prove the statement here (along with Heine Borel at the same time). For example, consider a covering of $[0,1]$ by open intervals and let $S$ be the set of $x$ so that $[0,x]$ is covered by a finite union of these intervals such that $\sum_{k=1}^n |I_k \cap (- \infty,x]| > x$. $S$ is nonempty since $0 \in S$, and the supremum of $S$ can't be less than $1$. The other proof of Heine Borel (shrinking closed intervals intersect) can also be adapted to give a direct proof of this fact (and countless other basic facts). $\endgroup$
    – Phil Isett
    Commented Aug 27, 2011 at 3:39
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$\mathbb R^n$ is not homeomorphic to $\mathbb R^m$ unless $m = n$.

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    $\begingroup$ 1+. I think this is the key point of the answer of Georges Elencwajg. It's not so really about one specific topological dimension and its computation, but rather that we can distinguish the affine spaces at all. I wonder how many decades (centuries?) mathematicians were convinced of this fact without having a proof? $\endgroup$
    – Martin Brandenburg
    Commented Jan 12, 2011 at 8:01
  • $\begingroup$ Until Brauwer provided a tremendous complicated proof in 1921 without using homology, I guess the mathmaticans believe it as sth geometrically trvial. $\endgroup$
    – Henry.L
    Commented May 4, 2013 at 4:07
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    $\begingroup$ The only proof I know is by puncturing each space and calculating relative homology. Is there a simpler proof? $\endgroup$
    – Uzu Lim
    Commented Aug 15, 2022 at 19:58
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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.

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    $\begingroup$ At least since the introduction of singular homology, there is no gap in the foundations of algebraic topology. But I agree that the Hauptvermutung has a feeling like being obvious. $\endgroup$
    – Lennart Meier
    Commented Jan 10, 2011 at 18:46
  • $\begingroup$ Answer editted to clarify what I meant. $\endgroup$
    – Daniel Moskovich
    Commented Jan 11, 2011 at 1:02
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    $\begingroup$ By "subdivision," I believe you mean PL-subdivision, making the Hauptvermutung a little less obvious. An essentially equivalent statement to its falsity in dimension 5 that I find very unintuitive is that there exists a 4-dimensional simplicial complex $K$ which is not a triangulation of a manifold, but whose suspension $\Sigma K$ is a triangulation of a 5-sphere. $\endgroup$
    – Richard Stanley
    Commented Jan 12, 2011 at 0:59
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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".)

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    $\begingroup$ This is my idea of an "obvious" answer that is hard to prove, i.e. one that has a physically persuasive justification. $\endgroup$
    – roy smith
    Commented Jan 10, 2011 at 5:17
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    $\begingroup$ The fact that bubbles are round only demonstrates a local, not gloabal, minimum, surely? The bubble can't explore arbitrary regions of phase space: it can only go downhill. $\endgroup$
    – Max
    Commented Jan 10, 2011 at 9:10
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    $\begingroup$ @Max: If it was only a local minimum then some bubbles would be round, and others would form different shapes. $\endgroup$
    – George Lowther
    Commented Jan 19, 2011 at 23:31
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    $\begingroup$ @George: suppose you have a double-well model with one well lower than the other. Also suppose that the physical states are (for whatever reason) always created in the higher well and that the barrier between the wells is higher than the fluctuations in the system at the given temperature. Surely then the physical system wouldn't see the global minimum. So this argument doesn't work even at the physical level (never mind rigorous treatment)$\ldots$ $\endgroup$
    – Marek
    Commented Jul 29, 2011 at 21:09
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    $\begingroup$ Ice 9 is a popular account of such a local min not being the global min $\endgroup$
    – Steven Gubkin
    Commented Mar 14, 2014 at 19:19
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The Jordan curve theorem! Of course in this case the real problem is the meaning of "closed curve".

