114
$\begingroup$

A while back I saw posted on someone's office door a statement attributed to some famous person, saying that it is an instance of the callousness of youth to think that a theorem is trivial because its proof is trivial.

I don't remember who said that, and the person whose door it was posted on didn't remember either.

This leads to two questions:

(1) Who was it? And where do I find it in print---something citable? (Let's call that one question.)

(2) What are examples of nontrivial theorems whose proofs are trivial? Here's a wild guess: let's say for example a theorem of Euclidean geometry has a trivial proof but doesn't hold in non-Euclidean spaces and its holding or not in a particular space has far-reaching consequences not all of which will be understood within the next 200 years. Could that be an example of what this was about? Or am I just missing the point?

$\endgroup$
  • 11
    $\begingroup$ This is anti-climactic since you (rather quickly!) chose an answer, but what is your definition of "nontrivial theorem"? For example, would Schur's Lemma or Maschke's theorem have counted? $\endgroup$ – Boyarsky Jun 20 '10 at 2:10
  • 5
    $\begingroup$ I also think this should be community wiki (since there really isn't a right answer). $\endgroup$ – Akhil Mathew Jun 20 '10 at 2:20
  • 6
    $\begingroup$ Very often non-trivial theorems become definitions, or new definitions are specifically chosen so that they become trivial. Thereafter, they have trivial proofs. For example, the fact that homology is invariant under homotopy is (almost) trivial once you know singular homology. Even more often, our whole way of viewing math changes so that we get used to some new amazing discovery (as in Joel's example of existence of uncountable sets below) $\endgroup$ – Ilya Grigoriev Jun 20 '10 at 3:03
  • 19
    $\begingroup$ "callousness" or "callowness"? $\endgroup$ – Yemon Choi Jun 20 '10 at 3:12
  • 7
    $\begingroup$ Something from Spivak's "Calculus on Manifolds" is particularly germane to this discussion: "There are good reasons why theorems should all be easy and the definitions hard [...] Definitions serve a twofold purpose: they are rigorous replacements for vague notions, and machinery for elegant proofs. [...] A fully evolved major theorem has three important attributes: (1) It is trivial. (2) It is trivial because the terms appearing in it have been properly defined. (3) It has significant consequences." $\endgroup$ – WetSavannaAnimal Jun 10 '11 at 0:37

43 Answers 43

6
$\begingroup$

The finite intersection property: If $C_\alpha$ (for $\alpha\in I$) are closed subsets in a compact space, and every finite intersection of $C_\alpha$-s is nonempty, then the whole intersection $\bigcap_{\alpha\in I}C_\alpha$ is nonempty.

Proof. Otherwise, the complement $\bigcup_{\alpha\in I}C_\alpha^c$ is an open cover of the space without a finite subcover.

You may prefer the version with the $C_\alpha$-s compact and no assumption on the space containing them, but this is the same since we can intersect all $C_\alpha$-s with some fixed $C_{\alpha_0}$.

To me, it is surprising that this trivial proof gives such a useful assertion.

One may argue that this boils down to De Morgan's Laws, which are also trivial but very useful!

$\endgroup$
  • 3
    $\begingroup$ Probably the nontrivial thing here is to isolate the notion of compactness and show its wide applicability. $\endgroup$ – Monroe Eskew May 9 '14 at 1:34
6
$\begingroup$

The ultimate example that I know of is the Central Limit Theorem, described by Tijms as ``the unofficial sovereign of probability theory''. Incidentally, its significance took time to sink in- it has been forgotten and reproved repeatedly throughout its history.

Classical CLT: Given iid random variables $X_1,X_2,\ldots$ of mean $0$ and variance $1$, the sequence of random variables $\frac{X_1+X_2+\cdots+ X_n}{\sqrt{n}}$ converges in distribution to a normal random variable with mean $0$ and variance $1$.

