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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.

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 "F--->F" or "F--->T" 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.

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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. –  Tom Goodwillie Nov 13 '10 at 20:06
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@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.) –  Peter LeFanu Lumsdaine Nov 14 '10 at 0:18
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@Michael Hardy: am I the only person who doesn't actually understand what the question being asked is? –  Qiaochu Yuan Nov 14 '10 at 0:24
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@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. –  Michael Hardy Nov 14 '10 at 1:31
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Oh, let's start flaming about $ 0^0 $ and the degree of the constant zero polynomial while we're here. –  Zsbán Ambrus Nov 14 '10 at 11:23

26 Answers 26

How many open covers does the empty topological space have? Not one, not none, but two: the empty cover $\varnothing$, since its union is $\bigcup\varnothing=\varnothing$, and the cover {$\varnothing$}, since its union is also $\bigcup${$\varnothing$}$\ =\varnothing$.

This comes up when using the Grothendieck plus-construction to sheafify a presheaf. Apply the construction to the (nonseparated) presheaf $P:\mathcal{O}(X)^{op}\to \mathrm{Set}$ sending every open set to the set $A$, with $|A|\geq 2$. Then the presheaf $P^+:\mathcal{O}(X)^{op}\to\mathrm{Set}$ agrees with $P$ on every open set except $\varnothing\subseteq X$, where $P^+(\varnothing)$ is now a one-element set {$*$}. This is because the matching families for the cover {$\varnothing$} of $\varnothing$ (of which there is one for each $a\in A$) are all set equal to the unique matching family for the refining cover $\varnothing\subseteq\ ${$\varnothing$} of $\varnothing$.

This elementary example comes from "Sheaves in Geometry and Logic", by Moerdijk and MacLane.

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Unfortunately there isn't a map from {\empty} to \empty, as you would expect for a cover. –  David Roberts Nov 15 '10 at 4:13
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Here I'm using the notation that if $S$ is a set (in particular, of sets) then $\bigcup S=\bigcup_{A\in S}A$. –  Owen Biesel Nov 15 '10 at 15:41
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I laughed out loud at this one! At least continuous functions patch easily over both open covers :) –  David MJC Nov 15 '10 at 21:45
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That is bizarre. –  Spice the Bird Sep 23 '12 at 6:16
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Well, yeah! Union being an operation $P(P(S)) \to P(S)$, i.e. a function from a 2-element set to a 1-element set if $S$ is empty. –  Todd Trimble Jul 20 '13 at 19:15

There is a big difference between statements such as, one the one hand "the empty sum is zero" or "0!=1" and on the other hand "1 is not a prime number". In my opinion, the latter does involve a convention (i.e., a choice) but the former does not.

The fist definition of a prime that comes to mind (and came historically, I guess) is "a natural number with no divisors except 1 and itself". This is a perfectly reasonable notion, but it leads to unpleasant contortions when one tries to state the prime decomposition theorem, including uniqueness. A similar phenomenon explains why an irreducible space is nonempty by definition. In these cases, the definition has been tailored to the need of getting cleaner statements. The question "is the empty space connected?" falls into the same category; I find it strange that the more common convention (which is yes) does not match the other two.

In the case of the empty sum, 0 is the only conceivable value, the other choice being "undefined": a mathematician hostile to the empty set might define finite (nonempty) sums by induction, starting from the one-term case and leaving the empty case meaningless. This would not lead to contradictions, only to lots of traps in proofs because whenever you take the sum of some finite set of numbers you first have to check that it is not empty, or treat the empty case separately.

And of course, if you run the inductive definition "backwards" from 1 term to 0 term you immediately find the right value for the empty sum. This is an efficient way to convince students.