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    $\begingroup$ Wasn't this already mentioned as the motivating example in the OP? $\endgroup$
    – Todd Trimble
    Commented Jan 9, 2011 at 16:10
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    $\begingroup$ @Sean: I think "easy" here should be taken to mean "without machinery." With enough machinery, everything is simple... $\endgroup$
    – Qiaochu Yuan
    Commented Jan 9, 2011 at 17:30
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    $\begingroup$ @Sean: Maybe they are hard just because the machinery to make it simple hasn't been invented yet :) $\endgroup$
    – José Figueroa-O'Farrill
    Commented Jan 9, 2011 at 18:14
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    $\begingroup$ It is not guaranteed by a silly counting argument that there are short simple statements with irreducibly hard proofs. $\endgroup$
    – gowers
    Commented Jan 10, 2011 at 7:50
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    $\begingroup$ But there aren't infinitely many that are readable within a human lifetime, the criterion you were using for proofs. $\endgroup$
    – gowers
    Commented Jan 10, 2011 at 20:13
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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.

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    $\begingroup$ I'm afraid you're right. The change of variables formula for multiple integrals is a notorious "Is it really this hard?" moment in mathematical exposition. $\endgroup$
    – Pete L. Clark
    Commented Jan 11, 2011 at 18:14
  • $\begingroup$ Did you mean the inverse function theorem or the implicit function theorem? I thought the former only required the multivariate chain rule. $\endgroup$
    – Qiaochu Yuan
    Commented Jan 12, 2011 at 20:44
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    $\begingroup$ The inverse and implicit function theorems are actually equivalent. $\endgroup$
    – Paul Siegel
    Commented Jan 15, 2011 at 15:57
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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."

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    $\begingroup$ I've always wondered about the next sentence: "It is used at least three times, in $\ast 113 \cdot 66$ and $\ast 120 \cdot 123 \cdot 472$." [my emphasis]. Does this express some sort of dry humour or is it meant seriously? $\endgroup$
    – Theo Buehler
    Commented Jan 12, 2011 at 3:58
  • $\begingroup$ The end of the proof is here: $\ast 110 \cdot 643$. quod.lib.umich.edu/cgi/t/text/… $\endgroup$
    – Theo Buehler
    Commented Jan 12, 2011 at 4:03
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    $\begingroup$ Then Poincare made a great comment to this book: To those who doesn't even know 1+1=2. $\endgroup$
    – Henry.L
    Commented May 4, 2013 at 4:09
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    $\begingroup$ Unlike other answers, that's an artifact of their particular formalization—modern formal proofs of 1+1=2 are immediate, at least in type theory. In fairness, it's cool enough they managed to formalize math at all. $\endgroup$
    – Blaisorblade
    Commented May 6, 2017 at 22:37
  • $\begingroup$ @Henry.L: Interesting quote. Reference? $\endgroup$
    – C.F.G
    Commented Oct 9, 2020 at 20:22
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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 $n$.

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    $\begingroup$ Isn't the problem how to <b>define</b> topological dimension? With the "right" definition, is the proof hard? $\endgroup$
    – Daniel Moskovich
    Commented Jan 9, 2011 at 15:21
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    $\begingroup$ Dear Daniel, indeed defining topological dimension required amazing ingenuity, from Lebesgue among others. He later explained that his intuition came from contemplating a brick wall and noticing that some points had to be covered by 3(=2+1) bricks! However even granting this, proving that $\mathbb R^n$ has dimension $n$ remains a difficult problem needing techniques of algebraic topology to be solved, even though the definition is on the level of general topology. (To be continued) $\endgroup$
    – Georges Elencwajg
    Commented Jan 9, 2011 at 16:11
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    $\begingroup$ (Continuation) This is confirmed by Munkres in his well-known text-book Topology. After 10 pages (§50) devoted to dimension theory, he concludes "we do not ask you to prove...that the topological dimension of an $m$-manifold is precisely $m$. And for good reason; the proof requires the tools of algebraic topology." Another indication of hardness is the "invariance of domain" theorem: non-empty open subsets of $\mathbb R^n$ and $\mathbb R^m$ are never homeomorphic unless $n=m$.This does not involve the definition of "dimension" but is quite a difficult theorem (first proved by Brouwer). $\endgroup$
    – Georges Elencwajg
    Commented Jan 9, 2011 at 16:32
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    $\begingroup$ It's easy to prove that $dim(\mathbb R^n)\le n$. The opposite inequality, $dim(\mathbb R^n)\ge n$, is equivalent and about as difficult to prove as the fixed point property for the n-ball (or n-cube $\mathbb I^n$). $\endgroup$
    – Włodzimierz Holsztyński
    Commented May 4, 2013 at 20:39
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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.

