106
$\begingroup$

I included this footnote in a paper in which I mentioned that the number of partitions of the empty set is 1 (every member of any partition is a non-empty set, and of course every member of the empty set is a non-empty set):

"Perhaps as a result of studying set theory, I was surprised when I learned that some respectable combinatorialists consider such things as this to be mere convention. One of them even said a case could be made for setting the number of partitions to 0 when $n=0$. By stark contrast, Gian-Carlo Rota wrote in \cite{Rota2}, p.~15, that 'the kind of mathematical reasoning that physicists find unbearably pedantic' leads not only to the conclusion that the elementary symmetric function in no variables is 1, but straight from there to the theory of the Euler characteristic, so that 'such reasoning does pay off.' The only other really sexy example I know is from applied statistics: the non-central chi-square distribution with zero degrees of freedom, unlike its 'central' counterpart, is non-trivial."

The cited paper was: G-C.~Rota, Geometric Probability, Mathematical Intelligencer, 20 (4), 1998, pp. 11--16. The paper in which my footnote appears is the first one you see here, doi: 10.37236/1027.

Question: What other really gaudy examples are there?

Some remarks:

  • From one point of view, the whole concept of vacuous truth is silly. It is a counterintuitive but true proposition that Minneapolis is at a higher latitude than Toronto. "Ex falso quodlibet" (or whatever the Latin phrase is) and so if you believe Toronto is a more northerly locale than Minneapolis, it will lead you into all sorts of mistakes like $2 + 2 = 5,$ etc. But that is nonsense.

  • From another point of view, in its proper mathematical context, it makes perfect sense.

  • People use examples like propositions about all volcanoes made of pure gold, etc. That's bad pedagogy and bad in other ways. What if I ask whether all cell phones in the classroom have been shut off? If there are no cell phones in the room (that is more realistic than volcanoes made of gold, isn't it??) then the correct answer is "yes". That's a good example, showing, if only in a small way, the utility of the concept when used properly.

  • I don't think it's mere convention that the number of partitions of the empty set is 1; it follows logically from some basic things in logic. Those don't make sense in some contexts (see "Minneapolis", "Toronto", etc., above) but in fact the only truth value that can be assigned to $\text{“}F\Longrightarrow F\text{''}$ or $\text{“}F\Longrightarrow T\text{''}$ that makes it possible to fill in the truth table without knowing the content of the false proposition (and satisfies the other desiderata?) is $T.$ That's a fact whose truth doesn't depend on conventions.

$\endgroup$
25
  • 9
    $\begingroup$ I agree that any standard definition of "partition" will give the empty set one partition. I suspect that the reason some people think of this conclusion as a mere convention is that they are reminded of some superficially similar situations. Is $1$ a prime number? Is $R$ a prime ideal in $R$? Is the empty space connected? Is a trivial module irreducible? Life is easier if you say "no" to all of these, even though this at first seems to call for a special convention. $\endgroup$ Nov 13, 2010 at 20:06
  • 24
    $\begingroup$ @some guy: “the quotient hasn’t any zero-divisors”: true, but nonetheless, it isn't an integral domain! …at least if you define that as a ring where “any product of non-zero elements is non-zero”, since in the trivial ring, the empty product of non-zero elements is zero :-) (I guess most people enjoying this question will agree that this is the “right” definition of integral domain, not the more common version which just considers binary products, but I think plenty of mathematicians might disagree.) $\endgroup$ Nov 14, 2010 at 0:18
  • 37
    $\begingroup$ @Michael Hardy: am I the only person who doesn't actually understand what the question being asked is? $\endgroup$ Nov 14, 2010 at 0:24
  • 10
    $\begingroup$ @Qiaochu Yuan: Maybe you are the only one. But I can't be sure of that. The question asks for other examples of non-trivial and interesting mathematics arising out of seemingly trivial instances of vacuity. $\endgroup$ Nov 14, 2010 at 1:31
  • 16
    $\begingroup$ Oh, let's start flaming about $ 0^0 $ and the degree of the constant zero polynomial while we're here. $\endgroup$ Nov 14, 2010 at 11:23

36 Answers 36

1
2
2
$\begingroup$

A The empty set is a covering map of any topological space. More generally, a covering map needn't be surjective (although many books claim just that). For example the inclusion of a closed and open subset of a space is a covering. Strangely, I would argue that this entails that the empty topological space, although connected, is not simply connected.

B Dually, given a field $K$, the zero algebra over $K$ is diagonal and in particular étale: the morphism of affine schemes $\varnothing \to \operatorname{Spec}(K) $ is étale. In the same vein, a nonzero constant polynomial over $K$ is separable (its nonexistent roots in an algebraic closure of $K$ are certainly distinct) . We may then say without any exception that the $K$-algebra $K[X]/(f(X))$ is étale iff $f(X)$ is a separable polynomial.