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Laurent, this is a very interesting and convincingly illustrated point of view. –  Georges Elencwajg Nov 14 '10 at 15:07
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I once heard it asserted that Euclid and his coevals did not consider 1 a prime number because they did not consider 1 a number. That seems consistent with our own colloquial usage when we say "A number of people commented on this idea." –  Michael Hardy Nov 14 '10 at 17:36
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@Micheal, in classical greek neither one nor two are properly numbers. –  Mariano Suárez-Alvarez Nov 15 '10 at 1:33
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@Laurent: From my years in Rennes, I remember a coffee-break conversation (back when the room was still on the 4th floor) where people were discussing whether it made sense to talk about the action of the null group on the empty set. (I don't think you were involved in that discussion, but I'm pretty sure Antoine Ducros was.) It stuck in my mind, I could never quite decide if it was very silly or too deep for me. –  Thierry Zell Nov 15 '10 at 16:25
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@Mariano. So the greeks instinctively knew that 2 is a very odd prime. –  David MJC Nov 15 '10 at 21:36

(1) The value of any sheaf on the empty set is the terminal object. (Consider the gluing condition for the empty open cover of the empty set.)

(2) If A→B is a morphism of sets, then we can define the factor set B/A. We have B/∅=B⊔*, where * is a one-element set. (Consider the left adjoint of the forgetful functor from the category of pointed sets to the category of morphisms of sets.)

(3) Sometimes the norm of a morphism of normed spaces f: X→Y is defined as sup_{x∈X: x≠0} ‖f(x)‖/‖x‖ or as sup_{x∈X: ‖x‖=1} ‖f(x)‖. This does not work for X=0. The correct definition is ‖f‖=sup_{x∈X: ‖x‖≤1} ‖f(x)‖. It also works for seminorms.

(4) The zero ring is the terminal object in the category of unital rings. It is not an integral domain, nor a local ring or a field.

(5) The empty manifold is not connected. Its number of connected components is 0. (Think of the following theorem: Every manifold is the coproduct of a unique family of connected manifolds. The cardinality of the family equals the number of connected components.)

(6) Examples in elementary mathematics abound. The zero vector space has an empty basis and a unique endomorphism A. The matrix of A in the unique basis is empty and the determinant of A is 1. There is exactly one function from the empty set to any other set (the empty function). Zero is a natural number, 0^0=1, the sum of the empty family of numbers is 0, the product of the empty family of numbers is 1, the product or the coproduct of an empty family of objects in a category is the terminal or the initial object of this category, the monoidal product of the empty family of objects in a monoidal category is the monoidal unit.

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I’ve long adored the fact that the determinant of the unique 0×0 matrix is one — even though it’s a zero matrix! –  Peter LeFanu Lumsdaine Nov 14 '10 at 0:24
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I'd say $0^0 =1$, not $0$ since it's an empty product. And the series expansion $e^z = \sum_{n=0}^\infty \frac{z^n}{n!}$ doesn't work when $z=0$ unless $0^0=1$ (consider the first term). (But that's not the whole story, since $0^0$ is an indeterminate form since $g(x)^{f(x)}$ can approach any nonnegative number as $f(x)$ and $g(x)$ approach 0, depending on what $f$ and $g$ are. However, if both are analytic, then the limit is 1. And in order to make $g(x)^{f(x)}$ approach anything but 1, the point $(g(x),f(x)$ must approach 0 along a curve with the $g$ or $f$ axis as a tangent line.) –  Michael Hardy Nov 14 '10 at 1:48
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@Michael: Oops, I meant 0^0=1, of course. This is a typo. –  Dmitri Pavlov Nov 14 '10 at 12:07
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I once ran into the issue mentioned in (3) when writing some foundational material about maps between modules for $p$-adic Banach algebras, but I came up with a different fix: because all these sets of norms were clearly living in the non-negative reals, I decreed that all sups should be taken in the set of non-negative reals. This fixed all problems at a stroke because the sup of the empty set was suddenly zero which is the answer one wants. –  Kevin Buzzard Nov 15 '10 at 8:00
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If one wants exponentiation to be continuous (as an analyst might) then $0^0$ has to be indeterminate. But I've never been motivated to make $0^x$ continuous at 0, because it becomes undefined just to the left of 0 (i.e., $x<0$) anyway. So I prefer to say $0^0=1$ and tolerate a little discontinuity. –  Andreas Blass Nov 15 '10 at 13:45

An elementary example, but pedagogically nice: a standard early induction proof example is that you can tile any $2^n \times 2^n$ square with one unit square removed, using L-shaped tiles of three unit squares each.