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    $\begingroup$ Right. Any Jordan curve the "man in the street" is thinking of is basically piecewise linear. $\endgroup$
    – Qiaochu Yuan
    Commented Jan 9, 2011 at 22:32
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    $\begingroup$ Well, unless they've never seen a circle, piecewise $C^\infty$ ... $\endgroup$
    – Yemon Choi
    Commented Jan 10, 2011 at 1:40
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    $\begingroup$ I don't draw PL-approximations to circles... $\endgroup$
    – Steven Gubkin
    Commented Jan 10, 2011 at 15:45
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    $\begingroup$ I don't even draw PL approximations to polygons! $\endgroup$
    – George Lowther
    Commented Jan 11, 2011 at 2:54
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    $\begingroup$ There is a trick of using your pinky knunckle as the point of a hand compass which allows you to draw "perfect" circles without the aid of a metal compass. I always impress students during office hours with this trick. $\endgroup$
    – Steven Gubkin
    Commented Jan 11, 2011 at 20:55
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The trefoil knot is knotted.

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    $\begingroup$ I don't think this is hard to prove, as long as you don't insist on a "useful" proof. See Tietze's 1908 proof. The moment you have any presentation of the fundamental group (Wirtinger/Dehn/whatever), you find a representation onto a non-abelian group and you finish. $\endgroup$
    – Daniel Moskovich
    Commented Jan 9, 2011 at 18:37
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    $\begingroup$ @Daniel: that seems like another example of hiding behind machinery to me. You at least need to know that fundamental groups are homotopy invariants and what the fundamental group of the circle is. If you wrote out the complete proofs of all the results you're depending on, would you still consider the resulting proof "easy"? $\endgroup$
    – Qiaochu Yuan
    Commented Jan 9, 2011 at 20:37
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    $\begingroup$ I strongly disagree with what everyone is saying. There's the formulation, then there's the proof. The formulation says (or at least immediately implies) that a certain 3-manifold (the trefoil complement) is not the solid torus. The proof is about as straightforward as any statement about 3-manifolds could possibly be. Moreover, the proof was found almost immediately when the techniques existed- it was never an "open problem". So yes- if I wrote out all the proofs of everything I use, I'd consider it long but easy. Indeed, I think the fundamental group proof is easiest. $\endgroup$
    – Daniel Moskovich
    Commented Jan 9, 2011 at 20:57
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    $\begingroup$ Modulo the Reidemeister moves I think an easier proof is noticing that the trefoil is $\mathbb{F}_2$ colorable and the unknot isn't. $\endgroup$
    – solbap
    Commented Jan 9, 2011 at 21:19
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    $\begingroup$ @Daniel: in the context of the OP's clarification in the comments ("I want theorems for which a plausible argument springs to mind at the level of sophistication required to understand the statement, but for which a proof requires a higher level of sophistication.") I think it is totally reasonable to argue that the statement of the result requires a much lower level of sophistication than any of its proofs. (The plausible argument here is something like "if you try it with a physical trefoil knot it's obviously knotted.") $\endgroup$
    – Qiaochu Yuan
    Commented Jan 9, 2011 at 23:48
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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.