The proof just Taylor's theorem and the definition of the exponential function (and Lévy's continuity theorem to confirm that the trivial proof indeed implies the theorem statement):

Proof: The Taylor expansion of the characteristic function $Ee^{itX}$ is: $1-t^2/2+o(t^2)$. Plugging in, the characteristic function of $\frac{X_1+X_2+\cdots+X_n}{\sqrt{n}}$ is $\left(1-t^2/2n+o(t^2/n)\right)^n$ which converges to $e^{-t^2/2}$. By Lévy's continuity theorem, convergence of characteristic functions implies convergence in distribution.QED
$\endgroup$
  • 6
    $\begingroup$ However, Levy's theorem that convergence in distribution is equivalent to convergence of the characteristic functions is less trivial. $\endgroup$ – Jochen Wengenroth Dec 1 '15 at 9:36
  • 1
    $\begingroup$ I agree, but given this result the proof is trivial. Editting to make my answer more honest. $\endgroup$ – Daniel Moskovich Dec 1 '15 at 11:41
5
$\begingroup$

I have looked through the answers but haven't come across Cantor's theorem, that the there is no surjection from a set $M$ on its power set $P(M)$.

I don't know whether the proof should be considered trivial, but it is short and easy to understand. The assumption of a surjection $f\colon M\rightarrow P(M)$ leads to the contradictory set $A_f:=\{m\in M\colon m\not\in f(m)\}$, the contradiction being $A_f\in A_f\longleftrightarrow A_f\not\in A_f$.

The implication of the theorem is that there is no "largest" set/infinitude!

$\endgroup$
  • $\begingroup$ Actually, Joel David Hamkins already gave this answer. (Technically he only gave the answer for the powerset of the naturals versus the naturals, but the proof is identical to the general case.) $\endgroup$ – Noah Schweber Jan 24 '17 at 20:55
  • $\begingroup$ In his first answer he addressed Russell's antinomy, didn't he? And didn't he show in his second answer that the interval (0,1) of reals is uncountable, not the power set of the natural numbers? $\endgroup$ – Joel Adler Jan 25 '17 at 16:06
  • $\begingroup$ Yes, but $(0,1)$ is immediately in bijection with $2^\mathbb{N}$; I wouldn't consider that a nontrivial fact. IMO the two results are "morally equivalent." (Of course, that's not a very strong opinion - I didn't downvote! I'm just saying the statement has arguably appeared already.) $\endgroup$ – Noah Schweber Jan 25 '17 at 16:09
  • $\begingroup$ I agree that the statement for the special case $\mathcal{N}$ has appeared already. But properties of the real numbers were used in its proof, and the general statement $|M|<|P(M)|$ does not follow. $\endgroup$ – Joel Adler Jan 26 '17 at 10:49
4
$\begingroup$

The Nielsen-Schreier Theorem : a subgroup of a free group is a free group.

The algebraic proofs are rather complicated, whereas the topological proof is trivial : a group is free if and only if it acts freely on a simplicial tree.

Of course the theory of covering spaces and fundamental groups is hidden somewhere.

$\endgroup$
4
$\begingroup$

The diamond lemma, a.k.a. Newman's lemma, which says that a terminating system is globally confluent if and only if it is locally confluent, has a very simple proof based on Noetherian induction. This proof was first published by Gérard Huet in 1980; see the Wikipedia article.

But this lemma has many nontrivial applications and, somewhat like the example of linearity of expectation mentioned elsewhere, it can often seem amazing that the possibly very complicated ways in which paths can diverge is irrelevant as long as local steps can be corrected.

$\endgroup$
3
$\begingroup$

Many theorems of finite group theory have such nature: they are non-trivial but their proofs are not so hard. But in the frame of infinite groups or finite loops, those are challenging problems. Below are some example:

1- a finite group with just two conjugacy classes is $\mathbb{Z}_2$.

2- a non-trivial finite $p$-group has non-trivial center.

3- finite groups have Lagrange property.

4- a finite group in which its all nontrivial proper subgroup have order a fixed prime $p$ has order $p^2$ and so is abelian.

Many theorems of finite dimensional vector spaces are also non-trivial with trivial proofs: the similar theorems are not true for modules or infinite dimensional cases or have hard proofs.

$\endgroup$
3
$\begingroup$

I am surprised that no one has mentioned Cantor-Schröder-Bernstein Theorem. It certainly is a non-trivial theorem until you see it for the first time. The proof I linked here, I believe, could be considered a trivial one if you draw "the picture" and observe how the constructed bijection maps the elements.

Another example could be Łoś's theorem. The proof is basically going through definition of ultraproducts and carrying out an induction on formulas. It is tedious to write down but at its core a trivial one. Though, I am reluctant to call the theorem itself trivial!