Surprisingly (to me), many textbooks take the base case as $n=2$. The better ones use $n=1$. But the version in The Book, though, surely starts at $n = 0$!

(Of course, I understand the pedagogy of not starting at 0: it’s usually best to make one point at a time. Trying to use this single example to teach about both induction and vacuity simultaneously would end up confusing most students. But when it’s not needed for the former, it does work nicely for the latter, I think!)

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When inductively calculating homology of spheres I like to take the ${-1}$-sphere (empty space) as base case. –  Tom Goodwillie Nov 14 '10 at 1:03
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Better take the unreduced suspension then! You do that to students? –  Tilman Nov 15 '10 at 8:54
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You wrote "The better ones use $n=1$. But the version in The Book surely starts at $n=0!$. But isn't $0!=1$? And yes, that totally on-topic pun is intended! –  Kevin O'Bryant Nov 15 '10 at 13:51
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Besides the $(-1)$-sphere, continued fractions are better defined from $n=-1$: $p_n/q_n=[a_0,a_1,\dots,a_n]$ so that $p_0=a_0$, $q_0=1$, but also $p_{-1}=1$, $q_{-1}=0$. –  Wadim Zudilin Dec 29 '10 at 14:33

Counting is a special case I think: the number of ways of doing nothing is always 1, because you do exactly that, nothing. The number of ways of doing something impossible is 0, because you can't do it. That's why we have: $$ \binom{n}{0}=1 \quad \text{but} \quad \binom{n}{n+1}=0.$$ So I don't think your partition example or the cell phone example are really about vacuous truth the same way the Minneapolis example is. Though if pressed I'm not sure how I would formulate precisely how to make the distinction.

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A zillion times I've had undergraduates ask me why 0! is 1, and that's not hard to answer by saying if you divide 4! by 4 you get 3!, and if you divide that by 3 you get 2!, and keep going and see where the pattern leads. But When they ask why $\binom{6}{0} = 1$, there's a bad way to answer that and a good way. The bad way talks about the empty set, etc. Here's the good way. Toss a coin 6 times; in how many ways can you get "heads" 3 times? The list of 20 includes HHHttt, HHtHtt, HHttHt, etc. Then do the same for 2 out of 6. Then 1 out of 6. Then 0 out of 6. Simple. –  Michael Hardy Nov 13 '10 at 20:42

If you've ever written code to convert an integer into a string of decimal digits, you may have come to the conclusion that the integer 0 should map not to the string 0, but to the empty string instead. Most algorithms I've seen need to introduce a kludge to make 0 come out right. After all, when we write 0 we are violating the usual rule of "no leading zeros".

A nice, natural recursive expression of the conversion process is

def itoa(n):
  if n==0: return ""
  return itoa(n/10) + chr(ord('0') + n%10)

which can be thought of as
The string representation of an integer consists of its leading digits (n/10) followed by its last digit (n%10).

Trying to fix this by returning "0" instead of "" would result in everything getting a superfluous leading zero.

On the other hand writing 0 as the empty string would be rather annoying.

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def itoa(n): (n>9 ? itoa(n/10) : "") + chr(ord('0') + n%10) –  Marc van Leeuwen Nov 16 '11 at 14:11

Over the reals, $\sup \emptyset = -\infty$ and $\inf \emptyset = \infty$.

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This is another one which comes from universal properties. One can regard a poset as a category where a \le b iff there is an arrow from a to b. Then the supremum is the colimit and the infimum is the limit. The empty supremum is the empty colimit, which is the initial object (if it exists), and the empty infimum is the empty limit, which is the final object (if it exists). And I guess you mean "extended reals." –  Qiaochu Yuan Nov 15 '10 at 21:10
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If one defines the distance between two points in a (suitable) space as the length of the shortest path, then this one implies the distance is infinite if there's no path. So that's one way this one is useful. –  Michael Hardy Nov 15 '10 at 21:30
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This is an example I would have given. In particular, in the extended reals, inf A is only less than or equal sup A if A is nonempty. The same logic explains why the zero polynomial has degree $-\infty$ and why the empty set has dimension $-\infty$. –  David MJC Nov 15 '10 at 22:03
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@David MJC: I disagree with your dimension convention. There are many situations where the empty set should not have a uniquely defined dimension. For example, one may consider a closed $n$-manifold as a cobordism between empty $n-1$-manifolds. –  S. Carnahan Nov 16 '10 at 2:54
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Scott, I think you are talking about the empty (n-1)-manifold there, not the empty set :) –  David MJC Nov 16 '10 at 8:59