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  • $\begingroup$ I doubt the algebraic proof is easier. $\endgroup$
    – Sean Tilson
    Commented Jan 9, 2011 at 23:03
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    $\begingroup$ The original algebraic proof (due to Nielsen) is not that bad. The basic idea is that you show that any subgroup has a generating set $S$ such that if $w$ is a reduced word in $S$, then when you plug the free group elements corresponding to elements of $S$ into $w$, at least one letter survives from each "letter" in $w$. The argument that you can do this is (slightly) similar to row reduction in matrices. There's a readable account of this towards the beginning of Lyndon and Schupp's classic book on combinatorial group theory that is well worth reading. $\endgroup$
    – Andy Putman
    Commented Jan 10, 2011 at 4:21
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    $\begingroup$ For me, the Theorem that subgroups of free groups are free, is not obvious at all. $\endgroup$
    – Martin Brandenburg
    Commented Jan 12, 2011 at 8:03
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    $\begingroup$ For me neither. The plausible argument above also "shows" that submonoids of a free monoid are free. $\endgroup$
    – Jan Weidner
    Commented Aug 3, 2011 at 16:01
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    $\begingroup$ Well, I didn't say it was right! $\endgroup$
    – Qiaochu Yuan
    Commented Aug 3, 2011 at 16:50
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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?)

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    $\begingroup$ Peano arithmetic doesn't have to be consistent in order for us to meaningfully do mathematics. It just has to have the property that the shortest contradiction is so long that it is impossible to write down before the end of the universe! $\endgroup$
    – Qiaochu Yuan
    Commented Jan 9, 2011 at 23:00
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    $\begingroup$ Wait a minute. If there is a contradiction, then everything is provable, right? So you can get nice short contradictions. I guess it could still be that the shortest proof of a contradiction must be very long. $\endgroup$
    – Jeff Strom
    Commented Jan 9, 2011 at 23:13
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    $\begingroup$ The consistency of Peano arithmetic is not "hard to prove" in the intended sense of "hard." Logical strength is not the same as psychological difficulty or length of proof or any plausible notion of the hardness of finding a proof. $\endgroup$
    – Timothy Chow
    Commented Jan 10, 2011 at 22:43
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    $\begingroup$ Gentzen's proof cheats as well; if you harbor serious doubts about the consistency of PA then you're likely to harbor doubts about such a strong induction principle. Cheating is inevitable. That doesn't mean that it's hard to cheat. $\endgroup$
    – Timothy Chow
    Commented Jan 11, 2011 at 14:56
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    $\begingroup$ The proof is extremely easy if done the way a non-logician would do it: point out that $\mathbb N$ is a model. What is hard is a SYNTACTIC proof. $\endgroup$
    – Christian Remling
    Commented Apr 12, 2014 at 18:54
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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?"

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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.

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    $\begingroup$ I would not view the ergodic theorem as obvious. $\endgroup$
    – Igor Rivin
    Commented Jan 9, 2011 at 21:22
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Chess is not a forced win for black.

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    $\begingroup$ @Professor Borcherds: if this is a theorem, could you give some indications of the proof? $\endgroup$
    – Pete L. Clark
    Commented Jan 9, 2011 at 17:40
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    $\begingroup$ My guess is that Prof. Borcherds said that is should be possible to prove this statement, and that is seems to be true, "obviously", but there is no such proof. As far as I know there is not even an idea of how one could prove such a statement. $\endgroup$
    – Tim van Beek
    Commented Jan 9, 2011 at 18:34
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    $\begingroup$ All the more surprising, since the game which is just like chess, but where each side gets to make two moves in a row is provably not a forced move for black (proof: exercise) $\endgroup$
    – Igor Rivin
    Commented Jan 9, 2011 at 21:24
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    $\begingroup$ Along the same lines is, "Chess is a forced win for black if White gives queen odds." $\endgroup$
    – Timothy Chow
    Commented Jan 9, 2011 at 21:49
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    $\begingroup$ @Tim van Beek: no, but there's an obvious reason white can at least draw. An example of a silly chess variant where white actually has a demonstrable winning strategy is when the first check wins the game. $\endgroup$
    – Chris Eagle
    Commented Jan 9, 2011 at 22:56
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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 :-)