$\endgroup$
16
  • 13
    $\begingroup$ Agreed, except that I strongly prefer not to call the empty space connected. How many components does it have? $\endgroup$ Nov 14, 2010 at 3:46
  • 22
    $\begingroup$ If we assume disjoint union is additive on the number of connected components, then the empty space should have zero connected components. I think this should correspond to the usual distinction between primes (connected spaces) and units (empty spaces) in situations where we have unique factorization. $\endgroup$
    – S. Carnahan
    Nov 14, 2010 at 10:22
  • 9
    $\begingroup$ I would define a space X to be connected if, whenever it is expressed as a topological disjoint sum (i.e., coproduct) of spaces, one of the summands is X itself. This leads to the conclusion that the empty space is not connected, because t is the topological disjoint sum of zero spaces. (Analogously, I would define a positive integer to be prime if, whenever it is expressed as a product, it equals one of the factors. That makes the empty product, 1, not prime.) $\endgroup$ Nov 14, 2010 at 22:49
  • 5
    $\begingroup$ @Georges: In the category of spaces, there can't be a universal covering. The theory of universal covering is a theory of pointed spaces: A pointed cover (Y,y) of a pointed space (X,x) is a universal covering if, for any pointed cover (Z,z) of (X,x), there is a unique X-morphism $f:(Y,y)\rightarrow (Z,z)$ of pointed covers. $\endgroup$
    – ACL
    Nov 14, 2010 at 23:48
  • 3
    $\begingroup$ Dear ACL: yes, I also like the idea of restricting the notion of (universal) covering space to pointed spaces. I vaguely remember reading that our Overlords (Deligne et al.) are quite categorical on this point. $\endgroup$ Nov 15, 2010 at 12:48
1
$\begingroup$

If we need to define the value of the Euler's function $\varphi$ at infinity, the best choice will be $\varphi(\infty)=2$ This is because $\varphi(n)$ is the number of generating elements in the cyclic group of order $n$, and so if $n\to \infty$, the the cyclic group tends to $\mathbb{Z}$ which has just two generating elements.

$0^0=1$ is another example, which is easy to prove for the natural zero, but it is not true for the real zero!

$\endgroup$
11
  • 4
    $\begingroup$ If the base is the real or complex zero and the exponent is the natural zero, then $0^0=1$. For example: $\displaystyle e^z=\sum_{n=0}^\infty\frac{z^n}{n!}$. If $z=0$, then the first term in this expansion is $0^0/0!=1$. $\endgroup$ Feb 10, 2014 at 16:58
  • 3
    $\begingroup$ @YemonChoi, I think that "the number of generating elements" means "the number of elements that form singleton generating sets", rather than "the minimum cardinality of a generating set". $\endgroup$
    – LSpice
    May 7, 2018 at 17:27
  • 2
    $\begingroup$ I think that's an argument for setting the value at zero — as it is the number of generating elements of $\mathbb Z/0\mathbb Z$. $\endgroup$ Jun 30, 2020 at 12:53
  • 1
    $\begingroup$ -1 for “It is not true for real zero.” I’d like to see an argument for this. Real exponentiation is an extension of natural exponentiation, i.e. it ought to have the same value wherever the latter is defined. $\endgroup$
    – user76284
    Jul 10, 2020 at 18:23
  • 1
    $\begingroup$ @user3840170 Of course it’s not continuous at 0. That doesn’t mean it has to be undefined at 0. The sign function is discontinuous at 0, but there are good reasons to let it be 0 there. $\endgroup$
    – user76284
    Dec 20, 2020 at 3:04
1
$\begingroup$

Maybe it is not serious, but then in a sense none of the examples in this thread are.

Jordan decomposition represents every operator on a finite-dimensional space in a unique way as $S+N$ where $S$ is a diagonalizable operator, $N$ is a nilpotent operator, and $SN=NS$.

The Jordan decomposition of the zero operator is $0+0$. It thus is the only operator which is diagonalizable and nilpotent at the same time.

Similarly, the zero Lie algebra is the only one which is both semisimple and nilpotent. And one might also argue that it is not simple. Or is it?...

