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A number of economists do not appreciate rigor in their usage of mathematics and I find it very discouraging.

One of the examples of rigor-lacking approach are proofs done via graphs or pictures without formalizing the reasoning. I would like thus to come up with a few examples of theorems (or other important results) which may be true in low dimensions (and are pretty intuitive graphically) but fail in higher dimensions.

By the way, these examples are directed towards people who do not have a strong mathematical background (some linear algebra and calculus), so avoiding technical statements would be appreciated.

Jordan-Schoenflies theorem could be such an example (though most economists are unfamiliar with the notion of a homeomorphism). Could you point me to any others?


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Proof by pictures still exists in higher math, for example see: . Although I agree that at the very least there is a rigor behind them. – crasic Jun 28 '11 at 22:21
Thanks a lot, guys! You provided me with some easy examples to quash these "but graphs are also proofs" statements.Since I can't accept all the answers, I just picked the one who would use some reputation points. But thanks to all of you! – johnny Jun 28 '11 at 22:22
I have converted the question to community wiki, as it's asking for a big list of examples and there is no single right answer. – Zev Chonoles Jun 29 '11 at 7:26
On MathOverflow: Results true in a dimension and false for higher dimensions – Rahul Apr 30 '15 at 21:00

12 Answers 12

up vote 15 down vote accepted

Here's an example that doesn't require too much mathematical knowledge, and the low-dimensional result is intuitive graphically:

We know that if a differentiable function $ f : \mathbb{R} \to \mathbb{R} $ has only one stationary point, which is a local minimum, then it must be a global minimum (this is intuitively obvious, and can be proved using Rolle's theorem). However, this result does not generalise to higher dimensions. An example would be $f : \mathbb{R}^2 \to \mathbb{R} $ with $ f(x,y) = x^2 + y^2(1-x)^3 $. This function has a unique stationary point at $ (0,0) $, which is a local minimum but not a global minimum (this can be seen by considering $ x >> 1 $). (Interactive 3D plot)

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How do you prove your first assertion? Try $x \mapsto (x^2+1)\cos x$. This is has its graph like a cosine whose oscillations are larger and larger as $|x| \to \infty$. $0$ is a local minima, and a stationary point, but is clearly not global minima, since for $x_n=2n\pi$ the values tend to $\infty$. – Beni Bogosel Jun 28 '11 at 21:47
@Beni: I think that "... has one stationary point which ..." is intended to mean "... has one stationary point, which ...". – John Bentin Jun 28 '11 at 22:05
@John: That doesn't make it correct, and my counter example is still sound. A stationary point is not necessarily a local minima. A local minima is a stationary point (if the function is differentiable), but not necessarily a global minima. You can easily picture what I am saying here. – Beni Bogosel Jun 28 '11 at 22:07
I guess the OP meant that the stationary point has to be unique which is not the case for your counter-example. And indeed, this is easily proved with Rolle's theorem. – johnny Jun 28 '11 at 22:18
I've edited to remove confusion. – Daniel Freedman Jun 28 '11 at 22:18

Simple symmetric random walks in 1 and 2 dimensions return to the origin infinitely many times, but not in 3 and higher dimensions.

That is, if you are on a number line (or coordinate plane) and repeatedly flip a coin to determine whether to take one step in the positive direction or one step in the negative direction (or do something to randomly choose 1 step in the positive or negative x or y direction), the probability that you'll get back to the origin is 1. If you do the same in 3 dimensions (or higher), where you're choosing randomly between 6 (or more) directions for taking 1 step, the probability of returning to the origin is less than 1.

edit: The probability $p(d)$ of a $d$-dimensional simple symmetric random walk returning to the origin is apparently called Pólya's Random Walk Constant. $p(1)=p(2)=1$, but $p(3)\approx 0.34$, $p(4)\approx 0.19$, $p(5)\approx 0.14$, $p(6)\approx 0.10$, $p(7)\approx 0.09$, and $p(8)\approx 0.07$ (from the linked MathWorld article).

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Which is exactly why you shouldn't wander around drunk... unless you live in the plane or on a line. – mathmath8128 Jun 28 '11 at 21:20
I can't wander vertically! – The Chaz 2.0 Jun 28 '11 at 21:23
@aengle: but do you really have 3 degrees of freedom when you're wandering around drunk? If so, I'd be interested to see where you do your drunk-wandering. That aside, there is the drunk-walk problem that a drunk on a sidewalk between a building and the street will end up in the street with probability 1... – Isaac Jun 28 '11 at 21:23
@Georges: Pólya's original article is here (in German). Here's a recent article in the Monthly by David Levin and Yuval Peres. Legend has it that Pólya was promenading on the Zürichberg-a hill, rather than a mountain-and kept walking into the same couple. Since the Zürichberg is rather a 2-dimensional gadget he was inspired to prove his theorem on recurrence to convince himself that he wasn't spying on that couple. The rest is history... Ich habe Deine Email nicht vergessen, ich bin noch nicht dazugekommen. – t.b. Jun 28 '11 at 21:57
Here is a more colloquial way to state the above result: A drunken man will find his way home, but a drunken bird may be lost forever. – Eric Naslund Oct 27 '11 at 9:50

"A non constant polynomial over $\mathbb R$ (or $\mathbb Q$, or an arbitrary field, according to your colleague's sophistication) which has no zero is irreducible". This is true in degrees $\leq3$ but false for higher degrees. Surprise your economist by asking if $x^4+4$ is irreducible and after a probably positive answer, calmly write down $$x^4+4=(x^2+2x+2)(x^2-2x+2)$$