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    $\begingroup$ It's straightforward to see that any maximal ideal with residue field C has this property, so the weak Nullstellensatz is equivalent to the statement that all residue fields are C. I don't think this is "obvious," exactly. Certainly similar obvious statements are false. For example, one might guess that all residue fields of an infinite direct product of copies of R are equal to R, and this is completely wrong. $\endgroup$
    – Qiaochu Yuan
    Commented Jan 9, 2011 at 20:42
  • $\begingroup$ There are some reasonably simple proofs of the result, my favorite being one that appeared in the Monthly in 2006 by Arrondo mat.ucm.es/~arrondo/monthly169-171-arrondo.pdf Still, they're never quite as obvious as one might think when first encountering the statement. $\endgroup$
    – Thierry Zell
    Commented Jan 10, 2011 at 1:37
  • $\begingroup$ -1. This is not an obvious fact at all. Of course, when you have learnt some classical algebraic geometry, you understand this Theorem and then also its various proofs ... but this does not show that it's obvious "a priori". $\endgroup$
    – Martin Brandenburg
    Commented Jan 12, 2011 at 8:06
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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.

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    $\begingroup$ Denis, thanks a lot for baiting me. I appreciate very much linquistic lessons I've learned. Of course, there are people to whom "the existence of non-trivial solutions to linear PDEs for constant coefficients operators" is obvious. Enjoy. $\endgroup$
    – Wadim Zudilin
    Commented Jan 9, 2011 at 14:07
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    $\begingroup$ I believe "H. Levy" should by "H. Lewy", c.f. en.wikipedia.org/wiki/Hans_Lewy. $\endgroup$
    – Pete L. Clark
    Commented Jan 9, 2011 at 17:38
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    $\begingroup$ @Denis: I don't doubt you for a moment, but can you explain a little bit better where the difficulty lies? Doesn't the Fourier transform reduce "analytic" question about linear constant coefficient differential operators to a corresponding "algebraic" question about polynomial multiplication operators on the transform side? That is, doesn't solving of $P(\partial/\partial x^i) f(x) = g(x)$ just reduce to the simple question of solving $P(\xi^i) \hat f(\xi) = \hat g(\xi)$ ? $\endgroup$
    – Dick Palais
    Commented Jan 9, 2011 at 19:57
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    $\begingroup$ @Dick: Yes, it basically does. But the problem is then to solve this and it is not obvious that there is an invers of this polynomial function in the space of (tempered) distributions. To prove this you need some tricks (a priori estimates to use Hahn-Banach / Bernstein-Sato-polynomials / an explicit formula for the inverse / ...) Also the idea to use the Fourier transformation is not obvious for every mathematician. $\endgroup$
    – Johannes Hahn
    Commented Jan 10, 2011 at 0:15
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    $\begingroup$ I don't think the existence of solutions to constant coefficient PDE is so obvious, even though the proof is conceptually straightforward -- my reason being that the statement is false on the torus. Maybe you would consider these obstructions obvious..? $\endgroup$
    – Phil Isett
    Commented Aug 27, 2011 at 3:45
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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.