The usual axiomatizations of set theory (without urelements) mean that every set in the entire set-theoretic universe is ultimately built from copies of the emptyset, in complex empty-box-in-a-box-in-a-box constructions.

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Amazing. I gave this answer (in terms of ordinals) four hours before and ot one vote only ! –  Denis Serre Nov 14 '10 at 8:10
    
Oh, sorry, I had't noticed the second part of your answer! Since the observation applies in ZFC not only to ordinals, but to all sets, I'll leave this answer up. –  Joel David Hamkins Nov 14 '10 at 11:23
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In a similar vein, Conway's theory of Games (which include the Surreal Numbers) gets quite a lot of mileage out of considering the "empty game" in which neither player has any move at all. Every game can be considered as a built from it, just as in Set theory. –  Eric Finster Nov 15 '10 at 8:43

The empty product in a group $G$ is the unit of $G$. This is the only way to avoid mistake in calculations.

Set theory begins by the construction of finite ordinals. The first one is $\emptyset$ and is denoted $0$. The next one is $\{\emptyset\}$, which is not empty ! It is denoted $1$. More generally, every finite ordinal is defined only in terms of the empty set recursively: $n+1:=n\cup\{ n \}$. Physicists (or administrators, politicians, whoever is asled to fund mathematics) might find this pedantic, but it is actually powerful.

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The idea of empty products and empty sums seems to make sense only if the operation involved is associative, and then whenever there's an identity element. When asked why the product of no numbers is 1, one can answer that multiplying something by no numbers just leaves it fixed, so it's the same as multiplying it by 1. But you need associativity for that to make sense. –  Michael Hardy Nov 13 '10 at 22:35
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....and so that's why it works in all groups. –  Michael Hardy Nov 13 '10 at 22:35
    
This was discussed in a question on math.stackexchange: math.stackexchange.com/questions/6832/… –  Peter LeFanu Lumsdaine Nov 14 '10 at 0:14

What about the two orientations of a point? $$*$$

(Pro trivialogia). I'd like to add some general remarks about the question you raised in the comment below, which seems to me a question of general interest. I see at least three general good reasons why it is worth dealing with trivial cases of mathematical notions.

  • Sometimes we simply do not know whether $x$ is a trivial object. Even if our main interest is in non-trivial cases, in the course of a proof or a computation we deal with unknown objects $x,\, y\dots,$ that may possibly degenerate. Therefore, we would like theorems, methods, rules, to hold with the minimum of assumptions, avoiding special separate treatments for degenerate cases (think to some classifications into unnecessary special cases, for algebraic equations, used at the beginning of algebra).

  • Abstraction. It is a great feature of modern mathematics the ability of translating a complicated notion belonging to a simple setting, into a simple notion belonging to a possibly more complicated setting (in many abstract contexts the cost of this operation is zero). Example: a limit or a colimit in a category, and in fact any universal construction, is just a zero-object in a suitable category (as an application, Freyd's theorem about existence of adjoint functors, &c.)

  • Constructions and proofs by induction. As soon as the induction step from $n$ to $n+1$ is suitably clarified, the validity of the general fact is reduced to that of the trivial case, that becomes the heart of the whole story. So for instance, the very reason of some facts about spheres is some (trivial, but important) fact about $\mathbb{S}^0$. This is of course also the case of constructions and operations with orientations.