[This is not about dimension but maybe it is in the spirit of your question: that lack of rigor can lead to mistakes, even if an assertion is true for some low integers]

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I would probably surprise you by saying that some economists couldn't even tell you what a polynomial is, let alone say which one is irreducible. Besides basic calculus, concave functions and Kuhn-Tucker, economics curriculum varies highly depending on the university. – johnny Jun 28 '11 at 22:37
@johnny: Although I'm not an economist, I nevertheless feel compelled to point out that real analysis is de-facto required for admission into any half-decent economics PhD program these days, so any academic economist would probably know quite a bit about polynomials... – Andrew Ho Feb 8 '15 at 2:29

Chaos cannot exist in one- or two-dimensional continuous dynamical systems. Among other things, this means that in one or two dimensions if you supply similar inputs you will get similar outputs.

However, in 3 or more dimensions the dynamics can be chaotic, meaning that similar inputs do not necessarily lead to similar outputs (you can witness this in the Lorenz equations, a crude model for atmospheric dynamics).

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+1: This seems like an appropriate fact to be absorbed by people attempting to make economic models. – yasmar Jun 29 '11 at 7:02
Chaos theory suddenly got a lot more interesting to me. – grautur Jun 30 '11 at 5:56
Does the statement in the first line have some formal statement? Namely, what definition of "chaos" do we need to take so that it cannot happen in 1D and 2D? – Wojowu Apr 3 at 10:05
Sensitivity to initial conditions, topological mixing and dense periodic orbits. – Chris Taylor Apr 4 at 7:23

May be this one. Every polygon has a triangulation but not all polyhedra can be tetrahedralized (Schönhardt polyhedron)

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A famous example in mathematical physics is the Ising model. This was invented by Wilhelm Lenz, who gave it as a thesis topic to his student Ernst Ising. Ising solved the problem in one dimension (which is rather easy), finding that there is no phase transition, and incorrectly concluded from this that there is no phase transition in three dimensions. In fact the model does have a phase transition in two or more dimensions, which is what makes it interesting.

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The Riemann mapping theorem only works for $\mathbb{C}^1.$ Also (alas), the only conformal automorphisms of open subsets of $R^n$ are Möbius transformations, when $n > 2.$

Another nice example (not a graphical example) is whether $x^n - 1$ always factors (over $\mathbb{Z}$) into polynomials whose coefficients are either -1, 0, or 1. The answer is no, but the first counterexample is in the early 100s.

Also, in Conway and Guy's Book of Numbers, the devious duo trick the reader into thinking that the maximum number of regions into which the plane can be divided by $n$ straight lines is $2^n.$

These two aren't examples of theorems that don't hold in higher dimensions, but they do demonstrate that tenable hypotheses still need to be proven.

Richard Guy has a nice article with examples like this:

"The strong law of large numbers," The American Mathematical Monthly 95.8 (1988): 697-712

As you can see, these might not be particularly motivational for economists. Perhaps the more important part of rigor to economists would be remembering the assumptions that theorems depend upon?

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For the factorization of $x^n-1$, the first $n$ where an irreducible factor has a coefficient besides $0$, $1$, or $-1$ is $n = 105$. The point is that the possible nonzero coefficients only depend up to sign on the odd prime factors of $n$ (not their multiplicities or any factors of 2), and for $n$ a product of two different odd primes the irred. factors have only coeff. $0$, $1$, and $-1$. So the first possible $n$ to be a counterexample would be $n = 3\cdot 5 \cdot 7 = 105$, and that is a counterexample. – KCd May 2 '12 at 17:29

Knots exist only in three-dimensional space.

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I feel like this is more of an example of things failing in higher codimensions, rather than in higher dimensions. – Robert Haraway Feb 9 '12 at 15:32
Cant' we define that an $n$-dimensional knot is an injective continuous mapping $S_n \rightarrow \mathbb{R}^{n+2}$? In this case, the usual definition is recovered in the case $n=1$. – goblin May 10 at 13:26

I think extending the real plane to the complex plane provides many, many surprising results.

Firstly, if a function is once differentiable in the complex plane, it is infinitely differentiable (these functions are infinitely smooth).

Building upon that, if you have a holomorphic function that is bounded in $\mathbb{C}$, it is a constant. This is Liouville's Theorem.

(Think of a smooth function like sine that is bounded in the reals but is certainly not a constant!)

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It is intuitively obvious that a continuous function should be differentiable everywhere except at a countable number of "kink" points!!!

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Are you trying to refer to Weierstrass function? – johnny Jun 28 '11 at 22:29
Yes, but what does this have to do with dimensionality? – Jesse Madnick Oct 11 '11 at 4:39
This is actually incorrect; the Cantor-Lebesgue function has the Cantor set as its set of critical points, and the Cantor set is uncountable. Also, this answer has nothing to do with dimensionality. – Robert Haraway Feb 22 '12 at 1:53
Can you cut an n-dimensional shape of diameter 1 into n+1 pieces all of diameter less then 1?
This is obviously true in n=2 and n=3, but its false for all $n>297$

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The opposite direction is also true. There are nice things that happen in higher dimension that do not happen in lower ones. For example, the complex function $f(z) = 1/z$ has a singularity at $0$ that cannot be removed (the limit of $f$ as $z$ goes to zero does not exist). from Hartog´s lemma (see wikipedia), one can always remove the isolated singularities of a complex differentiable funcion in two or more variables. In particular, the limits of these functions as they approach to the singular points always exist.

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