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    $\begingroup$ But see Milnor's proof of the Hairy Ball Theorem in the American Mathematical Monthly, July 1978, pp. 521-524. (I wrote about this here: topologicalmusings.wordpress.com/2008/07/22/… .) $\endgroup$
    – Todd Trimble
    Commented Jan 9, 2011 at 15:36
  • $\begingroup$ It is still not obvious, isn't it? Needs work I mean. $\endgroup$
    – Patrick I-Z
    Commented Jan 9, 2011 at 16:25
  • $\begingroup$ @Patrick: Yes, still takes work; I was really responding to "takes considerable machinery", presumably referring to homology theory. $\endgroup$
    – Todd Trimble
    Commented Jan 9, 2011 at 23:44
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    $\begingroup$ I misstated Lefschetz' easy argument. If the vector at p is non zero, then on a small disc around p all vectors are roughly parallel to that one. It is possible then to see that on the disc which is the complement of that one, the degree of the map defined by the vectors on the boundary is ±2. (Collapsing the external disc onto the inner disc, by deflating the balloon, reflects the vectors on the boundary circle in the tangent lines to that circle. This fixes the vectors at top and bottom, rotating the rest 180 deg every quarter circle. thus there is a zero outside the original disc.) $\endgroup$
    – roy smith
    Commented Jan 12, 2011 at 18:21
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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.]

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    $\begingroup$ This was my introduction to many sophisticated mathematical ideas. When teaching several variable calculus I asked my self how I could convince students that Stokes theorem was useful, and found that it could be used to prove this fact, hence also the fundamental theorem of algebra, Brouwer fix point theorem, etc... Equivalently, why is the one form "dtheta" locally exact but not exact? $\endgroup$
    – roy smith
    Commented Jan 20, 2011 at 6:04
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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: Embeddings and triangulations of real analytic varieties (many thanks again to Mohan and Benoit).

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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.

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  • $\begingroup$ That's a nice example! $\endgroup$
    – Alec Edgington
    Commented Jan 11, 2011 at 21:13
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    $\begingroup$ It seems to me far from obvious that that result should be true in all dimensions, especially after the dramatic disproof of the Borsuk conjecture due to Kahn and Kalai or the disproof of the Busemann-Petty conjecture. $\endgroup$
    – gowers
    Commented Jan 13, 2011 at 16:23
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"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.

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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.)

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    $\begingroup$ And the analogous statement in dimensions greater than 3 is true: they can always be straightened! $\mathbb{R}^3$ is the exception. $\endgroup$
    – Joseph O'Rourke
    Commented Aug 26, 2011 at 23:51
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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.

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    $\begingroup$ The obviousness of the impossibility of turning a sphere inside out really depends on what intermediate configurations are permissible. $\endgroup$
    – S. Carnahan ♦
    Commented Jan 9, 2011 at 17:45
  • $\begingroup$ Fair enough Scott. I was aware that a supposedly simple answer was becoming a complex comment. $\endgroup$
    – Mark Bennet
    Commented Jan 9, 2011 at 20:16
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    $\begingroup$ compactness is a n important concept. $\endgroup$
    – roy smith
    Commented Jan 10, 2011 at 3:58
  • $\begingroup$ Is it really hard to prove, though? One can say, "A point can go off to infinity on the plane but it can't on the sphere," which can be turned into a rigorous proof about sequential compactness. Right? $\endgroup$
    – Akiva Weinberger
    Commented Jun 2, 2016 at 18:39
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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!

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    $\begingroup$ -1: All kinds of stuff could go wrong! $\endgroup$
    – Autumn Kent
    Commented Jan 9, 2011 at 16:23
  • $\begingroup$ Like what? What could possibly go wrong? It's a PL disc, for crying out loud! $\endgroup$
    – Daniel Moskovich
    Commented Jan 9, 2011 at 19:25
  • $\begingroup$ The simple loop conjecture is also obvious, I take it? $\endgroup$
    – Autumn Kent
    Commented Jan 9, 2011 at 19:59
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    $\begingroup$ I don't know much about 3-manifolds, so perhaps my intuition is undeveloped--that said, when I found out about the sphere theorem, I did a double-take. Were someone to have asked me if I thought this was true before I ran across it, I would have guessed it was wildly optimistic. $\endgroup$
    – Daniel Litt
    Commented Jan 10, 2011 at 18:39
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    $\begingroup$ @Richard: Almost all undergraduates I know would answer neither "yes" nor "no" but would stare blankly, not understanding the question. $\endgroup$
    – Timothy Chow
    Commented Aug 26, 2011 at 18:26
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If $T_1, \dots, T_k$ are subtrees of a tree $T$ which pairwise intersect, then $\cap_{i=1}^k T_i$ is non-empty. This is a standard fact in graph theory and almost completely obvious, but it is surprisingly difficult to give a correct proof.