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Talking about a point $p$ as a connected real manifold $M=${p} of dimension $0$. It has two canonical orientations, $+1$ and $-1$, the generators of the top-degree exterior power $\Lambda^{top}(T_p M)=\mathbb{R}^1$. It seems to me another instance of a trivial case of a mathematical notion, as interesting and useful as the other ones (and slightly paradoxical as well). –  Pietro Majer Nov 14 '10 at 7:07
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And this is essential when considering 1-dimensional framed bordism. –  David Roberts Nov 15 '10 at 4:35
    
I don't know what a 1-dimensional framed bordism is, but I'm glad someone mentioned that there's a (nontrivial?) reason to think about the concept. –  Michael Hardy Nov 15 '10 at 4:48
    
I used to work with a notion of orientation (in a context of Fredholm pairs and Fredholm operators) and trivial cases where actually the key point of the whole story. But this question seems of general interest, so I add some remark on this point. –  Pietro Majer Nov 15 '10 at 14:10
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I wasn't convinced about orienting points -- as $GL(0)/SL(0)$ has one component, not two -- until I hit the statement "If $A = B\oplus C$, orienting any two uniquely determines an orientation on the third." –  Allen Knutson Nov 10 '11 at 4:44

$\bigcap \emptyset = V$

Unfortunately, I have read more than one philosophical comment on the "set theoretic depth" of this logical triviality.

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In some topology book I read that it's redundant to say the whole space is an open set after you've said the intersection of any finite set of open sets is open. –  Michael Hardy Nov 13 '10 at 20:45
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Certainly, there is nothing wrong with stating this in a class set theory like NBG, where it is correct. And you do not have to take the intersection of any two sets, since we are talkning about the empty set. Moreover, any two sets can be embedded in a set- the set containing exactly these two sets. –  Michael Greinecker Nov 14 '10 at 1:31
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Formally there's nothing wrong. The way $\bigcap$ is usually defined (in the language of set theory) is $(\forall x) x \in \bigcap A \Leftrightarrow ((\forall y) y \in A \Rightarrow x \in y)$ And, of course vacuously $(\forall x)(\forall y) y \in \emptyset \Rightarrow x \in y$. –  Peter Krautzberger Nov 14 '10 at 16:28
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@Harry: the basic POV of traditional set-theory based foundations considers all sets as once-and-for-all embedded in the class $V$. So then we really can consider intersections among them, and we do get $\bigcap \empty = V$. We're taking the limit of $\empty$ not in the category $\mathbf{Sets}$ (which gives $*$, as you say), but in partially-ordered class of subsets of $V$ (or equivalently, in $V$ itself ordered by $\subseteq$). Several authors in Algebraic Set Theory have considered the connection between these two structures. –  Peter LeFanu Lumsdaine Nov 15 '10 at 1:10
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The standard set theories are ZFC and NBG. When talking about sets without any qualifier, one talks about them. The usual definition of an intersection is the one given by Peter. That taking intersections over the empty set gives you the class of all sets is something you can prove in NBG. In ZFC, there are no classes, so the intersection over the empy set doesn't give you something in the theory. V is not a set, and this might be what Bourbaki was refering too. But even in ZFC, one can translate the property of being "in the intersection of the empty set" to a property that aplies to all sets –  Michael Greinecker Nov 15 '10 at 7:34

The terminal object of a category is the product over the empty set of objects.

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That's exactly the answer I would have given :-P –  Peter Arndt Nov 13 '10 at 22:04

The Generalized Continuum Hypothesis is the assertion that $2^\kappa=\kappa^+$ for all infinite cardinals $\kappa$, or in other words that the power set of a set of size $\kappa$ has the next larger cardinal size above $\kappa$.

If we consider all cardinals, rather than only the infinite cardinals, then the two provable instances of this equation occur in the following vacuous and near-vacuous facts:

  • The power set of a set with $0$ members has $1$ member.

  • The power set of a set with $1$ member has $2$ members.

All other instances of $2^\kappa=\kappa^+$, finite or infinite, are either false or independent of ZFC.

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I heard once that at a Logic Colloquium talk in Berkeley when Foreman or Woodin was speaking on their theorem concerning the consistency that the GCH fails everywhere, the speaker wrote "$2^\kappa\neq\kappa^+$ for all $\kappa$" on the blackboard. Leo Harrington reportedly went silently up to the blackboard, adding the words "except $\kappa=0$ or $1$." –  Joel David Hamkins Nov 14 '10 at 0:54

One is not a prime number, but zero is! It's different than the other prime numbers in $\mathbb{Z}$, though, because it has height zero rather than height one.