$\endgroup$
  • 6
    $\begingroup$ I agree about Łoś's theorem, but not the Cantor-Schröder-Bernstein Theorem. The latter seems like the opposite: The theorem is a trivial "squeeze theorem" about cardinal numbers, but it took 3 guys to get a correct proof. The proof is not that long, but it's clever and IMHO not "trivial." $\endgroup$ – Monroe Eskew May 9 '14 at 1:19
3
$\begingroup$

Farkas's Lemma and a variety of other theorems of alternatives are fundamental in the theory of optimization. The proof simply couples the Fundamental Theorem of Linear Algebra with the fact that a positive vector and a (nonzero) nonnegative vector in Euclidean space cannot be orthogonal.

$\endgroup$
2
$\begingroup$

My personal favorite is the Noether-Deuring theorem.

Let $K \subseteq L$ be fields with $L$ of finite dimension over $K$, $R$ a finite-dimensional algebra over $K$. Then $L \otimes R$ is naturally an algebra over $L$, thought of as "extending $R$ by scalars in $L$".

Let $U, V$ be finite-dimensional modules over $R$; then $L\otimes U$ and $L \otimes V$ are modules over $L \otimes R$. Obviously, any isomorphism between $U$ and $V$ as $R$-modules can be extended to an isomorphism between $L\otimes U$ and $L \otimes V$ as $L \otimes R$-modules.

Therefore, if $U \simeq V \in R-\text{mod}$, then $L \otimes U \simeq L \otimes V \in L \otimes R-\text{mod}$.

The Noether-Deuring theorem is that the converse is true. That is, if $L \otimes U \simeq L \otimes V \in L \otimes R-\text{mod}$, then $U \simeq V \in R-\text{mod}$.

Proof: If $L \otimes U \simeq L \otimes V \in L \otimes R-\text{mod}$, then it is also true in $R-\text{mod}$. But there, $L \otimes U \simeq U^n$, where $n = [L: K]$. Therefore, $U^n \simeq V^n \in R-\text{mod}$. By Krull-Schmidt, this is enough to show that $U \simeq V \in R-\text{mod}$.

Note that this isomorphism isn't natural; in order to get a natural choice of isomorphism, "descent data" is needed. But it's surprising that even without that "descent data", there is some isomorphism.

$\endgroup$
1
$\begingroup$

The proof that the deRham cohomology is equivalent to singular cohomology on a smooth manifold is in some sense trivial: one shows that the de Rham complex is a soft (hence cohomologically trivial) resolution of the constant sheaf, and it is not too hard to show that the cohomology of the constant sheaf is the same as singular cohomology. In a sense, it just follows from "abstract nonsense" about derived functors being computable from acyclic resolutions and the fact that soft resolutions are acyclic (a partition of unity argument). But it is certainly a nontrivial theorem.

$\endgroup$
  • 6
    $\begingroup$ So the proof of Serre's GAGA theorems is trivial, granting analytic and algebraic finiteness and ampleness theorems. The proof of Mordell-Weil is trivial, granting the theory of abelian varieties, height functions, and general finiteness theorems of Galois cohomology. If one does the hard foundational work beforehand, then yes, what remains is trivial. To make the deRham theorem useful, we need properties of the isomorphism. So a test: can you prove the integration map trivialization of top-degree cohomology on a compact oriented manifold matches the one defined in singular cohomology? $\endgroup$ – Boyarsky Jun 20 '10 at 2:51
  • 1
    $\begingroup$ Unfortunately not, because I know nothing about singular cohomology. $\endgroup$ – Akhil Mathew Jun 20 '10 at 3:26
  • 4
    $\begingroup$ Akhil, no worries: now you have a good exercise to keep in mind as you learn more about singular cohomology. You probably won't find it proved in any book (I never did), but eventually you'll figure it out for yourself. At least it gives you more appreciation for the subtlety of the deRham isomorphism. (By the way, the compatibility with cup products is another good one, but that is elegantly handled in Godement's sheaf theory book via "general nonsense" with pairings of resolutions.) There's also the matter of cohomology with compact supports... $\endgroup$ – Boyarsky Jun 20 '10 at 3:44
  • 2
    $\begingroup$ Wow, I thought I was the only person in the world who learned sheaf cohomology before learning any other type of cohomology theory! I remember that the first diagram I ever chased was the diagram giving the long exact sequence in cohomology associated to a short exact sequence of sheaves (my teacher told me that it would be more instructive to prove it by myself <grin>). $\endgroup$ – Andy Putman Jun 20 '10 at 19:22
  • 2
    $\begingroup$ Akhil, the book of Greenberg & Harper is short & sweet (with nice exercises). Their discussion of orientation is curious since use an "orientation sheaf" but don't have the general notion of sheaf and so get stuck in some contortions. Anyway, you can get a .djvu file of Munkres' book from our Russian friends at extracoder.com/genesis/0072.html (look for item 72583 in the numbering of the left column). And .djvu of Greenberg & Harper is at rapiddigger.com/download/… $\endgroup$ – Boyarsky Jun 21 '10 at 3:27
1
$\begingroup$