$\endgroup$
4
  • 1
    $\begingroup$ The question asks for other examples of non-trivial and interesting mathematics arising out of seemingly trivial instances of vacuity. So what is the non-trivial or interesting math arising from this one? $\endgroup$ Jul 9, 2020 at 19:48
  • 1
    $\begingroup$ @MichaelHardy I accept your objection. In fact the reason I decided to add this was that only recently in my work this seemingly vacuous circumstance caused quite nontrivial additional considerations. Unfortunately they are too technical to even attempt telling about it here. $\endgroup$ Jul 10, 2020 at 21:45
  • $\begingroup$ The proof will not fit into the margin. $\endgroup$ Jul 10, 2020 at 22:22
  • $\begingroup$ @MichaelHardy Exactly ;) $\endgroup$ Jul 10, 2020 at 22:24
1
$\begingroup$

I suppose another example would be those proofs by induction in which you don't need a basis, although no nice examples come to mind instantly. Here I mean proofs in which you show that if $P(m)$ for all $m < n$, then $P(n)$. You don't need to show $P(n)$ holds for the smallest value of $n$, since it is vacuously true that $P(m)$ holds for all smaller values, and therefore the thing proved in the inductive step entails the smallest instance as a special case.

This works not only for natural numbers, but for infinite ordinals.

$\endgroup$
8
  • 1
    $\begingroup$ “Integers” here should be “natural numbers”, right? (Not editing it myself in case you’re making a clever point that I’m missing.) $\endgroup$ Nov 14, 2010 at 4:13
  • $\begingroup$ @Peter: Yes. ......... $\endgroup$ Nov 14, 2010 at 5:29
  • 3
    $\begingroup$ Oh ? $\endgroup$ Nov 14, 2010 at 23:31
  • 1
    $\begingroup$ You can get around that by adding internal spaces $\endgroup$ Nov 15, 2010 at 0:45
  • 1
    $\begingroup$ see ? $\endgroup$ Nov 15, 2010 at 0:45
1
$\begingroup$

There are a number of examples from logic.

Many-sorted first-order logic with 0 sorts is propositional logic. A 0-ary function symbol is a constant symbol, and a 0-ary relation symbol is a proposition. These “degenerate” cases are actually quite interesting and important.

$\endgroup$
0
$\begingroup$

Is the span of the empty set in a vector space equal to $\lbrace 0\rbrace$, or does it have no span? The "correct" answer in my opinion is the latter. See Example 3.10.3 of http://math.mit.edu/~rstan/ec/ec1.pdf for a reason. On the other hand, a reason (which I find unconvincing) for the span to be $\lbrace 0\rbrace$ is given by PBRMEASAP at https://www.physicsforums.com/archive/index.php/t-84017.html. This site has a discussion of whether the empty set is a vector space. The correct answer is that it isn't, because one of the axioms is the existence of an additive identity 0.

Update. I agree with the comments that the span of the empty set is $\lbrace 0\rbrace$. What I said above was foolish.

$\endgroup$
6
  • 7
    $\begingroup$ Example 3.10.3 gives a formula for the number of spanning sets in an n-dimensional vector space over a finite field when n>0. The fact that the formula gives 1 rather than 2 when n=0 could be taken as evidence that the empty set does not span the 0-dimensional space, but I would much prefer to say that Example 3.10.3 gives a formula for the number of nonempty spanning sets (for all n). $\endgroup$ Nov 14, 2010 at 4:22
  • 14
    $\begingroup$ I'm afraid I have to disagree with this answer. All sorts of things would break otherwise. For instance, for any set $S$, we want the $K$-vector space $K^S$ of functions $S\to K$ to have a the standard basis $\{e_s | s\in S\}$, where $e_s$ is the delta-function at $s$. When $S$ is empty, this just says that the empty set (which has zero elements) is a basis for the zero vector space (which has dimension zero). It would be a big nuisance to have say that the empty subset does not have a span. Then you'd have to add exceptions to lots of theorems that would otherwise be true without restriction. $\endgroup$
    – JBorger
    Nov 14, 2010 at 6:29
  • 20
    $\begingroup$ Any reasonable definition of a span implies that the span of the empty set in a vector space is the zero subspace. No "convention" here: (1) It is the smallest vector subspace containing $\emptyset$. (2) It is the set of (finite) linear combinations of elements of $\emptyset$: the only such thing is the empty sum, which is 0. $\endgroup$ Nov 14, 2010 at 10:35
  • 14
    $\begingroup$ If you don't allow the empty set to span {0}, how on Earth are you going to define dim {0}? ({0} is not a basis for {0}, since 0 is a linear combination of elements of {0} in two ways.) $\endgroup$ Nov 14, 2010 at 22:35
  • 7
    $\begingroup$ There is a neat article by de Boor An empty exercise (doi), where he argues for the inclusion of $0\times n$ and $m \times 0$ matrices in Matlab. He discusses definitions of span and determinant etc. I found this from a link on the Wikipedia page for determinant. $\endgroup$
    – Ramsay
    May 5, 2011 at 11:38
1
2

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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