Zero is prime in any integral domain. Remember that the trivial ring is not an integral domain.

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I don't find this convincing. Yes, zero generates a prime ideal, which is an element of Spec, etc. But "prime number" has a specific definition that excludes $0$. –  Greg Martin Feb 9 at 19:20

I regard "negative thinking" in category theory as an example of cool vacuity: see e.g. http://ncatlab.org/nlab/show/negative+thinking. As category theory is not set theory, such vacuity does not necessarily involve the empty set directly, but the same principle of backwards generalization is used.

The fact that a set is uniquely determined by its elements (i.e., has no additional structure beyond the equality relation between its elements) is summarized by saying that a (-1)-category is a truth value: a morphism between two elements in a set is either true (the elements are the same) or false (they are not). So the morphisms in a 0-category (a set) are (-1)-categories (either true or false), just as the morphisms in a (1-)category are 0-categories (sets). This admits generalizations to situations where "truth" is a more subtle concept (e.g. parameter dependent).

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Zsbán Ambrus brings up an interesting example in the comments: the degree of the zero polynomial. The first time I was told about this issue I was told that it is largely a matter of convention. Well, maybe. Here is some evidence suggesting that $\deg 0 = \infty$:

  • The most basic one: the zero polynomial has infinitely many roots in an algebraic closure.

  • If one wants the degree of a polynomial to be a valuation, then we must define $\deg 0 = \infty$. This is the unique choice consistent with the requirements that $\deg fg = \deg f + \deg g$ and $\deg (f+g) \ge \text{min}(\deg f, \deg g)$, and it is necessary in order to make the corresponding absolute value nondegenerate.

  • One way to say the above geometrically when $F = \mathbb{C}$ is that the degree should describe the order of the pole of $f$ at infinity on the Riemann sphere. The function $0$ decays faster than the reciprocal of any polynomial in the neighborhood of infinity. In fact, the sequence of functions $x^n$ converges uniformly to it in a neighborhood of infinity as $n \to \infty$.

  • Another way to say the above is that, in the natural topology on $F[[x]]$, we have $x^n \to 0$. This can be appreciated even if you are, for example, a combinatorialist, because it allows you to say natural things about generating functions like $\frac{1 - x^n}{1 - x} \to \frac{1}{1 - x}$ as $n \to \infty$.

  • One can also define the degree of $f$ as $[F[x]/(f(x)) : F]$, in which case again we find that $\deg 0 = \infty$. This is just a fancier version of the first reason.

Edit: As James Borger points out, the middle ideas are mistaken. Corrected, they actually suggest that $\deg 0 = -\infty$:

  • $\deg 0 = -\infty$ is the unique choice consistent with the requirements that $\deg fg = \deg f + \deg g$ and $\deg (f + g) \ge \text{min}(\deg f, \deg g)$. With this definition, $|f| = 2^{\deg f}$ is now an absolute value.

  • Geometrically, when $F = \mathbb{C}$ the function $0$ has a zero of infinite order at infinity, hence a pole of order $-\infty$.

  • The relevant local ring here is really $F[[ \frac{1}{x} ]]$, and in the natural topology on this ring we have $\frac{1}{x^n} \to 0$.

  • Another reason to like this definition is that it gives a uniform statement of the division algorithm on $F[x]$.