The theorem that differential generalized cohomology is characterized by a differential cohomolgy exact hexagon -- originally asked/conjectured generally and proven for the ordinary case by (Simons-Sullivan 07) -- turns out to follow formally "by stable cohesion" (Bunke-Nikolaus-Völkl 13). A quick review is here: ncatlab.org/schreiber/show/IHP14.

$\endgroup$
1
$\begingroup$

$$ \int u\,dv = uv - \int v \, du. $$ The whole theory of generalized functions follows, as do lots of other things.

$\endgroup$
  • $\begingroup$ Someone down-voted this. Can someone explain why? $\endgroup$ – Michael Hardy Dec 1 '15 at 17:22
  • $\begingroup$ In fact, three have downvoted it, two have upvoted, currently. $\endgroup$ – Gerry Myerson Dec 1 '15 at 22:33
  • 5
    $\begingroup$ I downvoted it because I thought it was a wild exaggeration. $\endgroup$ – Todd Trimble Dec 2 '15 at 3:49
  • 2
    $\begingroup$ The principle of integration by parts occurs frequently and all over the place, no question about it, just as does, for example, completing the square or polarization identities, etc. etc. And it occurs in widely varying contexts. But the formula itself doesn't foreshadow all the objects or all the contexts to which it might apply, which vary in sophistication. So, if you say "the whole theory of generalized functions [distributions] follows", then I say no way, that's a wild exaggeration. Does some manifestation of it play an important role there? Of course! $\endgroup$ – Todd Trimble Dec 2 '15 at 21:26
  • 2
    $\begingroup$ Yet this seems to me a very nontrivial and important theorem with a trivial proof. $\endgroup$ – Pietro Majer Apr 4 '17 at 7:44
1
$\begingroup$

Assume the existence of a rank-into-rank cardinal. Then in the classical Laver tables $A_{n}$, we have $\mathrm{crit}((x*x)*y)\leq\mathrm{crit}(x*y)$ for all $x,y\in A_{n}$ (one can define $\mathrm{crit}(r)=\gcd(r,2^{n})$, so the idea of a critical point does not need to reference large cardinals).

The proof of this fact relies upon the observation that $\mathrm{crit}(j*k)=j(\mathrm{crit}(k))$ and the following easy Lemma.

$\mathbf{Lemma:}$ If $j:V_{\lambda}\rightarrow V_{\lambda}$ is an elementary embedding, then $(j*j)(\alpha)\leq j(\alpha).$

$\mathbf{Proof:}$ Let $\beta$ be the least ordinal such that $j(\beta)>\alpha$. Then $$V_{\lambda}\models\forall x<\beta,j(x)\leq\alpha,$$ so by elementarity, $$V_{\lambda}\models\forall x<j(\beta),(j*j)(x)\leq j(\alpha).$$ Therefore, since $\alpha<j(\beta)$, we have $(j*j)(\alpha)\leq j(\alpha).$

The fact that $\mathrm{crit}((x*x)*y)\leq\mathrm{crit}(x*y)$ is almost trivial when one assumes strong large cardinal hypotheses, but the fact that $\mathrm{crit}((x*x)*y)\leq\mathrm{crit}(x*y)$ has no known proof in ZFC.

$\endgroup$

protected by François G. Dorais May 8 '14 at 21:53

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).

Would you like to answer one of these unanswered questions instead?

Not the answer you're looking for? Browse other questions tagged or ask your own question.