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  • $\begingroup$ Perhaps if you use the word "connected", it becomes easier to give a correct proof. Gerhard "Likes Making Some Proofs Easier" Paseman, 2018.04.10. $\endgroup$
    – Gerhard Paseman
    Commented Apr 11, 2018 at 0:04
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    $\begingroup$ Is the following wrong? It's enough to do the case $k=3$: once we know this case, then $T_1,\ldots, T_{k-2}, T_{k-1}\cap T_k$ will satisfy the hypotheses of the theorem and we are done by induction. To do the case $k=3$, compute the Euler characteristic of $T_1 \cup T_2 \cup T_3$ using the "inclusion-exclusion" formula $\chi(X \cup Y) = \chi(X)+\chi(Y)-\chi(X \cap Y)$ valid for compact $X$ and $Y$. If $T_1 \cap T_2 \cap T_3 = \emptyset$ then the Euler characteristic is zero, which is impossible. $\endgroup$
    – Dan Petersen
    Commented Apr 11, 2018 at 8:31
  • $\begingroup$ Or more elementary, for k=3, assume $u\in T_1\cap T_2, v\in T_2\cap T_3$, then the unique path between u,v is completely in $T_2$. It must contain a vertice in $T_1\cap T_3$, else there is an edge divides $T_1$ and $T_3$ which makes $T_1\cap T_3 = \varnothing$ $\endgroup$
    – Reflecting_Ordinal
    Commented May 12 at 19:56
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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.

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    $\begingroup$ An equivalent example: The intermediate value theorem. $\endgroup$
    – Lennart Meier
    Commented Feb 16, 2013 at 0:22
  • $\begingroup$ Also equivalent: every path-connected space is connected. $\endgroup$
    – R. van Dobben de Bruyn
    Commented Apr 11, 2018 at 1:41
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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.

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    $\begingroup$ You need to put "smoothly" in front of "isotopic," otherwise the statement is false: all knots are isotopic (by shrinking all the "knottedness" to a point, a.k.a. bachelor's unknotting). Usually, one fixes this by defining knots to be equivalent if they are ambient isotopic, in which case you get the homeomorphism for free. (In the smooth case, the proof amounts to proving that the isotopy extends to the 3-sphere.) $\endgroup$
    – Autumn Kent
    Commented Jan 10, 2011 at 0:36
  • $\begingroup$ It's the definition of ambient isotopy! See e.g. Page 2 of Burde-Zieschang. $\endgroup$
    – Daniel Moskovich
    Commented Jan 10, 2011 at 14:30
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    $\begingroup$ That was a misleading passage in Wikipedia. I tried to clarify it. $\endgroup$
    – Douglas Zare
    Commented Jan 11, 2011 at 1:30
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    $\begingroup$ Glad to help, David. I agree with the assessment that the extension is obvious but not easy. $\endgroup$
    – Autumn Kent
    Commented Jan 11, 2011 at 4:22
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    $\begingroup$ @Daniel I disagree about which is the natural formalism. My physical intuition for why you can't untie a trefoil is that it would have to pass through itself, which I capture with the injectivity hypothesis. The fact that I would have to move space out of the way is completely irrelevant. After all, if I put a knotted rope in a vacuum, it is still knotted! $\endgroup$
    – David E Speyer
    Commented Jan 11, 2011 at 14:50
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In the same vein, chating: My advisor used to say "An interesting theorem is a theorem true which looks false". Well, tastes and colors... ;-)

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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.

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