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I was taught $0$ has no degree. –  Jeff Strom Nov 15 '10 at 1:47
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@Jeff: certainly one can make the case for this, but I don't remember a particular compelling argument in this direction. –  Qiaochu Yuan Nov 15 '10 at 2:41
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Dear Qiaochu, I'm not sure I agree with all this. First, it's not true that $\deg(f+g)\geq \min(\deg f,\deg g)$. Take $f=x^{100}$, $g=1-x^{100}$. It is true that $\deg(f+g)\leq \max(\deg f, \deg g)$. Second, 0 has a zero of infinite order at $x=\infty$, and hence a pole of order $-\infty$ there. Third, it seems beside the point to take convergence in $F[[x]]$, which is the local ring at $x=0$. I think the local ring at infinity $F[[x^{-1}]]$ is more relevant, and then $x^n \to 0$ as $n\to-\infty$. So I'd prefer to set $\deg 0 = -\infty$. –  JBorger Nov 15 '10 at 10:20
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$-\infty$ crops up here (and in other places such as the dimension of the empty set) because it is the supremum of the empty set: in this case it is the supremum of the set of powers with nonzero coefficients. –  David MJC Nov 15 '10 at 21:56
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So it's not true that if you differentiate a polynomial then its degree goes down by 1? (I suppose that even if you set deg(0)=-1 that rule is violated, so this isn't such a strong argument.) –  gowers Nov 20 '10 at 15:03

Zero is a limit ordinal, because it is the union of its elements.

Transfinite induction has two canonical statements. The "strong" statement, $$ (\forall \alpha)((\forall \beta)((\beta<\alpha) \rightarrow P(\beta)) \rightarrow P(\alpha))\rightarrow (\forall \alpha)P(\alpha), $$ doesn't split anything into cases. The version used most frequently in proofs says that any property preserved under unions and successors holds for all ordinals. Zero should rarely be a special case.

Also, "limit ordinals" should totally be called "colimit ordinals". The term "limit ordinal" refers to limit points in the order topology, thus excluding zero, but this is silly.

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Category theory does not have a monopoly on the word "limit". –  Tom Goodwillie Nov 19 '10 at 23:46
    
True :) I just think things should have names and definitions that reflect the way they're most often used. –  Gene S. Kopp Dec 3 '10 at 11:15

Recall that an abstract simplicial complex consists of a family of finite sets $K\subseteq 2^V$ such that $\sigma\in K$ and $\tau\subseteq\sigma$ implies $\tau\in K$. (Sometimes it is also assumed that $\{v\}\in K$ for all $v\in V$.) Elements of $K$ are called faces of $K$. Dimension of a face is its cardinality minus 1.

According to the above definition, if $S\not=\emptyset$, then the empty set $\emptyset$ is a face of the complex (of dimension $-1$, since it has $0$ elements). Applying the usual definition of simplicial homology gives us what is called reduced homology, which is often much better-behaved than the nonreduced one (obtained when we forget about the empty face).

In this context it is important to distinguish between the empty simplicial complex $K_e=\{\emptyset\}$ and the void simplicial complex $K_v=\{\}$, which may be understood as the $(-1)$-sphere and the $(-1)$-disk. (In particular $H_{-1}(K_e)=\mathbb{Z}$, $H_n(K_e)=0$ for all $n\not=0$, and $H_m(K_v)=0$ for all $m$.)

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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 http://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.

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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). –  Tom Goodwillie Nov 14 '10 at 4:22
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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. –  JBorger Nov 14 '10 at 6:29
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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. –  Laurent Moret-Bailly Nov 14 '10 at 10:35
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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.) –  Qiaochu Yuan Nov 14 '10 at 22:35
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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. –  Ramsay May 5 '11 at 11:38

The determinant of the $0\times 0$ matrix is $1$. First of all, it's the only way to make determinant of a direct sum of matrices be a product of determinants. Second, it is the sum of $0!$ terms each of which is a product of $0$ factors.

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Third, with this convention, Dogson's condensation formula for the determinant of a $2\times 2$ matrix boils down to the usual formula. –  Roland Bacher Feb 10 at 15:52
    
I guess I have to add that the sign of the permutation of $0$ elements is $1$, because the number of inversions is even. –  Lev Borisov Feb 10 at 17:54

'Silly' but I like them anyways:

$$\prod_{y\in\emptyset}\left(\sum_{x\in \emptyset}x\right)=1,$$

and

$$\sum_{y\in\emptyset}\left(\prod_{x\in \emptyset}x\right)=0.$$

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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 integers, but for infinite ordinals.

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“Integers” here should be “natural numbers”, right? (Not editing it myself in case you’re making a clever point that I’m missing.) –  Peter LeFanu Lumsdaine Nov 14 '10 at 4:13
    
@Peter: Yes. ......... –  Michael Hardy Nov 14 '10 at 5:29
    
(The software wouldn't let me post that simple response without those trailing dots because it was too short! Just as with the title of the thread.) –  Michael Hardy Nov 14 '10 at 5:30
    
Oh ? –  Joel David Hamkins Nov 14 '10 at 23:31
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see ? –  Steven Gubkin Nov 15 '10 at 0:45

Let $\otimes_{i \in I} M_i$ denote the tensor product of $R$-modules $M_i$. Then $\otimes_{i \in \emptyset} M_i$ is $R$.

(Reason: $\prod_{i \in \emptyset} M_i$ is the terminal object in the category of sets (see the answer of Eivind Dahl), i.e. a point, and multilinear maps on this to $N$ are just elements of $N$, i.e. homomorphisms $R \to N$.)

Another one: I know this is really silly and already contained somehow in the other answers, but anyway:

$\prod_{i \in \emptyset} 0 = 1$

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It makes if you consider what it would mean for a function be "0-linear". The empty product is as a module isomorphic to 0, but if we want to consider multilinear functions, we're not considering it as a module; as a set it's $\{()\}$. So we have functions from a one-element set, i.e. just any function, with the requirement that they be "0-linear", i.e. linear in all of the 0 arguments, which is a trivial requirement! So it does work after all. –  Harry Altman Nov 13 '10 at 23:10
    
This can be motivated either by the fact that tensor products represent multilinear maps, or (as per another answer) the fact that \otimes is an associative, unital operation, so “$k$-ary tensor product” with $k=0$ must give the unit object for $\otimes$. (All up to coherent natural isomorphism, of course!) –  Peter LeFanu Lumsdaine Nov 14 '10 at 0:11

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 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.

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Agreed, except that I strongly prefer not to call the empty space connected. How many components does it have? –  Tom Goodwillie Nov 14 '10 at 3:46
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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. –  S. Carnahan Nov 14 '10 at 10:22
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@Georges and ACL: No, the empty space is by no means connected, because many theorems containing the word “connected” will fail if it is. For example, every manifold is the coproduct of a unique family of connected manifolds. A space is connected (in the appropriate sense) if and only if its π_0 is a one-element set. The π_0 of the empty space is empty. –  Dmitri Pavlov Nov 14 '10 at 17:24
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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.) –  Andreas Blass Nov 14 '10 at 22:49
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@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. –  ACL Nov 14 '10 at 23:48

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!

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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 $0^0/0!=1$. –  Michael Hardy Feb 10 at 16:58
    
Yes you are right, only exponent should be Natural. –  M. Shahryari Feb 10 at 18:24

I'm always amused (and unpopular for it) when mathematicians claim that some terminology or notation is "true" rather than "convention", as if there is some God of Mathematics out there who hands us definitions that us mere humans are forbidden to tamper with. Actually all mathematical notation is convention.

We could, if we agreed as a body to do so, define "$m+n$" to mean what it used to mean, unless $m=n=1$ in which case it means 3. It isn't forbidden, because we invented "$+$"; it belongs to us and we can choose what it means. We won't do that, though, not because it is wrong according to some objective source of truth, but because it would cause us a whole heap of trouble. For example we would have to rewrite a huge number of theorems to add exceptional cases, and we hate exceptional cases. It isn't a question of "truth" since all of modern mathematics could be correctly stated using the new definition.

In the same way, we choose to set the empty product equal to 1, and the empty sum equal to 0, because those are the conventions that make our symbol manipulations so much easier and simpler than any alternative conventions would make them. It isn't really different, except in degree, to excluding 1 from being a prime. We do that because otherwise we would have to spend all day writing "let $p$ be a prime other than 1". Qualitatively the reason is the same: we like our theorems to be cleaner, with fewer preconditions and fewer subcases. Simplicity and generality is a fundamental aesthetic of mathematics.

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We mathematicians also seem to be hung up on repeatability and logical as well as otber forms of consistency. Why are we so hidebound, I wonder? –  The Masked Avenger Feb 10 at 16:59

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